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Thread: Maclaurin Series Help

  1. #1
    Len
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    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.
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  2. #2
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    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
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  3. #3
    Len
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    Quote Originally Posted by running-gag View Post
    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?
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  4. #4
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    $\displaystyle f(x)=e^x sin(x)$ 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'
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  5. #5
    Len
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    Quote Originally Posted by running-gag View Post
    $\displaystyle f(x)=e^x sin(x)$ 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?
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  6. #6
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    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
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  7. #7
    Len
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    Quote Originally Posted by running-gag View Post
    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?
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  8. #8
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    Quote Originally Posted by Len View Post
    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 $\displaystyle x^6$

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