[SPLIT] write the taylor series for f(x) = x e^x about x = 0

[Smily]

can someone help me? :roll:

write out the taylor series for f(x) = x e^x about x = 0

thank you,

 

[skeeter]

the maclaurin series for e^x is ...

\(\displaystyle \L e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... + \frac{x^n}{n!} + ...\)

so ... what would xe^x look like?

 

[Smily]

i need to do this problem using taylor series -
i did
f(x) = xe^x
f'(x) = (x + 1 )e^x
f''(x) = (x + 2 )e^x
f'''(x) = (x + 3 )e^x
f^n(x) = (x + n! )e^x

f(0) = 0
f'(0) = 1
f''(0) = 2
f'''(0) = 3

please, tell me what i need to do after it :roll:

 

[skeeter]

the Taylor series centered at 0 (also known as a Maclaurin series) for
f(x) = xe^x is ...

\(\displaystyle \L f(x) = x(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... + \frac{x^n}{n!} + ...)\)

\(\displaystyle \L f(x) = x + x^2 + \frac{x^3}{2!} + \frac{x^4}{3!} + ... + \frac{x^{n+1}}{n!} + ...\)

now ... to show you this is so, here's the more difficult method you want to do all laid out for you.

f(x) = xe^x ... f(0) = 0
f'(x) = (x+1)e^x ... f'(0) = 1
f"(x) = (x+2)e^x ... f"(0) = 2
f'''(x) = (x+3)e^x ... f'''(0) = 3
...
fⁿ(x) = (x+n)e^x ... fⁿ(0) = n

The general form for a Taylor series centered at 0 is ...

\(\displaystyle \L f(x) = \frac{f(0) \cdot x^0}{0!} + \frac{f'(0) \cdot x^1}{1!} + \frac{f"(0) \cdot x^2}{2!} + \frac{f'''(0) \cdot x^3}{3!} + ... + \frac{f^n(0) \cdot x^n}{n!}\)

sub in your values ...

\(\displaystyle \L f(x) = \frac{0 \cdot x^0}{0!} + \frac{1 \cdot x^1}{1!} + \frac{2 \cdot x^2}{2!} + \frac{3 \cdot x^3}{3!} + ... + \frac{n \cdot x^n}{n!}\)

simplify each term of the series ...

\(\displaystyle \L 0 + x + \frac{x^2}{1!} + \frac{x^3}{2!} + \frac{x^4}{3!} + ... + \frac{x^n}{(n-1)!} + \frac{x^{n+1}}{n!} + ...\)

now ... compare to the original series formed by multiplying x times the maclaurin series for e^x ...

\(\displaystyle \L f(x) = x + x^2 + \frac{x^3}{2!} + \frac{x^4}{3!} + ... + \frac{x^{n+1}}{n!} + ...\)