Integration by parts
The integral of u dv = uv - ( integral of vdu)
You have to be able to integrate 'dv' that you choose. If you can integrate both possible choices, then you want the 'u' to be the choice that would eventually differentiate to zero.
See the notes from last year about integrations by parts. There are more examples there.
Integrals involving powers of trig functions
More trig identities
Recall, sin^2 x + cos^2 x = 1 (pythagorean theorem)
sin^2 x = (1/2)(1 - cos2x)
cos^2 x = (1/2)(1 + cos2x)
Also, sin 2x = 2sinxcosx
cos2x = cos^2 x - sin^2 x
cos2x = 1 - 2sin^{2}x
cos2x = 2cos^{2}x - 1
Why would we want to address these trigonometric identities?
ex 1) integral sin^{2}x dx = (1/4)(2x - sin2x) + c
In the case of the second example, what you want to happen is have so that you only have one of the other trigonometric function. That's the goal- only one of the other trigonometric function that you have to integrate. This is because you are rewriting the problem so that you can more effectively approach it.
More examples of integrals involving powers of trigonometric functions
Today's nonsense:
- New cheerleader poster
- Little Dokata missing again
- Megan's genetically diabetes
- Katie and bee stings: "glad that they sting and die"
Example #3 - integral of sin^{2}x cos^{5}x dx. What you have to do here is convert the integral into either the sine or cosine. The "u" variable can be either of them. So then the du will get rid of the other one. There is no lone sine or cosine, because there's one squared and the other to the fifth. The pythagorean identity tells us that we can change two sines at the same time or two cosines at the same time. So we can take the one that has an odd exponent (in this case, cosine), and if they each had odd exponents take either one, and then separate it.
= integral sin^{2}x * cos^{2}x * cos^{2}x * cosx dx
= integral of sin^{2}x * (1-sin^{2}x)cosxdx
u=sinx
du = cosxdx
2007-09-06
Today's nonsense:
- Phoebe the desserter
- Cupcakes from Spiller
- Blatent dress code violations and the volleyball players
- Shagrin and Dakota ("much to your displeasure")
- Bill's subway and Spiller's four-spinach-leaf subway disaster
Tomorrow - we talk about trig substitution.
We try to get these problems to where there's only one sine or one cosine and then more of the other type of trigonometric functions. We can change two tangents into two secant squareds (sec^2 - 1). So in this first example of the day (see the notes), we want everything in terms of secant and with a sec-tan left over, or all in terms of tangent and that one secant. The "u" is going to have to be secant.
2007-09-07 - Trigonometric substitution
Today's nonsense
- Dakota bashing
- Next math club meeting: Tuesday, 8:10. Monday morning there's the "pick up your stuff" time.
- Children doing multiplication via "number trees" - what is this nonsense? "CSCOPE"- the district bought this, so for each of the classes up to algebra-2 and for science, it tells you what to teach each week and pretty much it takes away teacher individualism.
- Reading some notes that were being passed around- about "The Game."
-
sin^2 theta + cos^2 theta = 1
tan^2 theta + 1 = sec^2 theta
cos^2 theta = 1 - sin^2 theta -> a^2 - u^2 (constant squared - function squared)
sec^2 theta = 1 + tan^2 theta -> a^2 + u^2
tan^2 theta = sec^2 theta - 1 -> u^2 - a^2
The trick is that you will have to be able to identify your "u" and "a" in various problems in order to substitute the information.
Trig substitution
1. For integrals involving sqrt(a^2 - u^2) or it might just say a^2 - u^2, or it might be to the 3/2 power, but whatever, let u^2 = ... well, we want it to contain a^2. We're doing this different from our calculus book. The u^2 is going to be (a^2)*sin^2. If in place of u^2 I put (a^2)sin^2, you would be able to factor out everything until you are left with-- you factor out the a^2 and what do you have left? You will have 1-sin^2, and what's that? That's simply cos^2, so the sqrt(cos^2) is going to easily be cos, and so by using these substitutions and the trig identities we were able to get rid of a lot of work. So what would u be? u^2 = a^2 * sin^2. The variable u = a * sin (theta).
2. For integrals involving the sqrt(a^2 + u^2), let u^2 = (a^2)tan^2. By putting a^2 into the problem, that allows us to factor it out in the square-root. The value of u = a * tan theta.
3. For integrals involving sqrt(u^2 - a^2), So it's secant squared minus one, so we need to have an a^2, and then we need to get the secant^2.
u^2 = a^2 * sec^2 theta
Always start off with saying what "u^2" is. Remember to draw the triangle. Trigonometric substitution is used when u-substitution doesn't work.
Show using trigonometric substitution the integral of x^2 + 1.
Homework due today. 2007-09-10
Today's nonsense
- Calculus and physics test, Thursday and Friday
More examples
If you hate "u" don't worry about it, they are not in these problems. 2007-09-12
For #A6- be sure to draw your triangle.
Minilesson on completing the square
Let's take a trip back to why/when this completing the square stuff came from. Algebra 2, yes. So, let's say that y = x^2 + 6x - 7. This would be a parabola, and if you complete the square you'll get it in the form of y=a(x-h)^2 + k. By completing the square, you are putting it in that form. To complete the square, you want the x^2 to have a coefficient of 1. You're going to group your x^2 and x terms together: y = (x^2 + 6x ) - 7. You want to factor this thing so that it looks like y = ( )^2. So you want to factor it so that it's the same number, same term, you get to choose the number to put there. You have to look for two numbers that multiuply to give you the blank, and add to give you 6, so the numbers have to be the same, so in this case it's 3 & 3. And then square this "perfect square": 3^2 = 9. You can either add nine to both sides, or you can subtract nine as well on the end so it's y = (x^2 + 6x + 9) + 7 -9. Factor this, and you get (x+3)^2 - 16. The vertex of this parabola is (-3, -16). That's all you are doing when completing the square. We will two more of these problems. But what happens when you have a coefficient on your x^2? You get something like y = 3x^2 + 6x - 7, and then you get y = 3(x^2 + 2x + 1) - 7 - 3. You do not subtract 1, you have to remember that you had that giant coefficient. Now you can rewrite/factor and get something that looks better.
To find the vertex of a parabola, take the derivative and find where it is zero, and so you can find your x value, instead of remembering the quadratic equation.
2007-09-17 - Partical fractions
What you want to do in these scenarios is factor the denominator and then split the fraction up into A/first + B/first, and then set the numerator to A(x+1) + B(x+2) and then let x equals some values and then get...
2007-09-18
Some factoring is involved in the back of the sheet. The intent is to ask whether or not you know how to factor. To factor, the first step is to take out common factors, in order to binomialize.
When doing integration by partial fractions, and you have your A, B, and C, to figure out the equation where you say this=A(what)+B(what2) etc., you have to multiply each A/B/C variable by the factor that makes it equate to the common denominator that you were using earlier. Solving Separable Differential Equations
And if they are not separable, you're going to be taking a class called "differential equations." Solving differential equations means you're going to integrate, and separable equations means you're going to be able to separate and integrate. In some cases they will not already be separated. Solve each of the following:
1) (dy/dx) = x^2 - 3
2) (dy)/(dx) = y^2 + 3
1/(y^2 + 3) dy = dx
And this is, in actuality, an arctangent problem (trigonometric substitution).
3) (dy)/(dx) = 3xy - x
Factor out an x, move the 3y-1 over, and then you have a u-substitution, and it turns out to be du=3, and then you have a (1/3) integral (1/u)du. And to continue doing this problem, you have to remember that e^{a+b} is the same as e^{a}e^{b} so that you can better write this out.
4) y' = 2y/x
y = C_{2}x^{2}
Homework.
Separable differential equations
Math notebook on SDEs
Paul's online math notes re: SDEs
Wikipedia article (click for ODEs)
2007-09-20
Today's nonsense
- A kid going through Dakota's phone, changing the wallpaper, backgrounds, tones, calling people, etc.
- Administrators calling a "lock down" for paperclips or staples on the floor
- Xeroxing some slope fields
- PhysiCal
Homework due tomorrow. There are some slope fields to graph on #15 and #16.
Newton's Law of Cooling
So this is going to be about rates. The rate of change in the temperature of an object is proportional to the difference between the object's temperature and the temperature of the surrounding medium.
dT/dt = k(T-T_s) ===> ln|T-T_{s}| = C_{1}e^{kt}
If the temperature of an object is 100 degrees and is placed in a room held at 60 degrees (all Fahrenheit), how long will it take for the object to cool to 80 degrees Fahrenheit if it is 90 degrees after 10 minutes. How do you find the value of C_{1} in this case? You solve at t = 0, where you know that T = 100, and so you can find that the value of C_{1} is 40. Now you can use the other ordered pair, 90 degrees for when t = 10, so T=90 and now solve for k. And then the question wants you to solve for 't' when T=80, and so then you just need to solve normally.
C = initial difference between object and the surroundings
Let's try another one, an odd one from the same pages as on the homework page.
ex) The rate of change of y is proportional to y. When x = 0, y=4, and when x = 3, y= 10. Find y when x = 6.
dy/dx = ky
k = 0.3054
Fibanocci sequence- 1, 1, 2, 3, 5, 8, 13, 21
The rate of change of P varies jointly as J and P-3.
dP/dJ = kJ(P-3)
2007-09-21
Today's nonsense
- Bobby and Ketteman want to escape to the circus, to become jugglers, grow some woman-beards, and live happily ever-after
Exponential growth
For unrestrained population growth, the rate of change of population is directly proportional to the population.So that means dP/dT = kP. So this year we're going to be adding in growth restraints because infinite growth is inappropriate.
Logistics
One assumption for restrictive growth is that there is a certain maximum size, m, for the population. (For example, in ours, there was a maximum number of people that could contract the disease (117).) And the rate goes to zero (the slope approaches zero). So we started at zero and we ended at zero, so there were two different values that made our derivative zero, there were those two different values. That's what we are going to end up seeing. The rate goes to zero as the population approaches that size m. The rate of change of population is jointly proportional (we are going to have an = k(one)(two)) to the population and to the difference between the maximum and current populations.
dP/dT = kP(m-P)
So for any logistic graph, there's going to be a high horizontal asymptote at whatever x=m value is.
Let's solve for P.
