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How would one use numerical methods to find solutions to the following equations: $$\frac{2(b^3-a^3)}{3(b-a)^2}=e-1$$ $$\big(\sqrt{e^{2}+1}+\frac{1}{2}\ln(\sqrt{e^{2}+1}-1)-\frac{1}{2}\ln(\sqrt{e^{2}+1}+1)\big)-\big(\sqrt{2}+\frac{1}{2}\ln(\sqrt{2}-1)-\frac{1}{2}\ln(\sqrt{2}+1)\big)=\\ \frac{1}{4(b-a)}\bigg( \operatorname{arcsinh}(2b)+2b\sqrt{4b^2+1}-\operatorname{arcsinh}(2a)-2a\sqrt{4a^2+1}\bigg) $$

I understand it would be the intersect between to functions of $x$ and $y$ namely $f(x,y)=\frac{y^3-x^3}{y-x}$ and $f(x,y)=\frac{3}{2}(e-1)$. I can plug this into wolfram alpha to receive the solutions and the following enter image description here

I am told to evaluate this numerically it is necessary to create two multi variable functions out of the above two and minimise the function $\Phi(a,b)$. After this create a contour plot of $\Phi(a,b)$ and use the Newton Rhaphson method.

\begin{equation} f(a,b)= \frac{2(b^3-a^3)}{3(b-a)^2}-e+1=0 \end{equation} \begin{equation} \label{bigeq2} \begin{aligned} g(a,b)=\big(\sqrt{e^{2}+1}+\frac{1}{2}\ln(\sqrt{e^{2}+1}-1)-\frac{1}{2}\ln(\sqrt{e^{2}+1}+1)\big)-\big(\sqrt{2}+\frac{1}{2}\ln(\sqrt{2}-1)-\frac{1}{2}\ln(\sqrt{2}+1)\big)-\\ \frac{1}{4(b-a)}\bigg( \operatorname{arcsinh}(2b)+2b\sqrt{4b^2+1}-\operatorname{arcsinh}(2a)-2a\sqrt{4a^2+1}\bigg)=0 \end{aligned} \end{equation} \begin{equation} \Phi(a,b)=\big[f(a,b)\big]^2+\big[g(a,b\big]^2 \end{equation}

hwood87
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1 Answers1

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By the following

$$\frac{y^3-x^3}{y-x}=\frac{3}{2}(e-1)=a>0$$

by $y\neq x$ we obtain

$$y^2+xy+x^2=a$$

$$ \iff\left(\frac y {\sqrt a}\right)^2+\left(\frac y {\sqrt a}\right)\left(\frac x {\sqrt a}\right)+\left(\frac x {\sqrt a}\right)^2=1$$

which is an ellipse which, with exclusion of the points with $y=x$, is the solution.

user
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  • How would I find the solution with the second equation? – hwood87 Sep 22 '20 at 12:52
  • @hwood87 Do you mean with $y^2+xy+x^2=a$? – user Sep 22 '20 at 12:53
  • No sorry, with the equation: $$\big(\sqrt{e^{2}+1}+\frac{1}{2}\ln(\sqrt{e^{2}+1}-1)-\frac{1}{2}\ln(\sqrt{e^{2}+1}+1)\big)-\big(\sqrt{2}+\frac{1}{2}\ln(\sqrt{2}-1)-\frac{1}{2}\ln(\sqrt{2}+1)\big)=\ \frac{1}{4(b-a)}\bigg( \operatorname{arcsinh}(2b)+2b\sqrt{4b^2+1}-\operatorname{arcsinh}(2a)-2a\sqrt{4a^2+1}\bigg) $$ – hwood87 Sep 22 '20 at 12:55
  • @hwood87 : Could you explain how you got that expression, it might be an error in formulating the problem in the CAS. And even if it is correct, this is now a transcendental equation, not a simplification from the original algebraic equation. – Lutz Lehmann Sep 22 '20 at 16:10