Question

How do i expand f(x) = (2+x)/(3+x)*exp(-x) about -3? Taylor series seems to go to infinity. Wolfram alpha seems to come up with a series that doesn't go to infinity? The derivative of f(x) has a (3+x) term in the denominator so every term of f^(n)(a) goes to infinity. Is there another kind of series where this will work?

Answer 1

Can you link your result from WolframAlpha? I can just guess and assume that the computation by wolfram alpha does it via substitution, where it interprets \((2+x)/(3+x)\) as a polynomial and multiplies the series representation of \(\exp(-x)\) by it.

Answer 2

http://www.wolframalpha.com/input/?i=taylor+series++%282%2Bx%29%2F%283%2Bx%29*exp%28-x%29+about+-3

Answer 3

I'm pretty stumped so far as well

Answer 4

funny how the x+3 terms go up in exponent with each term in the series

Answer 5

is the answer at the bottom the correct series?

\[\sum _{n\geq 0} \frac{(-1)^n e^3 (3+x)^n (n!+(1+n)!)}{n! (1+n)!}+\sum _{1+n=0} -e^3 (3+x)^n\]

Answer 6

aha...

Answer 7

For when n = -1 in this series it would put the 3+x in the bottom only once. Again this makes no sense to me because at 1+n = 0 the term on the right is infinite.

Answer 8

dang it, my idea didn't pan out
something along the lines of rewriting it as\[(1-\frac1{3+x})e^{-x}\]is floating around in my head, but it's not getting me anywhere

Answer 9

I know they discuss this in MIT's OCW, I'm looking for the resource

Answer 10

look at problem 2A-12
http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-2-applications-of-differentiation/part-c-mean-value-theorem-antiderivatives-and-differential-equations/problem-set-5/MIT18_01SC_pset2prb.pdf

Answer 11

part e
here is the solution
http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-2-applications-of-differentiation/part-c-mean-value-theorem-antiderivatives-and-differential-equations/problem-set-5/MIT18_01SC_pset2sol.pdf

Answer 12

the trick they use is to substitute x=h+1

Answer 13

thanks but i dont even see the need to substitute h+1. i can differentiate and expand xln(x).

\[xln(x)\] \[f^{1}(x) = \frac{x}{x}+ln(x) =1+ ln(x) \]
\[f^{2}(x) = 1/x\]

and so on. i see no problem especially when u expand at 1. Maybe I'm missing something about taylor series

Answer 14

The professor totally mentions this exact situation in the lectures somewhere, but I just don't recall how he deals with it. If you feel like browsing yourself he mentions it somewhere in this series http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/index.htm

Answer 15

Thanks for trying turing. I appreciate it. I posted this question on math.stackexchange.com . Hopefully this is appropriate for there. They tend to get annoyed with me quickly on stack exchange websites lol.

Answer 16

I found the answer. See the post on math.stackexchange.com

http://math.stackexchange.com/questions/660181/taylor-series-about-3x-x4exp-x-expanded-at-x-4-how-do-i-replicate/660186#660186

Answer 17

Thanks for letting me know, it was driving me nuts.