Wikipedia - exponential growth
Meadows, Donella H., Dennis L. Meadows, Jørgen Randers, and William W. Behrens III. (1972) The Limits to Growth. New York: University Books. ISBN 0-87663-165-0
--- Good (dead) author, appears in a few other places.
Exponential and logistic growths
A neat visualization of exponential growth
Compound interest and exponential growth
Kurzweil's law of accelerating returns
Human population growth (Kimball's online biology book)
2007-09-25 - More logistics
Continuing our derivation from yesterday. This is related to logistical regressions, that tries to find a function that will fit the data. The final equation we came up with was: P = m / (1 + C_{2}e^{-kmT}). What's the limit as T->infinity? It approaches "m." So on the TI-84 calculators, go to [STAT] and then to [CALC] and then to the "Logistic" regressional analysis entry. So it will take some time ... and on the calculator the form that is produced is = C/(1 + ae^(-bx)). With the specific numbers that it found: (122.89)/(1+1137.113196e^(-.9378989775x)).
When was the growth the fastest? It'd be at your inflection point, which will be at the middle. We can do all of this mathematically, even if you forget these simple answers, well, you can derive it yourself. If we can remember, of course we might do it in physics instead of calculus, is some sample data and do some regressions. So if you're not in physics, sorry- trying to get a rise out of Ketteman but he doesn't seem to be paying attention.
Calculator version of logistic regresional analysis: y = C/(1+ae^{-bx}). So you notice, so then, yeah, Dakota pointed out that the B on the calculator is "m*k" on the version we did, and "C" is "m", so what is k? So "k" in our version is "b/c" in the calculator's version.
So we had some questions. (A) At what value of P is the growth (of the population of the people with the disease) the greatest? The equation for growth is dP/dT = kP(m-P). So when is that the greatest? So what is the P value at the inflection point? It'll be (1/2)m. Well, you're asked to find the maximum and minimum here, so you have to find the zeroes of the derivative, so if you want to know when dP/dT is the greatest, well, that translates to inflection points, but let's take the derivative of dP/dT, and what's the derivative? So (dP/dT)' = k(m-P) + -kP. So how do you find the maximums and minimums? Well, set it to zero and solve for P, what does P have to be to make that derivative of dP/dT to be zero? So: km=2kp, so P=km/2k = m/2. So there you go: the rate is the greatest when P is half of the maximum population. Once half has the disease, you start slowing down again. This is the general logistic assumption.
Study Taylor polynomails which involve factorials.
merc248
Conversation with merc248
(2007-09-25 18:05:41) kanzure: quick, what's the power series *really* about?
(2007-09-25 18:05:45) kanzure: Wikipedia doesn't explain much of it to me
(2007-09-25 18:25:25) merc248: the power series? so, as far as i understand, it's to represent functions with polynomials
(2007-09-25 18:25:45) merc248: which is very important in analysis since you have what's called "analytic functions," which are functions that are represented by power series within a given "interval of convergence"
(2007-09-25 18:26:07) kanzure: A guy in #not-math is giving a quick lecture on irc.freenode.net on how combinatorics and the power series are related, apparently the power series can be used for counting via the coefficients
(2007-09-25 18:26:13) kanzure: though I'm not entirely sure what's going on
(2007-09-25 18:26:15) merc248: the regular functions that you can express as a single term or a finite sum are a subset of the analytic functions; there's a bunch of other ones that can ONLY be expressed as a power series
(2007-09-25 18:26:27) kanzure: "interval of convergence" ?
(2007-09-25 18:26:30) merc248: hmm, seems prudent, i don't know mucha bout combinatorics though
(2007-09-25 18:27:05) merc248: yeah, say you have a sum from 0 to inf of say... x^2
(2007-09-25 18:27:19) merc248: it's whatever values of x that makes the entire sum finite
(2007-09-25 18:27:24) merc248: so in this case
(2007-09-25 18:27:37) merc248: you'd have the absolute values of x less than 1
(2007-09-25 18:27:54) merc248: because in any other case, you'll have a divergent series. make sense?
(2007-09-25 18:28:40) merc248: so in this case, you say that the radius of convergence
(2007-09-25 18:28:41) merc248: is 1
(2007-09-25 18:28:46) kanzure: what?
(2007-09-25 18:28:47) kanzure: wait
(2007-09-25 18:28:52) kanzure: if you're taking the sum from 0 to inf of x^2
(2007-09-25 18:29:03) kanzure: how are you going to find a value of x that will make the sum finite?
(2007-09-25 18:29:23) merc248: EVEN if say if x = 1 it diverges; it's the supremum of the values that DO converge, if that makes any sense
(2007-09-25 18:29:39) merc248: there's a couple of tests i think that you can do?
(2007-09-25 18:30:12) merc248: aww fuck, hold on, i forgot what they were, but they had to do with the root test and the ratio test
(2007-09-25 18:30:41) merc248: R = reciprocal of either the root test or the ratio test on the series
(2007-09-25 18:30:42) kanzure: No, that makes no sense
(2007-09-25 18:30:45) merc248: lol
(2007-09-25 18:30:48) merc248: yeah, i suck at explaining things
(2007-09-25 18:30:48) kanzure: can you explain divergence/convergence?
(2007-09-25 18:31:05) merc248: in a few words, divergence = values shooting off into infinity
(2007-09-25 18:31:10) merc248: convergence = finite values
(2007-09-25 18:31:19) merc248: typically when you're talking about divergence/convergence
(2007-09-25 18:31:20) kanzure: so
(2007-09-25 18:31:25) kanzure: if a sum is from some finite number -> infinity
(2007-09-25 18:31:27) merc248: you're talking about an infinite sum or product or some other infinite process
(2007-09-25 18:31:29) kanzure: how can it do anything but diverge?
(2007-09-25 18:31:45) merc248: ahh, check out... 1 / x^2
(2007-09-25 18:31:53) merc248: notice that if x > 1
(2007-09-25 18:31:57) merc248: you'll have a finite sum
(2007-09-25 18:32:22) merc248: because basically, the rate at which 1 / x^2 decreases is much faster than you can sum them up
(2007-09-25 18:32:29) kanzure: really?
(2007-09-25 18:32:32) merc248: yeah
(2007-09-25 18:32:37) kanzure: 1/(x^2) decreases faster than it would increase?
(2007-09-25 18:32:37) kanzure: so:
(2007-09-25 18:33:07) kanzure: sum of x^-2 from 0 -> infinity is faster than sum of x^-2 from 1 -> 0 ?
(2007-09-25 18:33:18) merc248: now, 1 / x is DIVERGENT. if you write out the first couple of terms, you can see why; you can group up i think 2^n terms together for any n and you always get 1/2 out of it, so the first 2^1 terms = 1/2, 2^2 terms = 1/2, etc.... so basically, you add 1/2 infinite amounts of times
(2007-09-25 18:33:25) merc248: well
(2007-09-25 18:33:28) merc248: x^-2?
(2007-09-25 18:33:30) merc248: or x^2
(2007-09-25 18:33:32) kanzure: yes,
(2007-09-25 18:33:37) kanzure: x^-2 == 1/x^2
(2007-09-25 18:33:42) merc248: err, ahh right :p
(2007-09-25 18:33:45) kanzure: I was playing with your example
(2007-09-25 18:33:56) merc248: yeah, pretty much, though i think
(2007-09-25 18:34:00) merc248: at least if you're talking about an integral
(2007-09-25 18:34:06) kanzure: no, I'm talking about a summation
(2007-09-25 18:34:06) kanzure: heh'
(2007-09-25 18:34:06) merc248: x^-2 would diverge from 1 -> 0
(2007-09-25 18:34:21) kanzure: well, integrals involve summations, yes
(2007-09-25 18:34:26) merc248: well yeah, though there's a thing called the "integral test" which ties the infinite series to its corresponding integral
(2007-09-25 18:34:42) merc248: basically, if the integral converges, the series converges
(2007-09-25 18:34:51) kanzure: so what would an integral 'converging' look like?
(2007-09-25 18:35:01) kanzure: I keep thinking of terms of limits approaching a value or not approaching anything
(2007-09-25 18:35:05) kanzure: I am not sure if that's the right way to think abotu this
(2007-09-25 18:35:05) kanzure: *about
(2007-09-25 18:35:19) merc248: ahh, okay, so if you have an integral from 0 to infinity of x^-2
(2007-09-25 18:35:24) kanzure: hmm
(2007-09-25 18:35:25) merc248: you have to have a limiting process
(2007-09-25 18:35:30) merc248: on the infinity part of the integral
(2007-09-25 18:35:32) kanzure: I have not evaluated any integrals from finite numbers to infinite numbers
(2007-09-25 18:35:36) kanzure: so ...
(2007-09-25 18:35:38) merc248: hmm
(2007-09-25 18:35:44) kanzure: maybe you can explain this as well?
(2007-09-25 18:35:44) kanzure: haha
(2007-09-25 18:35:57) merc248: well, bad example before, let's say integral from 1 to infinite of x^-2
(2007-09-25 18:36:11) merc248: you have an improper integral with infinity being the part where it fucks up
(2007-09-25 18:36:21) merc248: basically, you replace infinity with a variable T
(2007-09-25 18:36:23) merc248: and say
(2007-09-25 18:36:31) kanzure: improper integrals ... hmm. How do you work with improper integrals?
(2007-09-25 18:36:33) merc248: lim T-> inf of the integral from 1 -> T
(2007-09-25 18:36:39) kanzure: ah
(2007-09-25 18:36:47) merc248: then you evaluate it like you would a regular integral
(2007-09-25 18:36:58) kanzure: that's useful
(2007-09-25 18:37:10) kanzure: then you have lim T->inf [of your evaluated integral from 1->T] ?
