# Thread: maclaurin series

1. ## maclaurin series

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

2. 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

3. 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

4. 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?

5. 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

6. 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

7. how would you find the Maclaurin series for this then:
1/(1+x^5)

8. 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"?

9. Originally Posted by Jhevon
"pansy-butted"?
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

10. 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.

11. 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.

12. 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 $|x|<1$

13. Now me, I tend to forget that $\sum_{i = 1}^n 1 = n$. (I try not to memorize too much! )

-Dan

14. 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.

15. 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? $\sum_{n = 0}^{\infty}(-1)^n\left( x^5 \right)^n = \sum_{n = 0}^{\infty}(-1)^nx^{5n}$