Maclaurin Series for Expanding sin(2x)^2: Step-by-Step Guide

Click For Summary

Discussion Overview

The discussion revolves around expanding the function \( f(x) = (\sin(2x))^2 \) into a Maclaurin series. Participants explore various identities and series expansions related to trigonometric functions, focusing on the correct application of these identities and the implications of coefficients in the series.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses difficulty in expanding \( f(x) = (\sin(2x))^2 \) into a Maclaurin series and suggests that changing the form of the function is key.
  • Another participant challenges the initial transformation proposed, stating that \( (\sin(2x))^2 = 2\sin(2x^2) \) is incorrect due to the sign inconsistency.
  • A different identity is introduced: \( \sin^2 x = \frac{1}{2}(1 - \cos(2x)) \), which is then applied to \( \sin^2(2x) \) resulting in \( \sin^2(2x) = \frac{1}{2}(1 - \cos(4x)) \).
  • Concerns are raised about the coefficient \( 0.5 \) in the series expansion and whether it should be accounted for in the final result.
  • It is noted that the constant term of the series for \( \sin(2(2x)) \) is zero, and the \( 0.5 \) coefficient cancels out with the negative coefficient from the expansion of \( -0.5\cos(4x) \).

Areas of Agreement / Disagreement

Participants do not reach a consensus on the treatment of coefficients in the series expansion. There are competing views on the correctness of the initial transformation and the implications of the coefficients involved.

Contextual Notes

There are unresolved questions regarding the handling of coefficients in the series expansion and the implications of the transformations used. Participants rely on specific identities and series expansions without fully resolving the implications of their choices.

vabamyyr
Messages
65
Reaction score
0
i have trouble expandind f(x)= (sin2x)^2 into Maclaurin series
for sin(x) Maclaurin series is

[tex]\sum^{\infty}_{n=0} (-1)^n \frac{x^{2n+1}}{(2n+1)!}[/tex]

probably the key is to change (sin2x)^2 into new shape. I found that

(sin2x)^2=2sin(2x^2), but that coefficient 2 is bothering me, what to do?
 
Physics news on Phys.org
vabamyyr said:
(sin2x)^2=2sin(2x^2), but that coefficient 2 is bothering me, what to do?
That relation does not hold (the left side is always positive, the right side isn't).

But you can use another identity:

[tex]\sin^2 x=\frac{1}{2}(1-\cos 2x)[/tex]
 
[tex]\sin^2 2x=\frac{1}{2}(1-\cos 4x)=0.5-0.5cos4x[/tex]


and now if i apply for cosx maclaurin series expansion considering the function f(x)= -0,5cos4x i get the right answer but I am puzzled, where does the coefficient 0,5 go? i don't have to count that??
 
Last edited:
The constant term of the series for sin2(2x) is 0.
The 0.5 cancels the -0.5 from the expansion of -0.5*cos(4x).
 
Last edited:

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
8
Views
3K