Thread: find the maclaurin series

1. find the maclaurin series

i've gotten part of the way through into this problem and i'm just a little bit confused..

find the maclaurin series of f(x)= xe^x

and i can't solve it with a geometric series, i have to use the definition for a maclaurin series itself

so, i took the 1st, 2nd, & 3rd derivatives, and determined that the pattern was that the nth derivative will be e^x(n+x)

then i said that evaluated at zero, the nth derivative is e^0 (n+0) = n

so i plugged it into the formula of the nth derivative evaluated at zero times x^n all over n!

i said the first few terms were x + x^2 + 1/2 x^3 + 6x^4+...

now i'm sort of clueless as to where to go from here
please help!

2. Originally Posted by buttonbear
i've gotten part of the way through into this problem and i'm just a little bit confused..

find the maclaurin series of f(x)= xe^x

and i can't solve it with a geometric series, i have to use the definition for a maclaurin series itself

so, i took the 1st, 2nd, & 3rd derivatives, and determined that the pattern was that the nth derivative will be e^x(n+x)

then i said that evaluated at zero, the nth derivative is e^0 (n+0) = n

so i plugged it into the formula of the nth derivative evaluated at zero times x^n all over n!

i said the first few terms were x + x^2 + 1/2 x^3 + 6x^4+...

now i'm sort of clueless as to where to go from here
please help!
maclaurin series for $\displaystyle e^x$ :

$\displaystyle e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$

multiply by $\displaystyle x$

$\displaystyle xe^x = x + x^2 + \frac{x^3}{2!} + \frac{x^4}{3!} + ...$

3. i was told not to use anything that we proved in class (i.e. the maclaurin series for e^x) and to start from scratch, that's why i was doing all that work

4. $\displaystyle f(x) = xe^x$ ... $\displaystyle f(0) = 0$

$\displaystyle f'(x) = (x+1)e^x$ ... $\displaystyle f'(0) = 1$

$\displaystyle f''(x) = (x+2)e^x$ ... $\displaystyle f''(0) = 2$

$\displaystyle f'''(x) = (x+3)e^x$ ... $\displaystyle f'''(0) = 3$

$\displaystyle f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + ...$

$\displaystyle xe^x = 0 + x + \frac{2x^2}{2!} + \frac{3x^3}{3!} + \frac{4x^4}{4!} + ...$

$\displaystyle xe^x = 0 + x + x^2 + \frac{x^3}{2!} + \frac{x^4}{3!} + ...$

5. okay, so, this is the taylor polynomial for f(x) = xe^x?

and so then that means that it's true whenever the remainder is zero.. so i'm kind of confused as to how to show that

and how should i find the radius of convergence?

6. Originally Posted by buttonbear
okay, so, this is the taylor polynomial for f(x) = xe^x?

and so then that means that it's true whenever the remainder is zero.. so i'm kind of confused as to how to show that

and how should i find the radius of convergence?
Use the ratio test to get the interval of convergence