# Transcendental Equation Help

#### stoucha

Hi,

I am reading a paper where part of the solution to an equation of diffusion in a multicompartment system includes the "sum of all nonnegative roots kj of the following transcendental equation,

2*u*(cos(k)-cos(q))-k*sin(k) = 0.

Then the authors of the paper say: "Note that the roots periodically depend on the parameter q, and, for their analysis, it is sufficient ot consider 0<q<pi. In the case of small u (i.e. uMM1), all roots are close to j*pi:

k0 approximately equals 2*(u^0.5)*sin(q/2), j=0

kj approximately equals j*pi + ((2*u)/(j*pi))*(1-(-1)^j cos(q))

end quote.

Can somebody help me with this derivation? The whole statement about roots periodically depending on q is confusing because q is fixed constant of the system. Further it seems to me that if we consider the case were u<<1, then the equation simplifies to -ksin(k) = 0 whith the roots not having q or u in them at all.

What am I missing????

Thanks for any help.

#### fresh_42

Mentor
2018 Award
We have $2u\cos k -k\sin k = 2u \cos q$. The right hand side is periodic in $q$ and the left hand side doesn't depend on $q$ at all. Thus it is sufficient to consider all possible values of $\cos q$, that is $0 < q < \pi$, where I assume $q\neq 0$ for other reasons.

Next we have $\cos k -\dfrac{k}{2u}\sin k = \cos q$. If $u << 1$ then the quotient is large, i.e. $\sin k$ has to be close to zero, for otherwise we would leave the range $[-1,1]$ given by the right hand side. Now the zeroes of $\sin k$ are the points $j\pi$.

Since you haven't said what k0 and kj are, I can't explain those.

"Transcendental Equation Help"

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