As alternative approach You can consider that is solution of the DE...
(1)
From (1) You derive that...
...
... so that is...
(2)
Merry Christmas from Italy
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
Suppose , and , so that .
Observe that and , for all .
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]
. . . . . . . . . . .
Thus, as Chisigma found, by Taylor's theorem we have: