How to find the taylor for sin(x)^2 w/ sin(x), is this right?

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SUMMARY

The discussion focuses on finding the Taylor series for sin(x)^2 using the Taylor series expansion of sin(x). The correct approach involves squaring the entire series for sin(x), which is represented as the infinite sum sum((-1)^k * (x^(2k+1)/(2k+1)!)) from k=0 to infinity. Participants clarify that squaring the series requires a double sum to account for cross terms, rather than simply squaring each term individually. This method allows for the derivation of the first few terms of the series for sin(x)^2.

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Homework Statement



sin(x)= sum((-1)^k* (x^(2k+1)/(2k+1)!))k=0 to infinity

Homework Equations



so if i want to find sin(x)^2, (not sin(x^2), that would be easier though...)

The Attempt at a Solution


then...
do i square the whole thing, like this?

sum(((-1)^k* (x^(2k+1)/(2k+1)!))^2)k=0 to infinity

thanks a bunch!
 
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You have to square the whole series (x-x^3/3!+x^5/5!-...)*(x-x^3/3!+x^5/5!-...). It's not just the sum of the squares of each term. It's a double sum. There are cross terms. It's easy enough to find the first few terms that way.
 
Dick said:
You have to square ... that way.

ahhh thanks for answering both of my questions!

good night!
 

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