(2007-09-25 18:37:12) merc248: (NOTE! SUBSTITUTE IN T FOR INFINITY AS SOON AS YOU CAN BEFORE FOOLING AROUND WITH THE INTEGRAL)
(2007-09-25 18:37:21) merc248: (I LOST MANY POINTS ON A TEST BECAUSE I DIDN'T DO THAT :O)
(2007-09-25 18:37:26) kanzure: haha
(2007-09-25 18:37:33) merc248: yep, pretty much
(2007-09-25 18:37:34) kanzure: I lost many points on a test because I failed to learn trig substitution ;)
(2007-09-25 18:37:38) merc248: haha
(2007-09-25 18:37:52) merc248: so like, let's just say, integral from 1 -> T for now
(2007-09-25 18:37:53) kanzure: don't know my derivatives of inverse trigonometric functions well enough :(
(2007-09-25 18:37:54) merc248: of 1/x^2
(2007-09-25 18:37:54) kanzure: alright
(2007-09-25 18:38:17) merc248: then you get something like -x^-1 as your integral
(2007-09-25 18:38:20) merc248: evaluated from 1 to T
(2007-09-25 18:38:39) merc248: -1/T + 1
(2007-09-25 18:38:48) merc248: now, you do the limiting process: T -> infinite
(2007-09-25 18:38:50) merc248: err, infinity
(2007-09-25 18:38:55) merc248: -1/T -> 0
(2007-09-25 18:38:58) merc248: so you're left with 1
(2007-09-25 18:39:10) merc248: basically
(2007-09-25 18:39:18) kanzure: huh?
(2007-09-25 18:39:19) kanzure: what just happened?
(2007-09-25 18:39:23) merc248: hahaha
(2007-09-25 18:39:37) merc248: alright, so basically, we got our integral, which is -x^-1
(2007-09-25 18:39:41) kanzure: yes
(2007-09-25 18:39:44) merc248: from x^-2
(2007-09-25 18:39:48) merc248: then we evaluate from 1 to T
(2007-09-25 18:39:53) merc248: so we get -1/T + 1
(2007-09-25 18:40:02) kanzure: ookay
(2007-09-25 18:40:03) kanzure: *okay
(2007-09-25 18:40:11) merc248: and then, we bring in our limiting process: lim T-> infinity of -1/T + 1
(2007-09-25 18:40:24) merc248: T-> infinity implies that -1/T -> 0
(2007-09-25 18:40:38) merc248: so we're left with just our + 1 term
(2007-09-25 18:41:03) kanzure: yes
(2007-09-25 18:41:31) merc248: to tie that in to the whole integral test thing i was talking about, since the sum from 1 to infinity of x^-2 can be tested against the integral from 1 to infinity of x^-2
(2007-09-25 18:41:36) merc248: and since the integral has a finite value
(2007-09-25 18:41:42) merc248: the series has a finite value, therefore convergent
(2007-09-25 18:42:27) kanzure: so
(2007-09-25 18:42:32) kanzure: there are cases where the integral does not have a finite value?
(2007-09-25 18:42:38) merc248: oh, definitely
(2007-09-25 18:42:46) merc248: consider the integral from 0 to 1 of x^-2
(2007-09-25 18:42:53) kanzure: and if the integral does not have a finite value, can the series have an infinite value?
(2007-09-25 18:43:02) merc248: if you just draw the graph of x^-2, you can see that the left part from 0 to 1 has infinite area
(2007-09-25 18:43:07) kanzure: I mean, does the value of the integral determine the value of the series?
(2007-09-25 18:43:13) kanzure: (finiteness/infiniteness)
(2007-09-25 18:43:13) merc248: it doesn't determine the exact value
(2007-09-25 18:43:18) merc248: err
(2007-09-25 18:43:20) merc248: hmm
(2007-09-25 18:43:22) kanzure: but if one's finite, is the other necessarily finite?
(2007-09-25 18:43:32) merc248: that's a good question, i don't know if it goes both ways necessarily, let me check
(2007-09-25 18:43:38) kanzure: alright :)
(2007-09-25 18:44:01) merc248: ahh, yeah
(2007-09-25 18:44:02) merc248: it does
(2007-09-25 18:44:08) merc248: so if the integral diverges, the series diverges
(2007-09-25 18:44:21) kanzure: that's useful
(2007-09-25 18:44:49) kanzure: and in this specific instance, the term 'diverges' means that the integral does not approach a finite value
(2007-09-25 18:45:07) merc248: yep
(2007-09-25 18:45:22) merc248: in all cases, divergent means an infinite process does NOT reach a finite value
(2007-09-25 18:45:24) merc248: err
(2007-09-25 18:45:40) merc248: this might confuse you a bit, so kinda put it aside separate from what i just told you
(2007-09-25 18:45:47) merc248: (ie, don't mix it into the previous stuff i said)
(2007-09-25 18:45:52) merc248: but there's a way to sum divergent series
(2007-09-25 18:45:56) merc248: such that it has a finite value
(2007-09-25 18:46:01) merc248: check out cesaro sums on wikipedia
(2007-09-25 18:46:20) merc248: you can use cesaro sums to make 1 - 1 + 1 - 1 + 1... approach a finite value, that is, 1/2
(2007-09-25 18:46:28) kanzure: http://en.wikipedia.org/wiki/Cesàro_summation ?
(2007-09-25 18:46:35) kanzure: hahah
(2007-09-25 18:46:37) kanzure: that's wacky :)
(2007-09-25 18:46:47) merc248: (oh shit, okay, sorry, i should say that divergent means it is not convergent; convergence means it approaches a single finite value)
(2007-09-25 18:47:04) merc248: (in the last example, 1 - 1 + 1 - 1... definitely does not shoot off to infinity, but it doesn't converge to a single value)
(2007-09-25 18:47:14) merc248: yep :)
(2007-09-25 18:47:21) kanzure: == Definition ==
Let {''a''<sub>n</sub>} be a [[sequence]], and let
:<math>s_k = a_1 + \cdots + a_k</math>
be the ''k''th partial sum of the series
:<math>\sum_{n=1}^\infty a_n</math>.
The sequence {''a''<sub>n</sub>} is called '''Cesàro summable''', with Cesàro sum α, if
:<math>\lim_{n\to\infty} \frac{s_1 + \cdots + s_n}{n} = \alpha</math>.
(2007-09-25 18:48:19) merc248: ahhhh
(2007-09-25 18:48:25) kanzure: hard to read, but :)
(2007-09-25 18:48:27) merc248: i can't read fucking latex :O
(2007-09-25 18:48:31) merc248: at least not yet
(2007-09-25 18:48:32) kanzure: try typing it out in class
(2007-09-25 18:48:33) kanzure: haha
(2007-09-25 18:48:35) merc248: lol
(2007-09-25 18:48:38) kanzure: that'll be a quick way to learn to read it
(2007-09-25 18:48:43) merc248: oh man
(2007-09-25 18:48:43) kanzure: it doesn't render nicely with the eyes
(2007-09-25 18:48:49) kanzure: I've been trying to figure out a method to render it in nearly-real time
(2007-09-25 18:48:54) kanzure: so that maybe when I press enter it renders the last line
(2007-09-25 18:48:54) merc248: lol, good luck
(2007-09-25 18:48:55) kanzure: wouldn't that be nice?
(2007-09-25 18:49:02) merc248: best way to render it in real time: get a notepad, get pencil
(2007-09-25 18:49:03) kanzure: There's a way to do with emacs, there's this special "preview-latex" option
(2007-09-25 18:49:14) merc248: hmm
(2007-09-25 18:49:16) kanzure: but my Debian package manager is broken at the moment (have to fix that when I get some free time)
(2007-09-25 18:49:21) merc248: ahh shitty
(2007-09-25 18:49:24) kanzure: btw,
(2007-09-25 18:49:36) kanzure: http://heybryan.org/school/Calculus/index2.html
(2007-09-25 18:49:39) kanzure: My cal-2 class :)
(2007-09-25 18:50:43) merc248: aww yeah
(2007-09-25 18:50:44) merc248: rofl
(2007-09-25 18:51:12) merc248: dude, i gotta say btw, i appreciate you talking about math with me especially today
(2007-09-25 18:51:21) merc248: i was getting especially rusty over the past couple of weeks and have been lazy in reading math
(2007-09-25 18:51:30) merc248: and since i got fucking class starting tomorrow, i need all the refreshment as i can get
(2007-09-25 18:51:30) merc248: :O
(2007-09-25 18:51:52) kanzure: I am glad to have somebody to talk with in the first place
(2007-09-25 18:52:12) merc248: haha
(2007-09-25 18:52:18) merc248: man
(2007-09-25 18:52:24) kanzure: yeah, so in class right now
(2007-09-25 18:52:35) kanzure: we're going over limited exponential growth
(2007-09-25 18:52:55) merc248: aye
(2007-09-25 18:53:04) kanzure: where we solve a separable differential equation in order to find the exponentiated integrated form
(2007-09-25 18:53:34) kanzure: probably something to do with series, but we haven't been introduced to them "Formally"
(2007-09-25 18:55:11) merc248: ahhh right
(2007-09-25 18:55:36) merc248: you can generally solve ordinary diffeq's with series i think
(2007-09-25 18:55:39) merc248: if it's linear
(2007-09-25 18:55:45) kanzure: what do you mean?
(2007-09-25 18:55:50) kanzure: ODEs I know about
(2007-09-25 18:55:55) kanzure: but I am not sure of the connection with series
(2007-09-25 18:56:01) kanzure: I thought that you just simply solve ODEs through typical first-year methods
(2007-09-25 18:56:15) merc248: basically you ASSUME that you can solve it with power series; ie, summation 0-> inf of x^n
(2007-09-25 18:56:27) merc248: so you substitute that in for the y's in your equation
(2007-09-25 18:56:34) merc248: and do derivatives, etc.
(2007-09-25 18:56:46) merc248: then you do a bunch of summation end point manipulation, etc. etc.
(2007-09-25 18:57:20) merc248: that is, you want to make it so your sums all match up on the bottom end, so if you have sum 1-> inf, sum 0-> inf, etc., you want the second one to go from 1->inf so you can combine the two series together
(2007-09-25 18:58:07) merc248: (now, to do that, all you do is you take the 0th term out, like say if you have the sum 0->inf of a_n, it would be equal to a_0 + sum 1->inf of a_n)
(2007-09-25 18:58:45) merc248: now that you have, for example, sum 1-> inf a_n + sum 1-> inf b_n, you can combine the two into a single series: sum 1->inf a_n + b_n
(2007-09-25 18:59:19) merc248: and i _THINK_ what you do next is that you notice what values the coefficients have to be at in order to get whatever's now on the right side of the equation
(2007-09-25 18:59:27) merc248: (this is probably confusing the fuck out of you, so i'll stop :p)
(2007-09-25 19:00:09) kanzure: (sum from a->b of x) + (sum from a->b of y) == sum from a->b of (x + y) ?
