# maclaurin series

• July 23rd 2007, 03:49 PM
davecs77
maclaurin series
How would you find the maclaurin series for this
sin(x^5)
thanks.
• July 23rd 2007, 03:55 PM
Jhevon
Quote:

Originally Posted by davecs77
How would you find the maclaurin series for this
sin(x^5)
thanks.

recall that the Maclaurin series for sine is:

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

so for $\sin \left( x^5 \right)$, just replace $x$ with $x^5$ in the series above and simplify
• July 23rd 2007, 04:02 PM
topsquark
Quote:

Originally Posted by davecs77
How would you find the maclaurin series for this
sin(x^5)
thanks.

A McLaurin series is a Taylor series for x = 0. So we have:
$f(x) \approx f(0) + \frac{1}{1!}f^{\prime}(0)x +
\frac{1}{2!}f^{\prime \prime}(0)x^2 + ~ ...$

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

So
$f(0) = 0$

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

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

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

etc.

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

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

-Dan
• July 23rd 2007, 04:03 PM
davecs77
Quote:

Originally Posted by Jhevon
recall that the Maclaurin series for sine is:

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

so for $\sin \left( x^5 \right)$, just replace $x$ with $x^5$ in the series above and simplify

So you are just memorizing that for sin(x) or Maclaurn series?
Would that be the same for Taylor series?
• July 23rd 2007, 04:05 PM
Jhevon
Quote:

Originally Posted by davecs77
So you are just memorizing that for sin(x) or Maclaurn series?
Would that be the same for Taylor series?

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
• July 23rd 2007, 04:09 PM
topsquark
Quote:

Originally Posted by Jhevon
recall that the Maclaurin series for sine is:

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

so for $\sin \left( x^5 \right)$, just replace $x$ with $x^5$ in the series above and simplify

Or you can take Jhevon's pansy-butted approach and skip all the work. But why do that when you could do all that work? :)

-Dan
• July 23rd 2007, 04:11 PM
davecs77
how would you find the Maclaurin series for this then:
1/(1+x^5)
• July 23rd 2007, 04:13 PM
Jhevon
Quote:

Originally Posted by topsquark
But why do that when you could do all that work? :)

-Dan

because mathematicians don't like hard work.

there are some things you want to try once just to see if you can do them, and then afterwards you take the shortcut. finding Taylor/Maclaurin series expansions are such things.

"pansy-butted"?
• July 23rd 2007, 04:16 PM
topsquark
Quote:

Originally Posted by Jhevon
"pansy-butted"?

:o It was the best I could come up with on short notice. Hey, I'm a Mormon. I'm not supposed to swear so that limits my insulting remarks. :)

-Dan
• July 23rd 2007, 04:20 PM
davecs77
Quote:

Originally Posted by davecs77
how would you find the Maclaurin series for this then:
1/(1+x^5)

Does anyone know? I don't think you can memorize this one. Thanks.
• July 23rd 2007, 04:22 PM
ThePerfectHacker
Quote:

Originally Posted by davecs77
how would you find the Maclaurin series for this then:
1/(1+x^5)

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

You do the rest.

Note: You need to remember the most important power series as well as you name.
• July 23rd 2007, 04:29 PM
Jhevon
Quote:

Originally Posted by davecs77
Does anyone know? I don't think you can memorize this one. Thanks.

actually you can. this happens to be one of the easiest to memorize in fact. the appropriate one to remember is

$\frac {1}{1 - x} = \sum_{n = 0}^{\infty} x^n = 1 + x + x^2 + x^3 + ...$ for $|x|<1$
• July 23rd 2007, 04:35 PM
topsquark
Now me, I tend to forget that $\sum_{i = 1}^n 1 = n$. (I try not to memorize too much! :) )

-Dan
• July 23rd 2007, 04:35 PM
davecs77
Quote:

Originally Posted by ThePerfectHacker
$\frac{1}{1+x}=1-x+x^2-x^3+...$

You do the rest.

Note: You need to remember the most important power series as well as you name.

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.
• July 23rd 2007, 04:46 PM
Jhevon
Quote:

Originally Posted by davecs77
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.

he treated $\frac {1}{1 + x}$ as $\frac {1}{1 - (-x)}$ and then used the formula i gave you.

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