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Thread: How to do this maclaurin /a.n series

  1. #1
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    How to do this maclaurin /a.n series

    So i just took my final exam and felt confident about it except for one question that sorta stumped me.

    It is: find the first 6 a(sub)n numbers. Assume that the maclaurin series an*x^n is for (1+x+x^2)cosx.

    I wasn't sure how you are suppose to get numbers with the X's in that summation.

    I tried to write (1+x+x^2)cosx in the form of an*x^n and found that the numbers for a0 , a1, a2, a3 , a4 ,a5 = 1/2, 1/4!, 1/6!, 1/8!,1/10!
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  2. #2
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    Re: How to do this maclaurin /a.n series

    There are several different questions. You could use the basic definition of McLaurin series- that the coefficient of $\displaystyle x^n$ is the nth derivative or the function, evaluated at 0, divided by n!. And, of course, you would find those derivatives using the product rule.

    If you already know the McLaurin series for cos(x) it is much easier- the McLaurin series for cos(x) is $\displaystyle 1- x^2/2+ x^4/4!- x^6/6!+ x^8/8!- x^{10}/10!+\cdot\cdot\cdot$. Now just multiply that by $\displaystyle 1+ x+ x^2$.
    Last edited by HallsofIvy; May 3rd 2018 at 03:27 PM.
    Thanks from lc99
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    Re: How to do this maclaurin /a.n series

    Ahh. I was too lazy to actually multiply those two out. I didn't know that that was the correct way. gosh darn!
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