(2007-09-25 19:00:14) merc248: yep
(2007-09-25 19:00:49) merc248: now, under the context of diffeq's, you'll of course have a bunch of stuff where you'll have sum 0->inf x^n, sum 1->inf nx^(n-1), etc.
(2007-09-25 19:00:54) kanzure: I need to do some quick power browsing
(2007-09-25 19:01:15) merc248: so if you had sum 0->inf x^n + sum 1-> inf nx^(n-1), what would you do to get them under a single power series?
(2007-09-25 19:01:27) merc248: or i should say, a single summand
(2007-09-25 19:01:46) kanzure: http://en.wikipedia.org/wiki/Integral_test_for_convergence
(2007-09-25 19:01:58) kanzure: http://tutorial.math.lamar.edu/classes/calcII/IntegralTest.aspx
(2007-09-25 19:02:09) kanzure: http://jtaylor1142001.net/calcjat/Solutions/Series/IntTest/IntTest1/IntTest1Layers.htm (an online test?)
(2007-09-25 19:02:30) kanzure: http://pirate.shu.edu/projects/reals/numser/tests.html A list of convergence tests for series
(2007-09-25 19:02:46) merc248: ohhh man, good stuff
(2007-09-25 19:02:48) kanzure: http://www.sosmath.com/calculus/improper/series/series.html Short page on improper integrals
(2007-09-25 19:03:04) kanzure: What's a p-series?
(2007-09-25 19:03:13) merc248: you'll most likely learn comparison test, lim comparison, geometric series test, p series, root, ratio, alternating, and integral
(2007-09-25 19:03:21) merc248: i think it's x^p
(2007-09-25 19:03:22) merc248: i believe
(2007-09-25 19:03:29) kanzure: http://www.math.unh.edu/~jjp/radius/radius.html Excellent lecture on the convergence of infinite series
(2007-09-25 19:03:38) merc248: err wait, n^-p
(2007-09-25 19:03:49) merc248: aye
(2007-09-25 19:04:26) kanzure: http://planetmath.org/encyclopedia/IntegralTest.html
(2007-09-25 19:04:29) kanzure: don't know if that's any good
(2007-09-25 19:04:32) kanzure: looks kind of boring, but formal
2007-09-26
Improper integrals
Improper integral- an integral in which one limit of integration is infinite ("1 to infinity") or the integrand is unbounded. What if I am integrating 1/(x-1) from 1 to 7? There's an asymptote there at x=1, so the graph never hits there, so wouldn't there always be area there? So we would have to treat it as a limit.
- 1A. When the limit of integration is infinite, consider if you would: the integral from 1 to infinity of (1/x^2)dx. Well, we can't do this. We can't go to infinity. However, we need to approach this as a limit. So, infinity can't be there. So we're going to put something in front:
limit [as B->infinity] of integral [from 1 to B] of (1/x^2)dx. ===> limit [as B->infinity] of -x^{-1}. FROM 1 to B.
= Plug in the upper limit. That's lim [B->infinity] of (-(1/B) - (-1)). What does this equal as B->infinity? So the limit is 1.
We say that the improper integral (integral [from 1 -> infinity] of 1/x^2 dx) converges to 1.
-- You must rewrite the initial problem as a limit of an integral and consider as x approaches infinity. So on the AP test you have to include that. You have to have a setup to get your credit.
Improper integrals, sometimes can be evaluated in a sense, and sometimes they can't be. What's improper about them? That's what we need to look at.
- 1B. Now, consider the integral [from 1 -> infinity] of (1/x)dx. So that's:
lim [as B->infinity] of integral [from 1 to B] of (1/x)dx.
lim [as B->infinity] of ln(x) FROM 1 to B.
lim [as B->infinity] of (lnB - ln1).
-- So you have infinity-minus-zero and the limit of this, well, in this case, we say the improper integral diverges.
The P-series test
For the integral [from 1 to infinity] of (1/x^P)dx, what values of P will give a convergent integral? For P>1.
Learn different ways to evaluate series, and to see if a particular series with lots of sigmas converges or diverges.
Wikipedia - improper integral
Tests for convergence of improper integrals
Wikibooks - methods of integrating those improper integrals
[doc] Numerical methods to solving improper integrals
Improper integrals in real analysis
2007-09-27- More improper integrals
We can use any (finite) number, c, to define, so suppose we were trying to integrate from -infinity to infinity, how are we going to write that as a limit? Basically you split it up into two limits so that you have a constant for one limit of both of the integrals. If either one of them diverges, or both of them, then it diverges. If either or both of the new integrals diverge, the origianl integral diverges. If they both converge, then you add what they converge to, and that's what the original integral converges to.
When the integrand becomes infinite, and there's no infinity as the limits to integration, you'll still have to take the limit.
2007-10-11
Today's nonsense
- Trying to remember the Seinfeld references yesterday (Seven, Soda, ...)
- Spiller in New York with Mrs. Spiller, the mother-in-law, Brandon, and something about Danny Phantom being taken off the air.
- "Who took a bite out of my muffin? Where's the rest of my muffin?" - Ketteman's Seinfeld reference.
- Alyssa's absence, and is now has a truancy
Homework due: comparing improper integrals. Homework due Monday on Spiller's teacher website.
Sequences
Sequences talk about the individual terms. The series is the sum of all the terms of the sequence. Sequences can diverge and converge, just as well as series.
Sequences that have limits as n->infinity converge.
Sequences that do not have limit as n->infinity diverge.
Ex) Determine the convergence or divergence of the sequence with the given nth term. If the ... if it's false, give a reason why it's false, so if the sequence converges, find its limit.
an = (n) / (1-2n) ---------> converges to -0.5
a_n = 3+(-1)^n ---> diverges because it keeps flipping positive and negative there.
d/dx of a^u = a^u * lna * u'
a_n = (n^2) / (2^n(ln2)^2)
a_n = (1+(1/n))^n
Take the natural logarithm of each side.
A number close to 1 that is raised to infinity, approaches e??
What's the ratio test?
e) a_n = (-1)^n (1/n!)
f) a_n = (n+1)/n
g) a_n = (-1)^n (n/(n+1))
h) a_n = ( 3n^2 - n + 4 ) / (2n^2 + 1)
2007-10-12
Today's nonsense
- Broken bulb
Factorial reduction 2007-10-15 - Properties of sequences
Today's nonsense
- Let's do an unofficial count. Dakota has not seen The Princess Bride yet. Alright, we're done. Soon we're going to have to kick him out of the room, and with Katie, until they see the movie.
- Dakota: "You can't really cheat on a math test."
- "How do you do a capital 'b' in cursive?" - Dakota, week three on cursive note-writing.
- AntiAvery Day today
-
Definition of a monotonic sequence- A sequence {a_n} is monotonic if its terms are nondecreasing, so that a_1 <= a_2 <= a_3 ... <= a_n <= .... or if terms are nonincreasing (a+1 => a_2 => a_3 => ... => a_n => ...) Monotonic means you have no relative maximums and no relative minimums. So you can take the first derivative, and if there's no sign change, then that means that the sequence is monotonic. A parabola cannot be monotonic.
ex) Determine whether the sequences are monotonic.
a) a_n = 3 + (-1)^n ---> Not monotonic. You can't prove this just by showing the first four terms ... you have to show it based off of the equation of the sequence itself.
b) a_n = (2n)/(1+n) ---> There are two ways to prove that this is or is not motonic- either algebraic or by calculus. By algebraic, you compare 2n/(1+n) to 2(n+1)/(1+(n+1)) ... and then you show which one is greater and which one is less, so evaluate that out and cross multiply etc.
-- The other method: take the derivative. Yes, the very derivatives that we have spoken of in the past. My derivative comes out to be ... Lodehi - Hidelo all over lolo. So what's the numerator? What does the numerator come out to be? 2. Yes. Ahuh. Ahuh, oh yes you did. I understand. It's okay. You guys started off not talking, did you hear the griping about what we were doing? And then some people are paying attention. Do you see the derivative? "Who started the Princess Bride thing?" - Dakota. Anyway, if the derivative is always going to be increasing or decreasing, then you know it's monotonic. So, yes, it's monotonic, and we proved it two different ways.
Definition of a bounded sequence- A sequence {a_n} is bounded above if there is a real number M such that a_n <= M for all n ... the number M is called an upper bound of the sequence.
--- A sequence {a_n} is bounded below if there is a real number N such that N <= a_n for all n. The number N is called a lower bound of the sequence.
If there is a largest number in your sequence, then it's upper bound, and if there's a lowest number in the sequence, then there's a limit-bound that we don't cross. If it has an upper and lower bound, then it's a bounded sequence. You can find one of the bounds by taking the limit, in many cases.
Also, 3. A sequence {a_n} is bounded if it is bounded above and bounded below.
Bounded monotonic sequences-- If a sequence {a_n} is bounded and monotonic, then it converges. If you have an upper bound AND lower bound, and it's monotonic, then it converges to whatever that upper or lower bound is. But you can't say that "if it converges, then it's monotonic," which is not true. In the next chapter we will talk about series. Sequences is the numbering pattern. Series is talking about the sum of all of the terms. The sum of the sequence is the series.
ex) Determine whether each sequence is monotonic. Discuss the boundedness of the sequence.
1) a_n = 4 - (1/n) .... monotonic, so it's bounded (and therefore upper and lower) (just for yourself: both 3 and 4 are the bounds)
2) a_n = (cos n)/n ..... not monotonic because cosine goes up and down .. so it's bounded at 1/n and another one, so this is going to be a cosine graph that gets smaller and smaller. You'd have to plug in the first couple terms to figure out the actual bounds, of course.
3) a_n = (2n^2 - 1) / (n+3) .... monotonic and lower bounded. How do you know that it's monotonic? Bounded says whether or not there's a limit to how big or small your numbers get. The monotocity? Use the quotient rule.
