First three terms of Taylor series for xe^x at c=0

[vortex2008]

Hi everybody, I hope anyone could help

Homework Statement

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

http://dc12.arabsh.com/i/02388/kgybq4dwkug3.png

Homework 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! +...

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
http://dc12.arabsh.com/i/02388/bwk1diti9rjb.png

 

[micromass]

So the first derivative is $$f'(x)=e^x+xe^x$$. What happens if you differentiate again?

 

[vortex2008]

hmmmmmmmmmmm
i'm not sure, but let us say:
(e^x) remains as it is= (e^x)
and (xe^x) will be =(e^x + xe^x)

so the Ans. should be= e^x + e^x + xe^x

 

[micromass]

Why the sad face? Your answer is correct, so you can be happy!

 

[vortex2008]

reaaaaaally!
excuse me because I'm beginner

so,
f'(x)= e^x + xe^x
f''(x)= e^x + e^x + xe^x, and
f'''(x)= e^x + e^x + e^x + xe^x

is that correct? if so;
can i sum (e^x) with each other??
so it becomes
f''(x)=2e^x + xe^x
f'''(x)=3e^x + xe^x

 

[micromass]

Yes, that is all entirely correct.

So, in general, we have $$f^{(n)}(x)=ne^x+xe^x$$, but you don't need that here...

 

[vortex2008]

Thank you very very much :)

but, what should I do to find f(c),f'(c),f''(c), and f'''(c)?

the formula we have studied in the class looks like this:
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!+...

 

[micromass]

c=-1, right?

All you have to do, is substitute x with -1. So, for example, $$f(-1)=-e^{-1}$$.

 

[vortex2008]

I couldn't reach the same form of the Final answer :(that's what I did:

f(c)= -e^-1
f'(c)= e^-1 - e^-1 = 0
f''(c)= 2e^-1 - e^-1 = e^-1
f'''(c)= 3e^-1 - e^-1 = 2e^-1then I substitute in the formula:

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!+... and I got:

http://dc15.arabsh.com/i/02389/bg4i8qwon5a0.png

 

[micromass]

What is your final answer?

 

[vortex2008]

This is my answer:

and this is the answer in the back of the book:

 

[micromass]

Ah yes. So only the last part is different.

Your value of f'''(c) is correct: 2e^-1. But somehow, you wrote that as $$\frac{1}{2e}$$. This is incorrect, that should be $$\frac{2}{e}$$...

 

[Char. Limit]

Your error is this:

f'''(c)= 3e^-1 - e^-1 = 2e^-1

2e^-1 = 2/e, not 1/(2e)

EDIT: Looks like Micromass beat me...

 

[vortex2008]

ooooh I see!
thanks guys, but what about vectorial 3? (3!) it shouldn't be 6 ?

 

[Char. Limit]

ooooh I see!
thanks guys, but what about vectorial 3? (3!) it shouldn't be 6 ?

1. It's actually called a factorial... but whatever, I got what you meant.

2. You have 2/(3!e) = 2/(6e). How do you simplify that?

 

[micromass]

What is $$\frac{f(-1)}{3!}$$??

EDIT: now you've beat me to it

 

[vortex2008]

looooooooooool, I think I have too much misunderstanding!

I really fully appreciate your help
thanks (micromass), thanks (Char. Limit)

and I'm sorry for my bad English ^_^


good night to all :)