Using known Maclaurin series to approximate modification of original

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Homework Statement



Recall that the Maclaurin series for sin(x) is \sum\frac{(-1)^{k}x^{2k+1}}{(2k + 1)!}.

Use this formula to find the Maclaurin polynomial P5(x) for f(x)=xsin(x/2).

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The Attempt at a Solution



I know that to approximate sin(x/2) with the Maclaurin polynomial for sinx, I just substitute x/2 for x. But for xsinx, since the Maclaurin series is approximating sinx, can I just substitute the series for sinx so that I get x\sum\frac{(-1)^{k}x^{2k+1}}{(2k + 1)!}?
 
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yes you can :smile:
 


Yes. You can make that better if you now sweep the x inside the sum. You may also want to include the summation bounds as you can sometimes simplify further by shifting them.
 


Sorry, didn't see that I forgot the bounds. It's supposed to be from n=0 to infinity. Thanks for the help guys :)
 
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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