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I try to unterstand a derivation and need help.

There are given two functions $$ s=-cos(j\pi/n),s\in[-1,1] $$ and the nonlinear transformation $$ y(s)=C\tan[\frac{\pi(s+1)}{4}+\frac{s-1}{2}\arctan\frac{y^*}{C}]+y^*,y\in[0,\infty) $$ $y*$ and $C$ are constant parameters.

The derivation and its solution is $$ \frac{ds}{dy(s)}=4C/[\pi+2\arctan(y^*/C)]/[C^2+(y(s)-y^*)^2] $$

but i don't understand how to reach this solution.

K.defaoite
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Kingpot
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  • Is it actually $\frac{ds}{dy}$ you want to find? Since y is defined as a function of s, $\frac{dy}{ds}$ is more direct. But $\frac{ds}{dy}= \frac{1}{\frac{dy}{ds}}$ so start by finding $\frac{dy}{ds}$ anyway. – user247327 Jun 30 '20 at 12:39
  • Also, the first equation, $s= -cos(j\pi/n)$ is irrelevant since y is defined as a function of s. – user247327 Jun 30 '20 at 12:44
  • Yeah they are calculating $\frac{ds}{dy(s)}$ in the paper. When i am calculating $\frac{dy}{ds}$ for $\frac{1}{\frac{dy}{ds}}$, i don't get the same result. – Kingpot Jun 30 '20 at 12:54
  • The last line is ambiguous due to the multiple division symbols. – K.defaoite Jun 30 '20 at 13:24

1 Answers1

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Since y* and C are constants so is $\frac{1}{2}arctan(y*/C)$ so just call that C'. You want to differentiate $y= \tan\left(\frac{\pi(s+1)}{4}+ C'\frac{s-1}{2}\right)+ y*$.

The derivative of $tan(y(x))$ with respect to x is $sec^2(y(x))\frac{dy}{dx}$ so the derivative of $\tan\left(\frac{\pi(s+1)}{4}+ C'\frac{s-1}{2}\right)$ with respect to s is $\sec^2\left(\frac{\pi(s+1)}{4}+ C'\frac{s-1}{2}\right)$ times $\frac{\pi}{4}+ C'\frac{1}{2}$.

That is, $\frac{dy}{ds}= \left(\frac{\pi}{4}+ \frac{1}{4}arctan(y*/C)\frac{s-1}{2}\right)\sec^2\left(\frac{\pi(s+1)}{4}+ C'\frac{s-1}{2}\right)$.

To get $\frac{ds}{dy}$ take the reciprocal of that.

user247327
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  • You are totally right. I thought the result was not the same as in the paper, but it is just written different. – Kingpot Jun 30 '20 at 13:12