How would you find the maclaurin series for this

sin(x^5)

thanks.

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- Jul 23rd 2007, 03:49 PMdavecs77maclaurin series
How would you find the maclaurin series for this

sin(x^5)

thanks. - Jul 23rd 2007, 03:55 PMJhevon
recall that the

**M**aclaurin series for sine is:

$\displaystyle \sin x = \sum_{n = 0}^{\infty} \frac {(-1)^n}{(2n + 1)!} x^{2n + 1} = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} +...$ for$\displaystyle x$__ALL__

so for $\displaystyle \sin \left( x^5 \right)$, just replace $\displaystyle x$ with $\displaystyle x^5$ in the series above and simplify - Jul 23rd 2007, 04:02 PMtopsquark
A McLaurin series is a Taylor series for x = 0. So we have:

$\displaystyle f(x) \approx f(0) + \frac{1}{1!}f^{\prime}(0)x +

\frac{1}{2!}f^{\prime \prime}(0)x^2 + ~ ...$

$\displaystyle f(x) = sin(x^5)$

So

$\displaystyle f(0) = 0$

$\displaystyle f^{\prime}(x) = cos(x^5) \cdot 4x^5$ ==> $\displaystyle f^{\prime}(0) = 0$

$\displaystyle f^{\prime \prime}(x) = 20x^3 cos(x^5) - 25x^8sin(x^5)$ ==> $\displaystyle f^{\prime \prime}(0) = 0$

$\displaystyle f^{\prime \prime \prime}(x) = -125x^{12}cos(x^5) + 60x^2 cos(x^5) - 300x^7sin(x^5)$ ==> $\displaystyle f^{\prime \prime \prime}(0) = 0$

etc.

I'll give you a hint. After a lot of hard work on your part, you will find that $\displaystyle f^{(6)}(0)$ is the first non-zero term in the series.

(This one's just*nasty*to do by hand!)

-Dan - Jul 23rd 2007, 04:03 PMdavecs77
- Jul 23rd 2007, 04:05 PMJhevon
I don't like telling people to just memorize stuff, but yes, you should have the Maclaurin series for functions like sine, cosine, e^x, 1/(1 - x) ... memorized.

the Maclaurin series is just the Taylor series centered at zero. topsquark showed you how to derive it, but in this case, i think the memorization route is easiest. unless your professor requires you to derive it - Jul 23rd 2007, 04:09 PMtopsquark
- Jul 23rd 2007, 04:11 PMdavecs77
how would you find the Maclaurin series for this then:

1/(1+x^5) - Jul 23rd 2007, 04:13 PMJhevon
- Jul 23rd 2007, 04:16 PMtopsquark
- Jul 23rd 2007, 04:20 PMdavecs77
- Jul 23rd 2007, 04:22 PMThePerfectHacker
- Jul 23rd 2007, 04:29 PMJhevon
- Jul 23rd 2007, 04:35 PMtopsquark
Now me, I tend to forget that $\displaystyle \sum_{i = 1}^n 1 = n$. (I try not to memorize too much! :) )

-Dan - Jul 23rd 2007, 04:35 PMdavecs77
The answer in the book is summation from n = 1 to infinity of (-1)^n times x^(5n)

I see the (-1)^n since the values are changing from positive to negative every time, but where does the x^(5n) come from? Also how did you find your results... 1 - x + x^2 - x^3 ... I don't quite understand. Thank you. - Jul 23rd 2007, 04:46 PMJhevon
he treated $\displaystyle \frac {1}{1 + x}$ as $\displaystyle \frac {1}{1 - (-x)}$ and then used the formula i gave you.

where did the $\displaystyle x^{5n}$ come from? did you actually look at the series? $\displaystyle \sum_{n = 0}^{\infty}(-1)^n\left( x^5 \right)^n = \sum_{n = 0}^{\infty}(-1)^nx^{5n}$