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

1 Answer
Cesareo R.
Nov 24, 2016

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.

enter image source here

and a zoomed region

enter image source here

A sample of solutions

enter image source here

Related questions
See all questions in Absolute Value Equations
Impact of this question
4307 views around the world
You can reuse this answer
Creative Commons License