"Sensible Calculus" / Sequences (M. Flashman, 2005)
Math notes on sequences (Paul Dawkins, 2003-2007)
Short introduction to sequences (Mohamed A. Khamsi, 1996)
EquationSheet re: sequences & series
Wikipedia page on sequences
Relevant Wikibooks page on sequences
2007-10-16 - Introduction to series
Today's nonsense
- Today's lesson is dedicated to Bobby who rocks so there must also be paper and scizzors in here somewhere. First he had email troubles yesterday, and today he wasn't going to come because he's sick, but now he's here.
- Avery claims it is AntiAvery Day again, but that was clearly yesterday. What about the song that says that "short people have no reason to live," is that how it goes?
- Bingo next week.
- Math club on Monday morning? Student council is at that time. NHS is every other Tuesday.
- Homework due.
- The story of cornell and it starting to rain corn? (What's with this?)
- "The only constant in life is change."
- Do lakes have to have flowing water into it? Or are these only man-made lakes? (Katie's tangent ... blame her?)
- What was Spiller doing growing up in a Polish neighborhood?
A sequence is a number of terms in a series. So we're going to talk about series today, tagging.
Infinite series
sum from n=1 to infinity of a_n = a_1 + a_2 + a_3 + a_4 + ... + a_n + ...
These series go to infinity. So the question will be do the series, which are sigma which is summation, do they converge to a specific value or not? When you add up all the terms of the sequence, do they converge to a specific sum or not? Which are convergent and which aren't?
The terms have to be getting smaller for it to be convergent. Do they have to be monotonic? If it's monotonic and bounded, then it converges, but that's the sequence, so what about the series? Let's look at some partial sums.
Partial Sums
A partial sum means a sum of a finite number of terms. The first partial sum would be nothing but the first term of the sequence. The second partial sum would mean add the first two terms. So they might say, "estimate the sum of the series by using a seventh partial sum" meaning add the first seven terms, and so then you get an estimation of what the total sum is.
So in partial sums, what you do is you take the number of terms up to that nth partial sum, and find what that is, and after a few of these partial sums, you can start to look for a pattern, and then take the limit of that pattern to see if it converges. You have to be able to prove that a series converges to a particular number.
ex 2) the sum from n=1 to infinity of (1/n) - 1/(n+1)
- A telescoping series
ex 3) sum from n=1 to infinity of 2/(4n^2 - 1)
... Even though we are already using partial sums, we have to use partial fractions and this is because it's a telescoping series. We're going to want to call this "A over" and "B over" and what do those factor into? How do you factor 4n^2 - 1 ? That's 2n-1 and 2n+1. Our original problem can be rewritten as the same sum of 1/(2n-1) - 1/(2n+1) and this is in the form of the telescoping series. So now you have to find the sum, just say what the answer will be, because everything else drops out. List the first few terms and see what it is going to add up to be. So this is where you have the form where you have something canceling out, and quite often you will have to do partial fractions.
Now something that you dealt with in precalculus
Geometric series
Arithmetic: adding the numbers
Geometric series: where you are multiplying the terms.
Sum from n=0 to infinity of ar^n = a +
When n=0, what is the first term? In this case, each time in the geometric series, whatever is raised to the n, is the "r" or your common ratio. What is ar^2 divided by ar? It's r, and that's your common ratio. The same works for dividing the fourth by the third terms. This is a geometric series with ratio r.
Convergence of a geometric series
If you keep multiplying something by a number greater than one, is that thing ever going to converge? No. What if the r-value is 1? No, it'll be the same still. There has to be some kind of restriction on r in order to have a converging geometric series.
The geometric series diverges if |r| >= 1 so that means if |r| is below 1 then it will converge.
To converge, 0 <= |r| <= 1. The r-value can be negative (meaning the terms are alternating). In the case of 0 <= |r| <= 1, the geometric series converges to the sum (see below)
sum from n=0 to infinity of ar^n = a / (1-r) ---> and 0 <= |r| <= 1
We can prove that easily. Sometimes this is completely misunderstood ... some examples will now follow, to try to throw out that misunderstanding:
ex) Determine convergence/divergence. Find sum if convergent.
a) sum from n=0 to infinity of 3 (1/4)^n
Use the form that was given for the geometric series.
b) sum from n=0 to infinity of 4(5/4)^n
c) sum from n=1 to infinity of (2/3)(1/3)^n =
Paul's online math notes - convergence of series
Convergence tests for infinite series
BOX!
Videos - (Selwyn Hollis, 2005) (University of Houston)
2007-10-17 - More work with geometric/telescoping series and partial fractions
Today's nonsense
- Visitor! Mr. Warman? An awesome name.
Homework.
So far we have discussed: if it's telescoping, that's the one that looks like a fraction minus a fraction, and you write out the terms and they cancel, or partial fractions where we figure out the fractions and write out some terms and figure out the sum. The geometric formula could look misleading, because it's not the "a" in the formula but instead the first term (which may be defined differently, so check the summation symbol for that information). Let's try some more of "find the sum if it exists."
ex 1) sum from n=2 to infinity of 1/(n^2 -1) ... and so it's either a geometric series, telescoping series, or partial fractions, and so it's going to be partial fractions.
How do we apply this to repeating decimals? Say that we have 0.080808.... what then? You break it up into 8/10^2 and 8/10^4 and 8/10^6 etc.
Another box. Most of those homework problems will be geometric or telescoping because that's what we've covered so far.
Limit of n^th Term of a Convergent Series
If the series sum of a_n converges, then the sequence {a_n} converges to 0.
If the sum of all of the terms converges towards a specific value, then the sequence, the individual terms, converge to zero, so they are getting smaller and smaller. So the sequence is just what they end up at, but the series is what you do if you add them all up. If the series converges, that means the sequence, the terms themselves, must be getting closer to zero. If there is a sum that it heads towards, then the individual terms must be getting closer to zero.
-- If the sum converges to a specific value, then the terms get closer to zero.
First Test for Convergence
n^th Term Test for Divergence
If the sequence {a_n} does not converge to 0, then the series sum of a_n is going to have to diverge, if you're just adding + 1 + 1 + 1 etc., then what's that going to do? If it does not go to zero, then the series diverges. So take the limit of the sequence. You always start with the nth term test. If the limit is not zero, it diverges. If the limit is zero, you need to try another test.
A ball is dropped from 6 ft. The height of each bounce is three-fourths the present height. Find the total vertical distance traveled by the wall.
2007-10-18 - The Integral Test
Today's nonsense
- Seinfeld reference: we're all laughing at the little doodle from physics, but nobody seems to know what it's about.
- "Are you kidding me?" - Phoebe's senior quote. Officially.
- "Shift Happens" shirt v. "lie tangent to the curves shirt"
- Evidently I have to send an email out to Math club and thensome to Physics
If f is positive, continuous, and decreasing for x >= 1 and a_n=f(x), then the series that we are questioning (whether it diverges / converges) will do the exact same thing as the integral of the function does from 1 to infinity. So whatever the integral does, if you were to integrate from n=1 to n=infinity, whatever the integral does, the series does.
ex 1) Use the integral test to solve the sum from n=1 to infinity of n / (n^2 + 1)
-- Use u-substitution to solve the equivalent integral.
ex 2) sum from n=1 to infinity of 1 / (n^2 + 1)
-- This is an arctangent. Arctangent(infinity) is pi/2 and arctan(1) is pi/4 so the integral converges to pi/4 but we just need to know that the series/integral here converges.
All of these are playing off with what we did with integrals. So now we are going to talk about p-series.
P-series and Harmonic Series
We've mentioned this before, maybe not calling it P, but we've used the letter P.
The series from n=1 to infinity of 1/(n^p)
If we are going to infinity, if p is a certain number, the exponent on the variable in the denominator, well, it's going to converge with certain numbers. If p>1 it converges.
1. The p-series converges if p>1. Otherwise it's a natural log, and ln diverges.
2. The p-series diverges if 0 < p=<1. If p is negative, then it just flips back on top, and it diverges. So if it's negative, it's not really in the denominator.
The harmonic series is a special p-series. What possible special series could we have? One. Harmonic series is when p=1. So,
Series from n=1 to infinity of 1/n, and you know it converges to zero. So a harmonic series is a p-series, and it's when p=1. Harmonic tones are those with strings that have a certain length, and when played maybe their amplitudes all add up, musically? Whatever.
We know it converges both by the integral test and p-series. A general harmonic series is of the form:
series 1/(an+b)
The capital pi symbol is useful for multiplication, instead of the sigma symbol. Useful in combinations and permutations.
"Or a capital H with the top cut off." - Dakota.
Coffee table book about coffee tables. And then ether on the road, and then Numan's on fire, sleeping 15 minutes every 4 hours, ... Seinfeld reference overload.
ex 1) Sigma from n=1 to infinity of 1/n^2.
-- p series test because p>1. Or the integral test. Converges.
ex 2) Sigma from n=1 to infinity of 1 / (n ln(n) )
-- Diverges. Use u-substitution.
Homework due tomorrow.
17 min.
Concise statement of the integral test
Wikipedia: Integral test for convergence
Wolfram: Integral test
Math Refresher blog entry/article
Paul's notes on the integral test
Wikipedia: Harmonic series
Rearranging the alternating harmonic series (hey, look- some actual writing)
** "A Sweet Introduction to Series" (Dale Hoffman (no, not the other guy))
2007-10-22 - Review day
Today's nonsense
- Request to move the test back due to advisory period, Meg's baby shower, ..
- AIDS walk reference
- Avery's height, now officially at least 5'2''
- Avery's sandwhiches
- Magic land of the Trig == trigonometric substitution
- Test is now Wednesday/Thursday
- Camera battery charge is low
- Short shorts from decade day last Friday
#28 (Bobby) - See photograph
#3 (Phoebe) - She doesn't understand. See the photograph.
#4 - See the photograph. It converges because of the value of r, is less than one.
#5 - Phoebe - Area under the curve is integration.
#7 - Bobby - Determine which of the following converge. (A) diverges because p=1. (B) diverges because r>1 and it's geometric. (C) Diverges because p<1. (D) Diverges because p>1.
If the sequence converges to zero, you have to go on and try something else.
#21 - Phoebe - See photograph.
There is some integration by trig substitution. #26, #8, ... Bobby's #26.
#24 - Bobby - Determine if the following sequence converges/diverges. Sequence. Not series. So it's ((n+1)/n)^n. So we're just talking about the diverge. Which will it eventually converge or not? This is the whole "e" thing. It will probably be on the test.
