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Finding x for e^x = cos(x)

Date: 06/25/99 at 05:00:12
From: Matthew Lilley
Subject: How can I solve e^x = cos(x)

My friend recently gave me a problem he had adapted from a textbook: 
e^x = cos(x). (The original question was to integrate and find the 
area between the two curves. It stated explicitly not to attempt to 
find the solutions of x.)

This got me thinking - why can't we find the solutions? After a few 
attempts I found that there were definitely some problems.

I started by simply trying to rearrange, take the natural logarithm, 
etc., and kept getting back to my original equation.

I then thought of differentiating both sides, and got tan^-1(x) = -1. 
Of course this is incorrect, as you can't differentiate both sides. 
Can you?

I then asked my Physics teacher who is a genius at these sorts of 
problems. He tried a few rearrangements, and then came up with the 
idea of writing the two power series and equating them, canceling 
every second term. I tried that, but I don't really understand power 
series, and the result I get is a function with an infinite degree - 
hardly in my power to solve.

So I went to my Math teacher who replied that he could use a technique 
he called 'numerical analysis' to solve it by making the equations 
equal to zero, drawing a graph, and finding where they intersect. I 
said that this was fine, but that there were infinitely many 
solutions. This upset him and I think he's given up.

And so I turn to ask you - is there some sort of way to find the 
solutions to this seemingly simple problem?


Date: 06/25/99 at 07:30:12
From: Doctor Jerry
Subject: Re: How can I solve e^x = cos(x)

Hi Matthew,

Thanks for your question. I'm impressed with your energy in trying to 
resolve the question.

Most equations of the form f(x) = g(x) can't be solved algebraically 
or, as some say, in "closed form."  No finite process or expression 
will give finite expressions for the roots of this equation. If you 
want a particular root, then graphing and zooming will give you what 
you want or, better, a method like Newton's method will allow you to 
calculate the root to as much accuracy as you like. You can find 
Newton's method in most calculus books. It has a nice geometric 

     Let H(x) = f(x)-g(x).

Let a be the approximate location of a zero of H. A (generally) better 
approximation will be

     a* = a-H(a)/H'(a)

If a* is close enough to the zero, then stop. Otherwise, replace a by 
a* and repeat.

One zero of H is x = 0; the next one to the left is approx. -1.2.
Let a = -1.2 and apply the above algorithm. I get

     a* = -1.29695496481

Let a be a*. The next a* is


So, to at least 10 decimals, the first negative zero is 
-1.29269571937. This is, apart from continuing the algorithm to obtain 
more accuracy, the best anyone can do. 

- Doctor Jerry, The Math Forum   
Associated Topics:
High School Functions

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