# Math Help - Taylor Series Help

1. ## Taylor Series Help

Hi everybody, I hope anyone could help

1. The problem statement, all variables and given/known data

Find the first three terms of the Taylor series for f(x) at c.

2. Relevant equations

f(x)= f(c) + f'(c).(x-c)/1! + f"(c).(x-c)^2/2! + f'''(c).(x-c)^3/3! + ... + fn(c).(x-c)^n/n! + ...

3. The attempt at a solution

what I understand is that I have to find the followings:
f'(x), f''(x), f'''(c)
and
f'(c), f''(c), f'''(c)

is that right?

well, to find f'(x) I used the product rule

d/dx (uv) = u'v + uv'

u = x
u' = 1

v = e^x
v' = e^x

d/dx (xe^x) = e^x + xe^x
= e^x(1+x)

but unfortunately i couldn't go forward!!

the answer in the back of the book is

2. As alternative approach You can consider that $y(x)= x\ e^{x}$ is solution of the DE...

$\displaystyle y^{'} = y + e^{x} , y(-1)= -\frac{1}{e}$ (1)

From (1) You derive that...

$\displaystyle y(-1)= - \frac{1}{e}$

$\displaystyle y^{'} = y + e^{x} \implies y^{'}(-1)= 0$

$\displaystyle y^{''} = y^{'} + e^{x} \implies y^{''}(-1)= \frac{1}{e}$

$\displaystyle y^{(3)} = y^{''} + e^{x} \implies y^{(3)}(-1) = \frac{2}{e}$

...

$\displaystyle y^{(n)} = y^{(n-1)} + e^{x} \implies y^{(n)} (-1)= \frac{n-1}{e}$

... so that is...

$\displaystyle x\ e^{x} = \frac{1}{e} \sum_{n=0}^{\infty} \frac{n-1}{n!}\ (x+1)^{n}$ (2)

Merry Christmas from Italy

$\chi$ $\sigma$

3. Suppose $g(x) = x$, and $h(x) = e^{x}$, so that $f(x) = g(x)\cdot h(x)$.

Observe that $g^{n}(x) = \begin{cases}
1\mbox{ if } n = 1\\
0\mbox{ if } n\ge 2 \end{cases}$
and $h^n(x) = e^x$, for all $n \in\mathbb{N}$.

Using Leibniz's theorem, we get:

[tex]\displaystyle f^{n}(x)\bigg|_{x=-1} = \sum_{0 \le k \le n}\binom{n}{k}g^{n-k}(x)h^{k}(x)\bigg|_{x=-1} = \sum_{0 \le k \le n}\binom{n}{k}x^{n-k}{e}^{kx}\bigg|_{x=-1}[/Math]

. . . . . . . . . . $\displaystyle = \sum_{0 \le k \le n}\binom{n}{k}e^{x(n-k)}x^{k}\bigg|_{x=-1} = e^{x}\left(x+n\right)\bigg|_{x=-1} = \frac{1}{e}\left(n-1}\right)$.

Thus, as Chisigma found, by Taylor's theorem we have:

$\displaystyle f(x) = \sum_{n=0} ^{\infty} \frac {f^{(n)}(c)}{n!} \, (x-c)^{n} = \sum_{n=0} ^{\infty} \frac {\frac{1}{e}\left(n-1\right)}{n!} \, (x+1)^{n} = \frac{1}{e}\sum_{n=0} ^{\infty} \frac {\left(n-1\right)}{n!} \, (x+1)^{n}.$

4. Many, many thanks for all your help

Thank you (chisigma), and Merry Christmas from Saudi Arabia : )

Tkank you (
TheCoffeeMachine), your cooperation is highly appreciated