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Find a way to calculate $\begin{equation} f(x)= \frac{1-e^{-2x^{2}}}{x\sin x} \end{equation}$ correctly. Determine $f(0.00001)$ correctly to $12$ decimal places.

I try to find the solution with Taylor. After a long calculation I have: \begin{equation} 1-e^{-2x^{2}}=2x^2-2x^4+\frac{4x^6}{2}-\frac{2x^8}{3}+\cdots \end{equation} and \begin{equation} x\sin x=x^2-\frac{x^4}{6}+\frac{x^6}{120}-\frac{x^8}{5040}+\dot\cdots \end{equation} Then I try to... \begin{equation} (2x^2-2x^4+\frac{4x^6}{2}-\frac{2x^8}{3}+\cdots)=(ax+bx^2+cx^4+dx^6+\cdots)(x^2-\frac{x^4}{6}+\frac{x^6}{120}-\frac{x^8}{5040}+\cdots) \end{equation}

and to find $a,\,b,\,c,\,d$. Now I don't get any further.

J.G.
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Elenmel
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  • The ratio should start with a constant, not an $x$ term. It should be an even function, too. – J.G. Apr 22 '21 at 16:11
  • In this homework are you asked explicitly to use a Taylor expansion for that ? – Jean Marie Apr 22 '21 at 16:11
  • Divide the taylor expansions to match the function and substitute for x=.00001 until f(.00001)=2 is reached. This is because the answer should be 2=2.000... up to 11 zeroes plus the ones place digit. This was done with software. – Тyma Gaidash Apr 22 '21 at 16:12
  • No, I don't have to use Taylor expansion, but that was a hint. I don't know any other solution. – Elenmel Apr 22 '21 at 17:41

1 Answers1

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The ratio is asymptotic to $2$ for small $x$ and is even, so is $2+kx^2+o(x^2)$, and we need no more precision than that in this problem. Matching $x^4$ coefficients,$$\begin{align}(2+kx^2+o(x^2))(x^2-\tfrac16x^4+o(x^4)&)=2x^2-2x^4+o(x^4)\\\implies k-\tfrac13&=-2\\\implies k&=-\tfrac53.\end{align}$$So $f(10^{-5})\approx2-\tfrac{500}{3}10^{-12}\approx1.999999999833$ (note it's slightly less than $2$, contra @TymaGaidash's comment).

J.G.
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