find the maclaurin series

**buttonbear** · April 15th 2009, 04:44 PM

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!

**skeeter** · April 15th 2009, 05:11 PM

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 $e^x$ :

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

multiply by $x$

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

**buttonbear** · April 15th 2009, 05:15 PM

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

**skeeter** · April 15th 2009, 05:29 PM

$f(x) = xe^x$ ... $f(0) = 0$

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

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

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

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

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

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

**buttonbear** · April 16th 2009, 07:54 PM

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?

**matheagle** · April 16th 2009, 10:26 PM

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

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