#42 - Bobby - Determine the convergence or divergence of the series using the integral test: sum from n=1 to infinity of (1/n^2)cos(1/n)
This is u-substitution, and then you have -sin(1/n) from 1 to infinity, and if you plug in infinity, you get 0 - -sin(1). So if the integral converges, then the series converges. This is the integral test, or the "wall" thing that we do with our hands, it's similar to the wall, it's the integral version of the wall, it's the same idea except not.
#8 - Bobby - a trig problem (trig substitution)
Nth term test: take the limit as n->infinity, if you get a number anything other than 0, it diverges.
Some scribbling: what's on the review and what's on the test? If you're doing convergence/divergence of a series, now, the first thing you really want to look at, the nth term test, if the limit as n->infinity of your series does not equal zero, then the series is going to diverge. The nth term test determines your series. Sequence are the terms, series are the sums. Sequence means the pattern. What other things do we have to figure out divergence/convergence?
-- Geometric series test. How do you find the sum? It's the first term over 1-r. The absolute value of r has to be less than one in order to have a geometric sum work out here. So if r is greater than or equal to one, then it diverges, and there's no series to speak of.
-- Telescoping series. The telescoping series are the ones where you have to apply partial fractions. Because what you want to do is write out your first few terms and see what starts to cancel out. The ball bounce problem (2/3rds up every time, a geometric series, figure out the sum, double it, but then take away the first upwards distance because when you start to bounce it goes down once) is going to be on the test.
-- Integral test
-- P series (harmonic is just when it's 1/n or it's just when you integrate you get ln and so obviously with the harmonic it's going to diverge). This could even be 1/3n or 1/2n and obviously when you integrate this you are going to get a natural log. So more it's recognizing that it's a harmonic.
--
#15 - Ketteman -
#34 - Bobby/Dakota - This is the same sequence that you need to know. See one of the other photographs.
#38 - Same thing. The main concept.
#29 - Bill - you can prove divergence with integral test. If there was an integral test and p series test choice, then you could choose either one, so since you can do one of them you have to mark one of the available choices, even though you can think of another. This means that you have to know each type of the tests.
Comparison test with sequences instead of series, that's what we will be doing tomorrow now instead of the test.2007-10-23
Today's nonsense
- It shall be written that Ketteman was singing choir, and was thus offtopic, and loud enough for Spiller to hear it. Therefore we can now do physics whenever.
- "Because his battery is out of camera." - Katie (in a stroke of what might be brilliance).
Direct Comparison Test
Please do not blame Spiller if there is an answer choice of "direct comparison test" on the test. So you should forget this stuff on the test, but then remember them after this next test. Direct comparison, we're going to have to do, because you have some simple series, and what can you tell me about this series?
Series from n=0 to infinity of 1/2^n. It converges because r=1/2 and the series therefore converges to 2. It's the first term (in this case, 1), divided by one minus r, because the first term is at n=0.
-- Series from n=0 to infinity of n/2^n not geometric because you have a variable in your numerator. If you have any variables that are not a part of your "r" then it's not geometric. How in the world do we do this? We compare it to something that we do know how to do.
Series from n=1 to infinity of 1/n^3. The sequence converges to zero. It's a p-series because p>1 in this case (p=3) and in this case n in the denominator will converge.
-- There's also series fromn=1 to infinity of 1/(n^3 + 1), and it's not a p-series. We did some integration like that before, and well, we can compare that to 1/n^3, so 1/n^3 converges and 1/(n^3 + 1) is less than it, and so we're going to have the Wall theorem all over again.
Your series are going to start to look more complex.
a_n = n/(n^2 + 3)^2 is easy to integrate because it's u-substitution.
But ... b_n = (n^2) / (n^2 + 3)^2 ... and what is the magical method of integration? It's not u-sub. It's The Magical Land of Trig Substitution.
-- 3(tan theta)^2.
Direct Comparison Test (for real?)
The Wall Theorem for Series
Let 0 < = a_n < = b_n, for all n,
If Series B_n converges (because if the bigger one converges, then anything underneath it will converge it), then Series A_n converges.
If Series A_n diverges (because it's the smaller one, and anything above it will also diverge), then Series B_n (the bigger one) diverges.
So, the box is going to go: if this series converges, then this series converges.
Knowing that a_n is less than b_n, we can discuss the implications of the wall theorem.
- You could use this on the test tomorrow, but if it's an answer choice then don't choose it.
ex 1) Determine whether the series from n=1 to infinity of 1/(2+3^n) converges or diverges.
-- Ask whether or not 1/(3^n) converges or diverges. It converges to zero, and is bigger than the original, therefore the original converges as well.
Direct comparison test ... and then we will do the alternating series test next. And then the "great comparison test" (that's not what it's called). The ratio test is the way to go- the most powerful test, as you will find. Any questions on the test review?
Monotonic or nonmonotonic?
a_n = (3n+1) / n^2
So is it always nonincreasing or is it always nondecreasing? Take the derivative and prove that the derivative is always positive or always negative, and then it would be monotonic. The algebraic way says that, you take the first four terms (4, 7/4, ) and that get smaller, so you think the terms are getting smaller, so you want to prove that the a_n > = a_n+1 and so you try to show that it holds, so substitute it into the equation. This is the algebraic way of proving that it is monotonic. "n" is always going to be positive. If you cross multiply the comparison, you can write it out neatly to (3n+1)(n+1)^2 > = (3n+4)n^2. .... so let's look at the derivative version. 2007-10-26
Today's nonsense
- Homework due Monday. Posted just yesterday on the internet. Due Monday: pg. 628 #3-36
- Email to send out plus link ..
- Spiller was saluditorian, didn't know it until the end of his senior year in high school, lots of racial tension? From Detroit.
- Complaining about the problem on tension on the physics test.
- Spiller hit Ketteman with a ball and yelled "be quiet" (not absuive; not at all).
"Yesterday" we were working on the direct comparison test and now we are going to be continuing. And going on to the:
Limit Comparison Test
Suppose that a_n > 0, b_n > 0, and
limit from n -> infinity of ( a_n / b_n ) = L
where L is finite and positive. Then the two series (Series A_n) and (Series B_n) either both converge or both diverge.
ex 1) Use the limit comparison.test: series from n=1 to infinity of 1 / (2n+3).
- If it's Series of 1/n, then that diverges. So you say it's harmonic, so it diverges, because it's linear, well, let's look at this. The way this works, you're going to take the limit as you approach infinity of the ratio of the two, the one that you know is on the numerator and the one that is similar is on the denominator. What is the limit as n-> infinity of 1/(2n+3) all over (1/n) ... and essentially you look at the largest terms in the numerator and denominator and it's basically n/2n and that's 1/2. The limit comparison test says that if you compare the two functions, and you get a finite value, such as 1/2, then the two series that you are comparing are going to do the same thing. So if one of them you can figure out, and it diverges, then the other one has to diverge, so therefore both diverge in this case because you already know that 1/n diverges. So there you go.
2) Series from n=1 to infinity of (5root(n) + 6) / (2n - 11)
Compare to 1/sqrt(n) ... which diverges. And you can take the limit of the ratio which is something like 5/2. And since 1/sqrt(n) diverges, then our given problem also diverges.
3) Series from n=1 to infinity of 2/(3^n - 5)
Compare to 3^-n where r=1/3 and it's geometric and therefore it converges. The ratio is a finite and positive and nonzero number ... anything else means that it can't converge. You have to compare it with the same power-type function so that they will cancel it out and there will be a ratio of coefficients. And if you do it properly you can figure out that our original given problem (#3) converges.
4) Series from n=1 to infinity of sqrt(n) / (n^2 + 1).
- Compare to 1/( n ^ (3/2) )
5) Series from n=1 to infinity of n2^n / (4n^3 + 1)
27 min
2007-10-29
Today's nonsense
- Spiller's levels of pissed-offness, i.e. comparison of Katie v. Avery, which is kind of like a series test because we have to throw one against a wall.
- Video by the science teachers to the song of "Pink Cadillac." The neighbor-teacher took his old 66 Bewic, made fins out of cardboard, painted it pink, and drove around in Blues Brothers attire, with the sunglasses and everything, filmed it, etc. This guy had nothing better to do ... he used to make videos called "Mr. Science's Neighborhood."
Alternating Series
This is where the summation alternates between positive and negative terms in the series. How do we create this alternation? You do (-1)^n. And then you can start with to the power of n+1 and that's an odd number if you're starting at n=1 so that the original negative sign is preserved or not depending on what you're doing. Is that the only way to get alternating? No. You can use sine and cosine.
When is cosine=1? At 0 or at 2pi or 4pi or 6pi. And cosine is -1 at 1 etc.
Series from n=1 to infinity of a_n * cosine (pi * n) = - + - + - + - + ...
Series from n=1 to infinity of a_n * sin( pi/2 + pi * n ) = - + - + - + - + ...
And these are the only the alternating parts of the series that we are going to be dealing with. There are two steps that we are going to work with, and we combine them on the test, to determine diverging or converging.
Alternating Series Test
Let a_n > 0. The alternating series
Series from n=1 to infinity of (-1)^n * a_n
and
Series from n=1 to infinity of (-1)^(n+1) * a_n
converge, if the following two conditions are met.
1. It has to be nonincreasing- so: a_(n+1) < = a_n. Must stay the same or decrease.
2. n^th term test. If the limit as n-> of a_n = 0. This has nothing to do with the alternating sign, so the absolute value of the sequence has to be decreasing or staying the same.
Does it converge or diverge?
ex 1) Series from n=1 to infinity of (-1)^(n+1) * (1/n)
Can we just write converges or diverges? No. We have to acknowledge this stuff. So we have the alternating series test
Is a_n > = a_(n+1) ? Check. So you have to check if 1/n > = of 1/(n+1) ... and then you can see that n+1 > = n. And 1 > = 0, so yes, the first one works.
-- And does the limit as n->infinity of 1/n = 0? Check.
It converges by the alternating series test.
2) Series from n=1 to infinity of n / ( (-2)^(n-1) ) .......
So now plug this stuff in to the a_(n+1) < = a_n requirements and see if it qualifies. You can rewrite this series as (-1)^(n-1) * n/(2^(n-1)). So how about the alternating series tests? Can we check? Yes, checking is good. Can we try to verify that the first condition is met? Is 2n > = n + 1? Yes. As long as we are starting with n=1 or starting with n > 1. We have proved condition 1. So now we have to work with condition two ...
