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

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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?

\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}

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 :)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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