Finding the Taylor Series for y(x)=sin^2x

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Homework Help Overview

The discussion revolves around finding the Taylor series for the function y(x) = sin²(x). Participants are exploring methods to derive a general series representation without relying solely on repeated differentiation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster seeks a method to develop a general series for sin²(x) without performing extensive derivatives. Some participants suggest multiplying the Taylor series of sin(x) to derive the series for sin²(x). Others question the feasibility of using derivatives to reach the nth term.

Discussion Status

The discussion is active, with participants offering different approaches, such as series multiplication and derivative methods. There is no explicit consensus, but various lines of reasoning are being explored, including concerns about the remainder in the multiplication process.

Contextual Notes

Participants are navigating the constraints of deriving a series representation while considering the implications of using derivatives and series multiplication. The original poster expresses a need for a method that avoids extensive derivative calculations.

transgalactic
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how to find the taylor series for
[tex] y(x)=\sin^2 x[/tex]
i need to develop a general series which reaches to the n'th member
so i can't keep doing derivatives on this function till the n'th member

how to solve this??
 
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Multiply the two Taylor series together (or Maclaurin series if that's what you're using).
[tex]sin^2(x) = (x - x^3/3! + x^5/5! -+ ... + (-1)x^{2n + 1}/(2n + 1)! +...)((x - x^3/3! + x^5/5! -+ ... + (-1)x^{2n + 1}/(2n + 1)! +...)[/tex]

The first term will be x^2
 
Why can't you "keep doing derivatives on this function till the n'th member"?

y= sin^2(x)
y'= 2sin(x)cos(x)= sin(2x)
and the rest is easy.
 
what about the remainder ob the multiplication??
 

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