Question

expand sin (x-x^2) in power series until term containing x^4

Answer 1

The quick way would be to use the power series for \(\sin x\):
\[\sin x=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}\]
Substituting \(x-x^2\) for \(x\), you have
\[\sin (x-x^2)=\sum_{n=0}^\infty \frac{(-1)^n(x-x^2)^{2n+1}}{(2n+1)!}\]
Then plug in as many \(n\) as you need to get a term containing \(x^4\).

Answer 2

Thanks! Btw, can I express this by using limits?

Answer 3

There is a little problem with the answer above. different terms (for different n's) will contain "x" terms of same degree.

Answer 4

hm. I'm wrong actually, but the power series must be of the form \(\sum a_k x^k\). each \(k\) deals with only one \(x^k\).

Answer 5

can i express in the form of maclaurin series?

Answer 6

You need either to compute completely all derivatives of the function you are given, or use another "better trick".

use that: \(\sin(y) = y-y^3/3! + y^5/5! -...\) and put \(y=x-x^2\). After rearranging terms, you will have it in the form \(sin(x-x^2) = ..x + ..x^2 + ...x^3 + ...x^4 + ...\).

Answer 7

Actually we only need those 2 terms: \(y-y^3/3!+ -..\), with y=x-x^2, because \(y^5\) will give you \(x^5\) and above, and you don't need that.

here: \(y-y^3/3! = (x-x^2) - (x-x^2)^3/3! = x-x^2 - [x^3-3x^2x^2 + 3xx^4 - x^6]/3!\).
you can forget the \(x^5\) and the \(x^6\) term.

Answer 8

I'm a bit confuse. How to get this? \[\sin(y)=y−y3/3!+y5/5!−...\] Any name for this methods? Thanks :)

Answer 9

This is a "well known" formula. I would say that if you study Taylor studies, this is one you should remember, also the cos and the \(e^x\) function.

Answer 10

But you can obtain the series by using the taylor formula.. \(\sum_n \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n\)

Answer 11

Taylor series*

Answer 12

and the \(\frac{1}{1-x}\) series.

Answer 13

So all I need to do is to express \[y= (x-x^2)\] and sub into the series later? And can I directly use maclaurin series? since it's just a special case for taylor's series.

Answer 14

yes. this works well, but you need to be certain that you CANNOT obtain any terms x,x^2,x^3, or x^4 in the following terms. Here, we see easily that y^5 will NOT give any term lower than x^5, therefore we can forget this term and all the following ones.

Answer 15

maybe this is called substitution method. to be verified

Answer 16

there is an example of this technique on wiki https://en.wikipedia.org/wiki/Taylor_series#Calculation_of_Taylor_series

Answer 17

I try directly differentiating the value \[\sin (x-x^2)\] but the value i get is 0 for x^2 and x^4. I think your way is the best solution so far.