Then do the n^th term test. So what about the limit as n -> infinity of n / 2^(n-1) and the demonitor grows much faster than the numerator. So therefore it goes to 0. And so therefore this series converges by the alternating series test.
If it's an alternating series, now, alternating series means that every other term, so you can't go positive positive negative negative positive negative; every term is the opposite of the last one and the next one. If these two conditions of an alternating series are met, then it converges. 2007-10-30 - More on alt-series
Today's nonsense
- Gilson bashing. Apparently two students in Andrew's second-year algebra class have cried because of Gilson's death-ray eyes.
Today we are going to do more on alternate series.
Absolute and Conditional Convergence
If the series Sum of abs(a_n) converges, then the series Sum of a_n converges.
-- If some of the numbers are negative, then that's just going to cancel out in the first place. So if you have a series that has all positive terms, and if all of those terms somehow converge, and the series converges, then if the same series has some negative terms (even if it is not alternating) then that will also converge.
When you have an alternating series, it will not be good enough to say that it converges, you have to specify if it converges absolutely or conditionally. So if you do not specify anything, it means that it's absolutely convergent, but with alternate series there's also something called conditionally convergent that we will now deal with.
Definition of Absolute and Conditional Convergence
1. Series a_n is absolutely convergent if the Series of the absolute value of a_n is convergent.
2. Series a_n is conditionally convergent if the Series a_n converges but Series abs(a_n) is divergent.
ex) Determine if the following either converges absolutely or converges conditionally, or if it diverges.
a) Sum from n=1 to infinity of (-1)^n * 1/n ....
- The original series converges. However, the absolute value of the sequence diverges, so the original one converges conditionally.
b) Series from n=1 to infinity of (-1)^n * ( n^2 / (n+3)^2 )
- Diverges.
c) Series from n=1 to infinity of (-1)^n / (2n+1)!
- Absolutely convergent, but how can we show that the absolute value part is satisfied? We do not yet know the ratio test.
- Compare with n^2 ... and then you know that (2n+1)! > n^2 ... and n^-2 converges by the p-series test or the integral test.
Bonus: is there a way to make a series that has a negative term every third term? 2007-10-31 - Alternating series remainder theorem
Today's nonsense
Alternating Series Remainder
In a convergent alternating series, the absolute value of the remainder, R_N, involved in approximating the sum , s, by S_N is less than (or equal to) the first neglected term. That is, such that, just thought I'd throw in the such that, for a reminder, so Bill- this series goes on forever, we will do partial sums, the first partial sum is the first term, the fourth partial sum is the sum of the first four terms, okay, so partial sums, whatever numbers partial sums, that's just how many of the terms you've added up, and you can't add them up to infinity, and at some point the partial sums will stop, so like at 112, and the first neglected one is 113. So you're just approximating the sum. So the question is: what kind of approximation do you have? How accurate is it? Your approximation is within an accuracy of whatever your first neglected term is, in the case of alternating series.
|S - S_N| = R_N (the Remainder) < = a_(N+1)
Now, this remainder, this difference in your approximation and the actual answer, is going to be < = the next term, which has a_N+1.
So, say the next term was 7, and the next one is subtract 7, so that's why it can't be any bigger than the next term, so ..
ex 1) Approximate the sum of the following series by its first six terms.
Sigma from n=1 to infinity of (-1)^(n+1) * (1/n!)
-- Converges by AST.
91/144 - 1/5040 < = S < = 91/144 + 1/5040
ex) Approximate the sum of the series with an error of less than .001 ...
a) Series from n=1 to infinity of (-1)^(n+1) / 4^n
So we're looking for the first term less than 0.001, and then we add up all the terms prior to it, in this case that will be a total of four terms, and what do you get when you add up the four terms? And this sum is approximately .199 (becuase we are estimating to the nearest thousandths).
b) Series from n=1 to infinity of (-1)^n * e^(-n)
-- Have to add up seven terms. 0.732.
ex) Determine how many terms are needed to approximation the sum to within 0.001.
a) Series from n=0 to infinity of (-1)^n /(n+3)^2
- Just scroll through the tableset on the calculator in order to get to that point where you see the "within 0.001" and find that next neglected term. The first neglected term is term #30 where n=29, term #30 is the first term less than 0.001. So find the partial sum up to n=29.
b) Series from n=1 to infinity of 2(-1)^(n+1) / ( sqrt(x) * x^3 )
2007-11-02
Today's nonsense
- No bells because the administrators screwed up. Surprise?
- Homework due Tuesday.
1. Converges, Alt Series Test <-- you must tell whether or not it's conditional or absolute
2. Converges, nth Term Test <-- you can't have a convergence by an nth term test. If it's zero, you have to try another test.
3.Converges, Direct Comparions Test <-- What did you compare it to?
#7 Series from n=1 to infinity of sin(1/n) ... you were supposed to use the limit comparison to 1/n, and that gives you zero over zero, use L'Hopitals rule, and it is equal to 1, so it diverges. The sequence converges to 1, but the series diverges.
#10 Series from n=0 to infinity of (3^n + 2) / (2^n - 3)^2 .... which converges to 3^n / 4^n and that's geometric with r < 1 therefore it converges.
pg 645 #13-30, 37-70
Ratio Test
1. Series a_n converges absolutely if lim as n->infinity of |a_(n+1) / a_n | < 1. So this is just a geometric ratio where r is less than 1 and it converges.
2. Series a_n diverges if the lim as n->infinity of |a_(n+1) / a_n | > 1.
3. The Ratio Test is inconclusive if the lim as n->infinity of |a_(n+1) / a_n | = 1.
- So go try something else. Sorry, the ratio test couldn't help you today.
- See photograph of the Box.
ex 1) Determine convergence or divergence of Series from n=>0 to infinity of 2^n / n!
So take the limit as n->infinity of | ( 2^(n+1) / (n+1)! ) * n! / 2^n | = lim as n->infinity of | 2 / (n+1) | = 0 and that's less than 1, yes, so, this thing (series) converges.
- We multiplied by the reciprocal instead of just dividing like the Ratio Test box suggests.
2) Series from n=0 to infinity of n^2 * 2^(n+1) / 3^n
- The limit is (2/3). So this thing converges by the ratio test.
3) Series from n=1 to infinity of n^n / n!
- Use the ratio test when you have a factorial or an "n" exponent, so this is a good candidate to use the ratio test on.
- The lim as n->infinity of | (n+1)^(n+1) / (n+1)! * n!/n^n | .... and this is a really, really really good one to simplify.
lim as n->infinity of | (n+1)^(n+1) / (n+1) * 1/n^n |
lim as n->infinity of | (n+1)^n / n^n | = lim as n->infinity of | ((n+1)/n )^n |
lim as n->infinity of | (1+ 1/n)^n | = e>1 So it diverges because e is greater than 1.
Later we will do more with the ratio test. After the test.
4) Series from n=1 to infinity of (-1)^n * sqrt(n) / (n+1)
- Recognize that this is an alternating series too. But we'll be using the ratio test.
---> lim as n->infinity of | sqrt(n+1) / (n+2) * (n+1) / sqrt(n) | =
And that's basically n^(3/2) to the n^(3/2) ... so it's practically 1 and it's inconclusive, so moving on ...
- By AST it converges conditionally, because 1/n^(1/2) by the p-series, 1/2 is less than 1, so ...
-- Also look up the root test.
If you take away the power, you take the nth root, then the thing you have left has to be less than 1, and that's sort of what we've been talking about all along.
2007-11-05
Today's nonsense
- Signed up the Math Club to the mailing list.
- Sean Rodgers
- Homework due tomorrow. We haven't talked about the pitiful root test.
- "No. We were in the valley. We were supposed to be roofing it but instead we were sticking around on the roof, so we shot nail guns at other roofs, and we could hear that the nails went far, you can ask Tyler Cromwell, they were up there with me, Aaron Mash, they were all down there working, we had to work, ..." - Ketteman's random story.
- Evidently we have notes from the Math Club. Thinking about a soup kitchen night, and taking the month of December off. Thinking about doing some boring activity while watching the movie on November 15th.
- Do the rabbits have tattoos in their ears and the pigs the tag? They squeal. You have to cut goat horns, they start bleeding, then they bleed for the whole day if you're name is Phoebe and you're freaking out. Spiller thinks that goats are psychotic.
-
The Pitiful Root Test so that we can make it an even number
Let Series a_n be a series with nonzero terms.
1. Series a_n converges absolutely if limit as n->infinity of (|a_n|)^(1/n) < 1
2. Series a_n diverges if lim as n->infinity of (|a_(n)|)^(1/n) > 1.
3. The Root Test is inconclusive if
lim as n->infinity of (|a_(n)|)^(1/n) = 1
ex 1) Series from n=1 to infinity of e^(2n) / n^(n) === Series from n=1 to infinity of (e^(2) / n )^(n)
- So this test is all about figuring out what is being raised to what power. "This is the exact same as the geometric series test." - Dakota's revelation. Your "r" is now a specific value, and here it is much like an r, and if that thing much like an r is approaching a number less than one, you raise it to an nth root and it's already raised, so when you raise it to the nth root you're just taking away the n, so you're just looking at the inside part. So what we are asking in particular is:
limit as n->infinity of e^2 / n ... and 0 < 1 so it will converges absolutely.
Geometric series ... diverges if it is greater than or equal to 1. But in this case, of the root test, if it's already equal to 1 then the ratio test is inconclusive.
2) Series from n=1 to infinity of ( (n+1)/(2n+1) )^(n) ...
- Goes to 1/2 .. converges. Geometric series test would point out that 1/2 is less than 1, so that's where geometric tests says it converges.
2007-11-12
Today's nonsense
- 32 hours of labor.
- "Find the one with the best review, the best review, cheapest best reviews. Alright. Shhh. Elven? Thirteen, fourteen, counting- ? I'm saying, counting, but I'm counting everyone that is on the roll. Yeah, we said that about the. Alright, are you ready?"
