Taylor series with summation notation

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Homework Statement



f(x) = \frac{1-cos(X^2)}{x^3}

which identity shoud i use?
and tips on this type of questions? once i can separate them, then i'll be good


thanks!
 
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Do you know a Taylor series for \cos x?
 
benorin said:
Do you know a Taylor series for \cos x?

yeah, but there's a x^3 on the bottom...
 
Sure, but the summation isn't over x so you can put the x in the sum or outside the sum.
 
Example:

\frac{1-\sin 2x^3}{x}=\frac{1-\sum_{k=0}^{\infty}\frac{\left( 2x^3\right)^{2k+1}}{(2k+1)!}}{x} = {\scriptstyle \frac{1}{x}}-{\scriptstyle \frac{1}{x}}\sum_{k=0}^{\infty}\frac{2^{2k-1}x^{6k+3}}{(2k+1)!}
= {\scriptstyle \frac{1}{x}}-\sum_{k=0}^{\infty}\frac{2^{2k-1}x^{6k+2}}{(2k+1)!}[/tex]​
 
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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