# Taylor Series/Limits/Convergence Test

• May 28th 2009, 12:04 PM
km5305
Taylor Series/Limits/Convergence Test
Hey everyone,

I am having some trouble working through these problems, and I was hoping that someone can help me out, or at least put me on the right track. I managed to get through all of them but I'm pretty sure I messed them up.

#1
Define the function f(x) in two pieces by
f(x) = 0 if x ≤ 0, and f(x) = e^(-1/x^2) if x > 0.
Show that the function f(x) is continuous at x = 0. Furthermore, show that f`(0) exists. Your life will be made easier if rather than considering
lim d/dx e^(-1/x^2)
x->0+
you explain why we may instead consider
lim d/dt e^(-t^2)
t-> ∞
Then show that f(ⁿ)(0) exists for all n.
Write the Taylor Series for f(x) centered at 0. Show that this series converges everywhere, but for x > 0 does not converge to f(x).

#2
Suppose that you have the series

an,
n = 1
and assume that
lim |an|^(1/n) = L.
n->∞
Show that if L < 1, then the series converges (absolutely), and if L > 1, the series diverges. Finally, give a specific example of a series where L = 1 and the series converges, and another example where L = 1 and the series diverges.

#3
Assume that the Taylor series for e^z converges for not just for all real numbers z, but for all complex numbers z as well. By setting z = ix, where i^2 = −1 and x is real, show that:
e^ix = cos(x) + i sin(x).

Use this formula to derive Euler’s formula:
e^iπ + 1 = 0.

I apologize that this doesn't look very nice, as well. Not really sure how to do the symbols properly in a text form.
• May 28th 2009, 12:37 PM
Any chance you could maybe tell us where you got stuck so we don't go over stuff you've got sorted. Just small notes below...

1. $t=\frac{1}{x}$ as x->0, $t->\infty$. Hence $e^{-t^2} -> 0$ so conclude convergence using that etc.
For $f^n(0)$ you could differentiate a few times to get a general formula on how its going to turn out. Should always have $e^{\frac{-1}{x^2}}$ in each term so hence each term -> 0.

2. There's a proof of that here Root test - Wikipedia, the free encyclopedia

What are you stuck on?
• May 28th 2009, 12:48 PM
km5305
Well, do I start the last one using the Taylor Series for e^z, and substitute z=ix into that series? Like:

[(ix)^n]/n!
n=1

I guess turning that into cos(x) + i sin(x) is kinda confusing to me right now. It makes me think that I need to use the Fourier Series. After spending so much time on it, I am feeling a little brain dead.

As for the part about Euler's Formula, is that just as simple as substituting x=π?
• May 28th 2009, 01:22 PM
expand the taylor series of $e^{iz}$ in the same way you would $e^x$ to get...

$e^{iz} = 1 + iz + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} +$ $\frac{(iz)^5}{5!} + \cdots$

$= 1 + iz - \frac{z^2}{2!} - \frac{iz^3}{3!} + \frac{z^4}{4!} + \frac{iz^5}{5!} - \cdots$

$= (1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots) + i( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots )$

Which are the expansions of cos(z) and sin(z)

And for the second part yes just sub in $\pi$
• May 28th 2009, 01:33 PM
km5305
Wow, I guess I was trying to make that last one wayyy too complex. As soon I saw the "expand it out" it clicked. Thanks for your help!