What is the Taylor expansion of x/sin(ax)?

In summary, the Taylor expansion of $$f(x) =\frac{x}{sin(ax)}$$ around $$x_0 = 0$$ is $$x/\sin(x) = 1 + \frac{x^2}{6} + \frac{7x^4}{360} + \mathscr O(x^6)$$. This can be found by using the series form of sine and setting ##y = \sin(x)/x - 1##, then plugging in the series of ##\sin(x)/x## into the series of ##1/(y+1)## and neglecting all terms of degree six or higher. Keep track of these neglected terms using the ##\mathscr O## notation.
  • #1
RedDwarf
10
0
Hey everyone
1. Homework Statement

I want to compute the Taylor expansion (the first four terms) of $$f(x) =x/sin(ax)$$ around $$x_0 = 0$$. I am working in the space of complex numbers here.

Homework Equations


function: $$f(x) = \frac{x}{\sin (ax)}$$
Taylor expansion: $$ f(x) = \sum _{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$

The Attempt at a Solution


I thought I could use the series form of sine:
$$sin(ax) = \sum (-1)^n \frac{(ax)^{2n+1}}{(2n+1)!} $$
$$x/sin(ax) = \sum (-1)^n \frac{ (2n+1)! } { a^{2n+1} }x^{-2n}$$
While this is in fact a series, this doesn't look like a Taylor expansion at all. Is there a clever way of seing the Taylor expansion without actually calculating all the derivatives by hand?
Wolfram Alpha gives a rather neat result, but I have no clue how one gets there.
 
Physics news on Phys.org
  • #2
That is not how division works. You have essentially assumed that 2=(1+3)/2 = 2/1 + 2/3 = 8/3.

I suggest you do the following:
- What is the expansion of sin(x)/x?
- What is the expansion of 1/(1+y)?
- Let y = sin(x)/x - 1.
- Remember that you only need to keep terms up to ##x^3## as you are only interested in the four first terms.
 
  • Like
Likes RedDwarf
  • #3
Orodruin said:
That is not how division works. You have essentially assumed that 2=(1+3)/2 = 2/1 + 2/3 = 8/3.

I suggest you do the following:
- What is the expansion of sin(x)/x?
- What is the expansion of 1/(1+y)?
- Let y = sin(x)/x - 1.
- Remember that you only need to keep terms up to ##x^3## as you are only interested in the four first terms.
The expansion of ##\sin (x)/x = 1-x^2/3!+x^4/5!-... = \sum (-1)^n \frac{x^{2n}}{(2n+1)!}##
The expansion of ## 1/(1+y) = 1-y+y^2-y^3+... = \sum (-1)^n y^n##
Then, if I set ##y = \sin (x)/x -1##:
##x/sin(x) = 1- (sin(x)/x-1)+(sin(x)/x-1)^2-(sin(x)/x-1)^3+... =3-sin(x)/x + sin^2(x)/x^2- 2sin(x)/x - (sin(x)/x-1)^3+...
= 3 - 3 sin(x)/x + sin^2(x)/x^2-...##

I am not sure how to continue from here. How can I plug in the series of sin(x)/x into the series of 1/(y+1) without everything blowing up immediately? And what do I do with the terms ##sin^2(x)/x^2## and those of higher order, where I am basically supposed to calculate the square of an infinite sum?
Thanks for your help though, I feel much closer to an answer already!
 
  • #4
As already stated, you do not need to keep the entire infinite sum as you are only interested in the first terms. You can write (for example)
$$
\sin(x)/x = 1 - \frac{x^2}{6} + \frac{x^4}{120} +\mathscr O(x^6).
$$
The ##\mathscr O## will help you keep track of the order of the terms you have neglected.
 
  • #5
Orodruin said:
As already stated, you do not need to keep the entire infinite sum as you are only interested in the first terms. You can write (for example)
$$
\sin(x)/x = 1 - \frac{x^2}{6} + \frac{x^4}{120} +\mathscr O(x^6).
$$
The ##\mathscr O## will help you keep track of the order of the terms you have neglected.
Let's see if I can do this:
If ##\sin(x)/x = 1 - \frac{x^2}{6} + \frac{x^4}{120} +\mathscr O(x^6)##, then
$$1/(y+1) = 1-1 + \frac{x^2}{6} - \frac{x^4}{120} +1 + (-\frac{x^2}{6} + \frac{x^4}{120})^2
= 1+\frac{x^2}{6} - \frac{x^4}{120} +\frac{x^4}{36} - 2\frac{x^6}{6*120}+ \frac{x^8}{120})^2$$
Neglecting all terms with ##x^6## or higher:
$$x/\sin(x) = 1 + \frac{x^2}{6} + \frac{7x^4}{360} + \mathscr O(x^6)$$
Yeah, that looks like what I have seen before. Thanks a million!
 
  • #6
You really should be writing out the ##\mathscr O## in every step or your equalities will not be correct. I also would not bother computing the higher order terms explicitly as they will be eaten by the ##\mathscr O## anyway and it already contains unknown coefficients. Just shove them into it right away. The general approach is valid though.
 

Related to What is the Taylor expansion of x/sin(ax)?

1. What is the Taylor expansion of x/sin(ax)?

The Taylor expansion of x/sin(ax) is a mathematical series that represents the function in terms of its derivatives evaluated at a specific point. In this case, it is expanded around the point x=0.

2. How is the Taylor expansion of x/sin(ax) derived?

The Taylor expansion of x/sin(ax) can be derived using the Taylor series formula, which involves taking derivatives of the function and plugging in the desired point. In this case, the function is expanded around the point x=0, so all the derivatives are evaluated at that point.

3. What is the purpose of the Taylor expansion of x/sin(ax)?

The Taylor expansion of x/sin(ax) is used to approximate the function at values near the expansion point. It can also be used to find the values of the function at points where it may be difficult to evaluate directly.

4. How accurate is the Taylor expansion of x/sin(ax)?

The accuracy of the Taylor expansion of x/sin(ax) depends on the number of terms included in the series. The more terms that are included, the more accurate the approximation will be. However, adding too many terms can also lead to computational errors.

5. In what situations is the Taylor expansion of x/sin(ax) commonly used?

The Taylor expansion of x/sin(ax) is commonly used in engineering and physics applications, particularly in problems involving oscillatory functions. It is also used in numerical analysis to approximate functions and solve differential equations.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
979
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
548
Replies
12
Views
940
  • Calculus and Beyond Homework Help
Replies
1
Views
455
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
206
  • Calculus and Beyond Homework Help
Replies
27
Views
2K
  • Calculus and Beyond Homework Help
Replies
17
Views
795
Replies
2
Views
2K
Back
Top