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Solving Transcendental Equation

Posted 10 years ago

Dear Mathematica users,

I have a equation which looks like:

x^2 + Cosh[xt]+1/Sqrt[x^2+1]=0

I would like to plot different real roots of x as a function of t (0<t<4) (if there are two real roots than I want two graphs). How can I do that using Mathematica ? This is a transcendental equation problem. Thank you.

POSTED BY: H Paudel
2 Replies
Posted 10 years ago

Hi,

Yes you are right.....Here is the actual problem. My equation is,

{-(8.367455073574008*^-50 (9.623881668444158*^35 + \ (1.894701703474944*^32 q^2 (0. + ArcCosh[ 1/q1.5873015873015871*^28 (4.8^-20 - 1.0500000000000003*^-34 \[Omega])] - ArcCosh[ 1/q1.5873015873015871^28 (4.8*^-20 + 1.0500000000000003*^-34 \ [Omega])]))/(Sqrt[-360000000000 q^2 + [Omega]^2]))^2)/(q^2 (500 \ Cosh[(3 q)/500000000] + 10004 Sinh[(3 q)/ 500000000])^2) + (1 - (2.892655367231639*^-25 \ (9.623881668444158*^35 + (1.894701703474944*^32 q^2 (0. + ArcCosh[ 1/q1.5873015873015871*^28 (4.8^-20 - 1.0500000000000003*^-34 \[Omega])] - ArcCosh[ 1/q1.5873015873015871^28 (4.8*^-20 + 1.0500000000000003*^-34 \ [Omega])]))/(Sqrt[-360000000000 q^2 + [Omega]^2])) (Cosh[(3 q)/ 500000000] + 1/100 Sinh[(3 q)/500000000]))/(q (500 Cosh[(3 q)/ 500000000] + 10004 Sinh[(3 q)/ 500000000]))) (1 - (2.892655367231639*^-25 \ (9.623881668444158*^35 + (1.894701703474944*^32 q^2 (0. + ArcCosh[ 1/q1.5873015873015871*^28 (4.8^-20 - 1.0500000000000003*^-34 \[Omega])] - ArcCosh[ 1/q1.5873015873015871^28 (4.8*^-20 + 1.0500000000000003*^-34 \ [Omega])]))/(Sqrt[-360000000000 q^2 + [Omega]^2])) (Cosh[(3 q)/ 500000000] + 1/25 Sinh[(3 q)/500000000]))/(q (500 Cosh[(3 q)/ 500000000] + 10004 Sinh[(3 q)/500000000])))}=0

This equation must have four roots of Omega. I would like to find the real roots and plot as a function of q. Here 3.003*^8<q<4.003*^9

Thank you.

POSTED BY: H Paudel

Are you sure that that is the correct function?

Each of the three terms is non-negative:

Show[Plot[1/Sqrt[x^2 + 1], {x, -5, 5}, PlotRange -> {All, {0, 2}}], Plot[Cosh[x], {x, -5, 5}], Plot[x^2, {x, -5, 5}]]

enter image description here

One of them is always larger or equal one. (In fact apart from x=0 all three terms are always positive. At x=0 the sum is 2.) So I would expect the sum to be always larger than 0. Hence there are no real zeros. When t=0 the Cosh is one for all x. For t>0, Cos[t x]>=1, so t does not help here.

Cheers,

M.

POSTED BY: Marco Thiel
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