I’m looking for options of an equation in two variables
n. Alternatively, I’d even be okay with a plot which reveals the variation of 1 versus one other.
Nnorm[n_] := Sqrt*(1 - 2/(n + 1/6))^(1/4)*(1 - 2/n)^(-1/ 2)*(1 - 2/(n + 1/12))^(half); F[u_, L_, k_, w_] := u^2/(eight Pi*L*Sqrt[1 + (L/(2 k))^2])*Cos[2*w*k*ArcSinh[L/(2 k)]];
The worth of parameters
w are recognized. After utilizing
FullSimplify[F[0.01, L, k, 10]]and my requirement between relation of
F = -0.00003*Nnorm I’ll get hold of an equation like this :
(7.957747154594767`*^-6 Cos[20 k ArcCsch[(2 k)/L]])/( L Sqrt[4 + L^2/k^2]) + (0.00003*( 2 n (1 - 2/(1/12 + n)) Sqrt[1 - 2/(1/6 + n)])/(-2 + n)) == 0
That is what I wish to remedy or plot, an equation in two variables. The one factor that partially works is the next :
F[0.001, 100, 200, 10]; (* acquiring the worth of F right here after which manually substituting as follows : *) Remedy[-0.00003*2*(1 - 2/(n + 1/6))^(1/ 2)*(1 - 2/n)*(1 - 2/(n + 1/12)) == 3.55881271708, n]
This does not all the time work so I can not put this in a loop or an
I’ve tried utilizing
InverseFunctionto no avail.
I can not use
FindRoot by taking some worth for both
n and utilizing it on the opposite as a result of I merely don’t know what to present for beginning worth for the
I’ve additionally checked out this, this and this on the very least.
The principle perpetrator right here, I feel, is
ArcSinh with out which
Remedy will most likely do the job. I attempted to remove
ArcSinh by evaluating the values of
okay however once more nothing.
Please give me some assistance on easy methods to deal with this situation.