How many real solutions do the equation #sin(e^x)-5^x-5^(-x)=0# have? Thank you!

1 Answer
Nov 24, 2016

Answer:

No real solutions.

Explanation:

The equation is #sin(e^x)=2(e^(lambda x)+e^(-lambda x))/2=2cosh(lambda x)#

with #lambda = log_e5# but

#2cosh(lambda x) ge 2# and #-1 le sin(e^x) le 1# so no real solution is possible.

The equation #sin(e^x)=2cosh(lambdax)# surely has infinite complex solutions arranged in a capricious pattern.

You can obtain complex roots proceeding as follows.

(1) Define the complex function #f(z)=sin(e^(x+iy))-2cosh(lambda(x+iy))=0#
(2) Determine the real and imaginary components. In this case we have

#Re(f(z))=r(x,y)=-2 Cos(lambda x) Cosh(lambda y) + Cosh(e^x Siny)Sin(e^x Cosy)#

#Im(f(z))=s(x,y)=-2 Sin(lambda x) Sinh(lambda y) + Cos(e^x Cosy)Sinh(e^x Siny)#
(3) Using an iterative procedure such as Newton-Raphson, determine the solutions for

#{(r(x,y)=0),(s(x,y)=0):}#

To have an idea about the roots placement there is an attached plot showing in blue the contour for #r(x,y)=0# and in red the contour for #s(x,y)=0#

The complex roots lay into the intersection of both curves.

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and a zoomed region

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A sample of solutions

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