Proving Pn(x^2) as the 4n+2-nd Taylor Polynomial of sin(x^2) using Rn(x) Limits

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The discussion centers on proving that Pn(x^2) is the 4n+2-nd Taylor polynomial of sin(x^2) by demonstrating that the limit of R2n+1(x^2) approaches zero as n approaches infinity. The remainder term Rn(x) is crucial in this proof, as it indicates the accuracy of the Taylor polynomial approximation. Participants emphasize understanding the definitions of Pn(x^2) and R2n+1(x^2) to effectively tackle the limit proof required for the conclusion.

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Show that Pn(x^2) is the 4n+2-nd Taylor polynomial of sin(x^2) by showing that
[tex]\lim_{n\rightarrow infinity}[/tex] R2n+1(x^2) = 0.

note that Rn(x) represents the remainder

I'm stuck on this question, can anyone help me please?
 
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On what are you stuck? Do you know what Pn(x^2) is? Do you know what R2n+1(x^2) is? Do you know how to prove its limit goes to zero? Do you know why that limit would imply what you're trying to prove?
 
I'm stuck on the entire question, I know what P and R are but i don't know how to prove the question. I also do not know why the limit of R helps solve this question.
 
Last edited:

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