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? Thanks, Matt
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 explanation. 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 -1.29270289753 -1.29269571939 -1.29269571937 -1.29269571937 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 http://mathforum.org/dr.math/
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