2007-11-19
Today's nonsense
- Homework due today and tomorrow.
- Pencil led crises
Use Taylor's theorem to obtain an upper bound for the error of the approximation and then calculate the exact value of the error.
arcsin(0.4) is approx. 0.4 + (0.4)^3 / 3! ......
R_(3) < = | f^(4) of (z) * x^(4) all over 4! | ...... the fourth derivative is -3x(2x^2 + 3) / sqrt(1-x^2)(x^(2) - 1)^(3)
So when is this thing going to be at a maximum? To find a max or min take the next derivative. Plug in 0.4 and you get 7.334 for that derivative. So, 7.334 * (0.4)^4 all over 4! ... and our remainder has to be less than or equal to that. So what does that evaluate to? R_(3) < = 0.00782 ....
(#57) f(x) = e^(x) is approx. 1 + x + x^(2) / 2! + x^(3) / 3!
The error cannot exceed 0.001.
R_(3) < = | e^(z) * x^(4) all over 4! | < = 0.001.
When is 1/e^(z) its largest? At 1. So then we have x^(4) / 4! which we want to be < = 0.001.
So: |x^(4)| < = 0.024.
|x| < = 0.3936 but we know that x must be less than zero, therefore x goes from -.3936 < = x < = 0 because they told me it had to be less than zero. So the absolute value is the part that is telling me ...
(#59) f(x) = cos(x) is approx. 1 - x^2 / 2! + x^(4) / 4!
We are trying to do the same thing. On alternating series, it's the next neglected term, so in this case you do not do R_(5) but instead R_(6)(x) .... otherwise it does not work out.
Examples
... continuing from long ago.
Find the interval of convergence
2) Series from n=0 to infinity of (-1)^(n) (x+1)^(n) / 2^(n)
Start with the ratio test: if the problem has anything to do with intervals and radii, then start with the ratio test.
Limit as n->infinity of | (x+1)^(n+1) * 2^(n) / ( (2^(n+1)) * (x+1)^(n) ) | = | (x+1) / 2 | < 1
c = -1 (what value of x will make it zero? x+1 is your x - c) ...
Now what is the radius?
|x -c| < R
So, x -c would be x+1 therefore you're centered at -1, and to get that by itself, you have to do what to get it by itself? Multiply by two. Therefore, our radius is two.
So what is our interval, state the end points but don't say if we include them or not yet, so it goes from -3 to 1, you're centered at -1 so go two to the left and two to the right.
If you get a limit of infinite, then the radius is zero; if the limit is zero, then the radius is infinite.
Plug in x=-3 and what does this problem become? Series from n=0 to infinity of 1. The series diverges. The nth term test: the nth term is one, so it diverges. Since this diverges, there's a parenthesis on the left side, so it's (-3, 1 ....
Plug in x=1 and what does this problem become? Series from n=0 to infinity of (-1)^n and the next term is not always less than the previous term, so it diverges. So your interval is from (-3, 1) with nonopen ends.
Do the ratio test. Check the endpoints. This isn't all that bad. I'm sure you could say that. And if you had a question, I hear that spoons, somebody who sits at the counter tomorrow morning would have questions.
ex 3) Series from n=1 to infinity of x^(n) / n^(2)
Show that you are checking each endpoint. |x-c| < R.
[-1,1]
ex 4) f(x) = Series from n=1 to infinity of x^(n) / n = x + x^(2) / 2 + x^(3) / 3 + ......
Find the three different intervals of convergence. The first one is:
(a) Integral of f(x)dx
Series from n=1 to infinity of x^(n+1) / ( n(n+1) )= (1/2)x^(2) + (1/6)x^(3) + (1/12)x^(4) + (1/20)x^(5)
Now use the ratio test. You just get x. |x| < 1 .... So it's centered at 0, and the radius is one, so now we have to check the end points. c = 0 and R = 1.
(c) f'(x)
Taking the derivative of a power series. There's a very fine point on the test that you have to be careful with.
f'(x) = Series from n=1 to infinity of x^(n-1)
f''(x) = Series from n=1 to infinity of (n-1)x^(n-2) ....