1. ## Maclaurin Series Help

Find the terms through $\displaystyle x^6$ in the Maclaurin series for f(x)

$\displaystyle f(x)=e^x sin(x)$

I'm not sure where to start this problem. I also have

$\displaystyle f(x)=\dfrac{1}{1+x+x^2}$

but perhaps with help to the first I will be able to solve the second. If you have time tho help/ hints would be appreciated. Thank you.

2. Hi

Mac Laurin series of f is

$\displaystyle \sum^{+\infty}_{n=0} \frac{f^{(n)}(0)}{n!}\:x^n$

where $\displaystyle f^{(n)}$ is the n-th derivative of f

You have to calculate the derivatives of f up to the 6th one and use the formula

3. Originally Posted by running-gag
Hi

Mac Laurin series of f is

$\displaystyle \sum^{+\infty}_{n=0} \frac{f^{(n)}(0)}{n!}\:x^n$

where $\displaystyle f^{(n)}$ is the n-th derivative of f

You have to calculate the derivatives of f up to the 6th one and use the formula
Could you or someone do the first derivative just so I can see how the formula is used?

4. $\displaystyle f(x)=e^x sin(x)$ is a product of 2 functions

$\displaystyle u(x)=e^x$ and $\displaystyle v(x)=sin(x)$

You need to use the formula of derivation of a product : f'=u'v+uv'

5. Originally Posted by running-gag
$\displaystyle f(x)=e^x sin(x)$ is a product of 2 functions

$\displaystyle u(x)=e^x$ and $\displaystyle v(x)=sin(x)$

You need to use the formula of derivation of a product : f'=u'v+uv'
So

$\displaystyle f'(x)cosx e^x + sinx e^x$

So when I get the derivative what do I do?

6. Yes but it is better to factor

$\displaystyle f'(x)=e^x (\cos x + \sin x)$

You have to calculate all the derivatives up to the 6th one

7. Originally Posted by running-gag
Yes but it is better to factor

$\displaystyle f'(x)=e^x (\cos x + \sin x)$

You have to calculate all the derivatives up to the 6th one
I'm not sure I understand this and I appreciate your on-going help. The derivatives are as follows:

$\displaystyle f''(x)= 2cosx e^x$
$\displaystyle f'''(x) = -2sinx e^x + 2cosx e^x$
$\displaystyle f''''(x)=-4sinx e^x$
$\displaystyle f'''''(x) = -4sinx e^x - 4cosx e^x$
$\displaystyle f''''''(x)= -8cosx e^x$

What is the next step?

8. Originally Posted by Len
I'm not sure I understand this and I appreciate your on-going help. The derivatives are as follows:

$\displaystyle f''(x)= 2cosx e^x$
$\displaystyle f'''(x) = -2sinx e^x + 2cosx e^x$
$\displaystyle f''''(x)=-4sinx e^x$
$\displaystyle f'''''(x) = -4sinx e^x - 4cosx e^x$
$\displaystyle f''''''(x)= -8cosx e^x$
OK

The next step is to apply the formula

$\displaystyle \sum^{+\infty}_{n=0} \frac{f^{(n)}(0)}{n!}\:x^n$

where $\displaystyle f^{(n)}$ is the n-th derivative of f

You are asked only the terms up to $\displaystyle x^6$

$\displaystyle \sum^{6}_{n=0} \frac{f^{(n)}(0)}{n!}\:x^n$