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I tried to expand $f(x)=\frac{1}{\sqrt{1-x^2}}$ about $x=0$ with Taylor series directly $$ f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n \tag{1}\label{eq1}$$ and calculated the following derivatives: $$\begin{align} f' (x) &= -\frac{1}{2}(1-x^2)^{-1.5}(-2x) = x(1-x^2)^{-1.5}\\ f'' (x) &= (1-x^2)^{-1.5}+3x^2(1-x^2)^{-2.5}\\ f''' (x) &=3x(1-x^2)^{-2.5}+3[2x(1-x^2)^{-2.5}+5x^3(1-x^2)^{-3.5}]\\ &=9x((1-x^2)^{-2.5}+15x^3(1-x^2)^{-3.5} \\ f'''' (x) &=9[(1-x^2)^{-2.5}+5x(1-x^2)^{-3.5}]+15[3x^2(1-x^2)^{-3.5}+7x^4(1-x^2)^{-4.5}]\\ &=9(1-x^2)^{-2.5}+45x(1+x)(1-x^2)^{-3.5}+105x^4(1-x^2)^{-4.5} \end{align}$$ Hence $f'(0)=0$, $f''(0)=1$, $f'''(0)=0$ and $f''''(0)=9$. Substituting into \eqref{eq1} yields $$f(x)=\frac{1}{\sqrt{1-x^2}}\approx1+\frac{1}{2}x^2+\frac{3}{8}x^4\tag{2}\label{eq2}$$ Then I realized that I can do change of variable $y=x^2$ and expand $f(y)=\frac{1}{\sqrt{1-y}}$ about $y=0$ using \eqref{eq1}, which is much more computational efficient: $$f(y)=\frac{1}{\sqrt{1-y}}\approx 1+\frac{1}{2}y+\frac{3}{8}y^2\tag{3}\label{eq3}$$ Substitute $x^2$ back to y: $$f(x)=\frac{1}{\sqrt{1-x^2}}\approx 1+\frac{1}{2}x^2+\frac{3}{8}x^4 \tag{4}\label{eq4}$$ Since \eqref{eq2} and \eqref{eq4} are equal to at least the 4-th order, both approaches give us the identical Taylor series. By the way, this can be applied to Einstein's mass energy equation: $$E_{total}=\frac{mc^2}{\sqrt{1-{(\frac{v}{c})}^2}}\approx mc^2+\frac{1}{2}mv^2[1+\frac{3}{4}(\frac{v}{c})^2]\tag{5}\label{eq5}$$ The first term in \eqref{eq5} is the energy that the particle of mass m has when at rest or the rest energy and the second term is its kinetic energy.

Gary
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Leon Chang
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1 Answers1

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From a table of Maclaurin series: $$\arcsin x=\sum_{n=0}^\infty\frac{\Gamma(n+\tfrac12)}{\sqrt\pi\,(2n+1) n!}x^{2n+1}$$

Taking derivative of both sides we have $$\frac{1}{\sqrt{1-x^2}}=\sum_{n=0}^\infty\frac{\Gamma(n+\tfrac12)}{\sqrt\pi\,n!}x^{2n}$$ which was not in the table.

Bob Dobbs
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    I made a part of OP a tik tok post İ hope it is wellcome. Physics part important. https://vm.tiktok.com/ZM67fHkEw/ – Bob Dobbs Feb 06 '24 at 07:56
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    The way one derives the Maclaurin series of $\arcsin$ is by integrating the binomial series for $(1-x^2)^{-1/2}$. So this is a circular reasoning. – Gary Feb 06 '24 at 08:18
  • My purpose is just to show the use a table of information in mathematics. – Bob Dobbs Feb 06 '24 at 08:26
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    @BobDobbs You used $\Gamma$ function in the expansion. I am not familiar with it. But wiki shows $\Gamma (n+\frac{1}{2})=\frac{(2n)!}{4^n n!}\sqrt\pi$, so it does match the regular formula. Is there a reason to use $\Gamma$ and introduce $\pi$? – Leon Chang Feb 09 '24 at 23:06
  • @Gary Just to make sure, you mean $\sin^-1x=\int \frac{1}{\sqrt{1-x^2)} dx$ $\frac{d}{dx} \sin^-1 x=\frac{1}{\sqrt{1-x^2)}. correct? – Leon Chang Feb 09 '24 at 23:18
  • @Gary (edited) Just to make sure, you mean $\sin^{-1}x=\int \frac{1}{\sqrt{1-x^2}} dx$ thus $\frac{d}{dx} \sin^{-1} x=\frac{1}{\sqrt{1-x^2}}$. correct? – Leon Chang Feb 09 '24 at 23:28
  • @LeonChang I borrowed the formula from the list. It was given in this form. – Bob Dobbs Feb 10 '24 at 02:24
  • Taylor expansion of $f(x)= \frac{1}{\sqrt{1-x^2}}$ is based on Binomial series $f(x)= \frac{1}{\sqrt{1+x}}$ which results in a double factorial: $(2n-1)!! =\frac{(2n)!}{2^n n!}$. That shows in the final result of $\displaystyle f(x)=\sum_{n=0}^\infty \frac{(2n)!}{4^n (n!)^2}x^{2n}$ – Leon Chang Feb 11 '24 at 06:17