This example shows an interesting graphical approach to find out whether e^pi is greater than pi^e or not.

The question is: which is greater, e^pi or pi^e? The easy way to find out is to type it directly at the MATLAB® command prompt. But it motivates a more interesting question. What is the shape of the function z = x^y-y^x? Here is a plot of z.

% Define the mesh x = 0:0.16:5; y = 0:0.16:5; [xx,yy] = meshgrid(x,y); % The plot zz = xx.^yy-yy.^xx; h = surf(x,y,zz); % Set the properties of the plot h.EdgeColor = [0.7 0.7 0.7]; view(20,50); colormap(hsv); title('z = x^y-y^x'); xlabel('x'); ylabel('y'); hold on;

It turns out that the solution of the equation x^y-y^x = 0 has a very interesting shape. Because interesting things happen near e and pi, our original question is not easily solved by inspection. Here is a plot of that equation shown in black.

c = contourc(x,y,zz,[0 0]); list1Len = c(2,1); xContour = [c(1,2:1+list1Len) NaN c(1,3+list1Len:size(c,2))]; yContour = [c(2,2:1+list1Len) NaN c(2,3+list1Len:size(c,2))]; % Note that the NAN above prevents the end of the first contour line from being % connected to the beginning of the second line line(xContour,yContour,'Color','k');

Here is a plot of the integer solutions to the equation x^y-y^x = 0. Notice 2^4 = 4^2 is the ONLY integer solution where x ~= y. So, what is the intersection point of the two curves that define where x^y = y^x?

plot([0:5 2 4],[0:5 4 2],'r.','MarkerSize',25);

Finally, we can see that e^pi is indeed larger than pi^e (though not by much) by plotting these points on our surface.

e = exp(1); plot([e pi],[pi e],'r.','MarkerSize',25); plot([e pi],[pi e],'y.','MarkerSize',10); text(e,3.3,'(e,pi)','Color','k', ... 'HorizontalAlignment','left','VerticalAlignment','bottom'); text(3.3,e,'(pi,e)','Color','k','HorizontalAlignment','left',... 'VerticalAlignment','bottom'); hold off;

Here is a verification of this fact.

e = exp(1); e^pi pi^e

ans = 23.1407 ans = 22.4592

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