Finding the Taylor Polynomial f4 for sin(2x) at x=pi/4.

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Homework Help Overview

The discussion revolves around finding the fourth Taylor polynomial for the function f(x) = sin(2x) centered at a = π/4. Participants are exploring the Taylor series expansion and its application to this specific function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to replace x with 2x in the Taylor series formula for sin(x) and question if this transformation is correct. There is also mention of centering the polynomial at π/4 and deriving the function accordingly.

Discussion Status

The discussion is active, with participants sharing their thoughts on the correct approach to derive the Taylor polynomial. Some guidance has been offered regarding the formula for Taylor series, but there is no explicit consensus on the steps to take next.

Contextual Notes

There is a noted requirement to center the polynomial at π/4, which raises questions about the derivation process from the standard Taylor series centered at 0. Participants are also referencing external resources for further clarification.

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Homework Statement


find the taylor polynomial f4 for f(x)=sin(2x) and a=pi/4


Homework Equations


sin(x)=((-1)^n)(x^(2n+1))/((2n+1)!)



The Attempt at a Solution



so replace x with 2x?
you get ((-1)^n)(2x)^(2n+1)/(2n+1)!)

is this right?
 
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seto6 said:

Homework Statement


find the taylor polynomial f4 for f(x)=sin(2x) and a=pi/4


Homework Equations


sin(x)=((-1)^n)(x^(2n+1))/((2n+1)!)



The Attempt at a Solution



so replace x with 2x?
you get ((-1)^n)(2x)^(2n+1)/(2n+1)!)

is this right?

Its a good thing to present it the right way...

You know the formula right?

[tex]\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} \cdot (x-a)^{n} = \sum_{n=0}^{\infty} \frac{f^{(4)}(\frac{\pi}{4})}{4!} \cdot (x-\frac{\pi}{4})^{4}[/tex]

which you simply expand as show in your Calculus book which then inturn gives the right answer!
 
i see i have to center it at pi/4 but this is ((-1)^n)(2x)^(2n+1)/(2n+1)!) centered at 0 so i have to derive it sin(2x) using center pi/4 right

thanks man
 
seto6 said:
i see i have to center it at pi/4 but this is ((-1)^n)(2x)^(2n+1)/(2n+1)!) centered at 0 so i have to derive it sin(2x) using center pi/4 right

thanks man

you welcome :) Generally do you have the Edwards and Penny Calculus Bible? You find the whole definition and example in there :)
 

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