- Dakota's rear thermometer
Homework due Wednesday: pg. 656 #1-4,13,16,17,19-21,25,28-30
Polynomial Approximations
You did this last year, but only in the form of a linear approximation.
ex 1) For f(x) = e^(x), find a first degree polynomial, in other words you are finding the equation of the tangent line. Sincei t's first degree, we're going to say P_(1) = a_(1)x + a_(0) whose value and slope agree with the value and slope of f at x=0. So check f(0) and now d[f(0)]. So P_(1)(x) = x+1. The slope is 1, so it's 1x + something, but when you plug in zero, you should get one, and so what do you have to do to get the constant, so ... the constant is one.
First degree polynomial = linear. There are no fractional exponents in any polynomial.
Second degree polynomial = quadratic.
2) Find a second degree approximation for f(x)=e^(x), P_(2) = a_(2)x^(2) + a_(1)x + a_(0).
So we want all of these conditions to be the same. So we need f(x), f'(x), and f''(x). We want f(0), f'(0) and f''(0) to all match for e^(x) and our approximation. In the linear approximation they have the same point and the same slope. The second derivatives must equal each other, so 2a_(2) = f''(0). When I plug 0 into P' what does it equal = a_(1), so a_(1) must be 1. If I plug 0 into P, what do you get? a_0 = 1. So P_(2) = (1/2)x^2 + x + 1. So look at the pattern ... the cubic approximation would be (1/6)x^(3) + (1/2)x^2 ... etc. It's a factorial for that first coefficient. 24 = 4!. The exponent matches the 1/the_factorial.
Taylor and Maclaurin Polynomials
Definition of n^th Taylor polynomial and Maclaurin polynomial.
These are polynomials that will approximate a function. There is in fact a pattern. You were totally oblivious to what you were doing, and then at least you will believe that there is a pattern.
If f has n derivatives at c, then the polynomial:
P_(n)(x) = f(c) + f'(c)(x-c) + f''(c) /2! * (x-c)^2 + ... + fn(c)(x-c)^(n) / n!
And this is known as the nth Taylor polynomial for f at c.
If c=0, then it's a Maclaurin polynomial and it reduces down to
P_(n) = f(0) + f'(0)x + f''(0) x^(2) / 2! + f'''(0)x^(3) / 3! + .... + f(n)(0)x_(n) / n!
is called the nth Maclaurin polynomial for f.
2007-11-13
Today's nonsense
Not all Taylor are Macluarins, but all Maclaurins are Taylors. If I wanted a polynomial of whatever degree you wanted, now, it fits the pattern, yesterday a reference was made to "the zeroth derivative". You see, you take your first, and, yeah, the old one was bothering me, yeah, until somebody comes up and writes on the overhead and mashes it, but it happens, alright, so, what is my first term going to be? Well. Remember. Not f of x, because a Macluarin is centered at zero, so f of 0, and technically this still fits the pattern, what derivative is this? The zeroth. So this is the same as f(0)x^o all over 0! and there's no extra term out front. If you take the first derivative out front, then that'd be x to the one over one factorial, and so whatever factorial I'm dealing with that's the exponent on the x and that's the factorial that you are dividing by, and the next one ... yeah. x to the second divided by 2 factorial and it just keeps going on and on and on like that, so the third derivative is x^3 and over 3! and that's all that we are going to do today, so for a fourth degree polynomial we will have to do four derivatives, in most cases, there's a little trick to help you save a little time, and it's a little trick (yeah, right), so let's do example one. Find the I forget the terminology that they use let's do they eighth degree polynomial approximation, I should have just put P sub eight, in the future I will, for f(x)=cos(x) centered at 0. So let's look at how we can rewrite this question and you will come to understand it. I could have written: find let's say the Maclaurin. What if I had just wrote that? Macluarin tells you it's centered at zero, and P_8 means to the eighth degree, you will need eight derivatives, and 8 derivatives is normally annoying, and since it's cosine it just flips back and forth sine cosine sine cosine with pluses and minuses. To do this problem we are going to start off with F(x) = cos(x) and then we are going to find f prime, f double prime, f triple prime, the fourth, fifth, sixth, seventh, and eighth derivative, and most of them aren't this long but like I said this one is quick because so what is F prime and F double prime? F double prime should be - cosine(x) and f prime x should be -sin(x). So the fourth derivative is cos(x), and the sixth one is negative, and the eighth one is positive. So this is positive sine and this is negative sign and this is positive sign. So you take that many derivatives that you need that degree. Now what do we do? Okay. You're going to plug in whereever you are centered you are going to plug that in to that derivative because that is going to give us our coefficients. When we wrote out f prime of zero and f triple prime of zero those are actual numbers and those numbers come from evaluating the derivative. So what is f prime of zero, so what is f double prime of zero, and of course you might want to abbreviate but the fact that this is ... 1, 0, -1, 0, 1, 0, -1, 0, 1, so there aren't things that get much easier than this, the only thing that does get much easier is if you were doing e^x and the derivative is e^x and e^x and e^x alright? So now we just write out our polynomial, if you got that now we just write out our pattern. I don't have room .. I have some terms that go away, don't I? What's the first thing that I am going to write. One. And that one is our original x, and technically that's .. .and all of that is one, so it's just the number one. The next term is zero, I don't have to write it, but if you wanted to, it'd be zero times x^1 divided by x factorial, and the next one would be minus one times x to the 2, so it's times x^2 over 2 factorial, and the next term would be zero x cubed over 3 factorial, the next term would be 1x^4 over 4! Oh yeah there's a pattern, yeah? You can leave the factorials, if there happen to be other coefficients, or yeah you could simplify it out, she's saying if you want to leave it like that our multiply it out and put 24, you could keep going or you will notice as Bryan pointed out you'll start to see the pattern. If I asked for P sub 20 then it's not so much work you start to see the pattern and then the next term would be .... so if I wanted to approximate what's the cosine of 0.1 radians? I could plug 0.1 into this polynomial, so the more terms you have .. you could graph cosine x and type this into y1 and you will see that for a little while they look like they overlap each other. Notice that this is actually an alternating series. Somebody made a comment about the exponents, they are all even, and remember talking about the odd and even functions? Cosine is even, so all of the exponents on the polynomial should be even too. Without thinking too hard, what is the polynomial for sine? X over 1 factorial minus x cubed over 3! and then add x^5 over 5! and then subtract x^7 over 7! and e^x? Well, e^x happens to be alright this gets very strange. Do you know what the first few terms of e^x are? Do you know that this is one half x squared? This is x and this is one. So try this. What do you get if you take the absolute value of this plus the absolute value of this? Take away the alternating part ... can you write the polynomial as a sigma or summation? You would write Sum from n=0 to infinity of something like x^2n over (2n)! and then you need your alternating portion so something like (-1)^(n+1)... could you write sine as a summation? It would just be x^(2n+1) , and if I do 2n+1 what is n starting with? So the sum would be from n=1 to infinity of x^(2n+1) / (2n+1)! (the exponent factorial is in the denominator). Once you start to see the pattern you can just write the sigma. So it starts to get a little amazing and then we want to do e^-3x and that's what we are going to do so let's try another example. If you take away the alternating and add them together, and if we can do e, and e is very easy, and you start to wonder, e is defined as how far you have to go over from 1/x starting from 1 so that you have an area of 1, and then you find that this infinite summation has a limit of e, then you have (1+(1/n))^n and that's e, and then there's some kind of pattern to this as if nature has an alignment to it. Well, no, e is based on natural log. Bryan can actually throw us in something that makes us laugh. I thought it was funny too. I wish he would interject more. I know, that's what happens, you get the reputation of being punny and people start looking for a fun and then they don't get it. So! Does everybody know what happened today in third period when you left? I saw Bryan and Luisa and myself in the lab and there was a lunch and I was contemplating ... you had the Doritos, some kind of snack, yogurt? It's a smudge on the overhead, and then a comma, so this is a fourth degree polynomial, here's what you were thinking, if you haven't any clue what's going on, you have to take up to the fourth derivative, and then you're going to evaluate each of those derivatives at, zero because it's a Maclaurin, so it's very systematic, and it's just, systematic. So here's where if you mess up and put 1/2 e^2x because you are getting derivatives and integration confused, it's 2e^2x and .... so. Derivative of e^u is e^u or u'e^u, and oh yeah, derivatives aren't gone. F = e^2x and F prime = 2e^2x and F double prime = 4e^2x. Uh oh. Look at him typing away. So you said this is one, two, four, eight, sixteen. Just write the polynomial. P_4 of (x). Oh I don't know. An dtypically they put parenthesis around that. Ah-huh. Well. Yesterday we were just showing where the pattern came from, and pretty much that was the only, yeah, yeah, yeah, no, no, this is what we're doing right now, all you need is, it's just a pattern. So I can write this. It's just the answer now. It's as simple as, and this is where (ARM MOVEMENT), one. Plus. 2x^1 / 1! which is just 2x and I am not going to write the extra stuff if you want to you can. I can see where this is going. I will write your answer down here. See you are starting to reduce and that's going too far and it's like the second example they've done and it's not very fun, and it's not like they've done the homework already. 4/2! which is known as 4/2 which is known as 2. And 8/3! reduces to 4/3 .. and then 16/4! reduces to 2/3 times x^4. Now. If you were to write this in summation notation, summation notation is going to be pretty difficult looking at Luisa's reduced form, and not on this homework, no, sometimes it's just nice to see this so that you can see the pattern. Oh. Let's see if you can figure out what the pattern is. Can you write this in summation notation? Sum from n=0 to infinity of what's the bottom going to be? Oh you haven't figured out the exponent? Is the exponent n? Okay, the exponent is n, and the bottom is n factorial? What you might not recognize is that this is 2x paranthesis cubed, so it's just 2x. Do you know what e^x is in summation notation? You start taking mighty shortcuts when you learn this. X^n over n! so then you start to say so could I if I knew e^u which would be u^n instead of x^n if I wanted e^2x would I just put e everywhere there is an u or everywhere there is an x? Let me show you one shortcut, so I am going to do a problem that is very very similar. So now when you go back they are going to charge you a signature and a planner. You are not as smart as we ... are you tired? Did you close last night? Oh that's why you look tired. You were the one who brought it up, I didn't ask if you were manager last night. Amafooze? Sonic. No. She doesn't skate. So. Do you have to skate? Do new employees have to skate? You know, every single perso