<p>Can anyone solve this without a calculator:</p>
<p>e^(x)= -x</p>
<p>My calc gives me an answer, but I cannot figure out how it got it.</p>
<p>newton's method?
lol sorry i was afk on aim</p>
<p>using fixed-point iteration:
denote x-zero as x[0] and x sub 1 as x[1]
x=-e^x
say x[0]=0
then
x[n]=e^x[n-1]
x[1]=-e^(0)=-1
x[2]=-e^(-1)=-0.36788
x[3]=-e^(-0.368)=-0.6922
keep repeating...
x[inf]=-0.5671...
well, thats one way of doing it.</p>
<p>you could also use newton-raphson iteration, but i cant be bothered to ascii-math explain that here.</p>
<p>i would said around -.5 just from the intersection of graphs of said functions</p>
<p>huh... couldn't u just say -e^x=x... take ln of both sides... -x=lnx? that seems a whole lot easier then doing that... the intersection would be what the calculator gave you...</p>
<p>Hmm, Cujoe is a little confused. The ln of -e^x is certainly not -x. But the idea is along the right lines.</p>
<p>ln [ exp(x) ] = ln(-x)</p>
<p>x = ln(-x)</p>
<p>Anyway, that's a pretty unenlightening restatement of the original problem. This equation has no algebraic/explicit solution because it involves both transcendental and polynomial expressions in x. So solving it via calculator or Newton's method is the best you can do.</p>
<p>no that's not what i implied!!!! lol... actually it is isn't it... oops lets call it a slip up then... :)</p>
<p>Work it out by hand by iterating and using a slide-rule and log books if necessary. (I still have my log books somewhere)</p>
<p>yeah, and i have my dad's sliderule somewhere in my desk :)
it has the most glorious engrish manual, too.</p>
<p>My father sold his slide rule. :( And it was log-log as well!</p>
<p>-x=y</p>
<p>so e^(-y)=y, (Iny)/(y)=-1 so In(y^-y)=-1
y^(-y)=e^(-1)</p>
<p>y^y=e, it is impossible to solve for an exact value of y.</p>
<p>Without checking your calculation to get there, it would be a slight lie to say y^y=e implies it is impossible to solve for an exact value of y directly. It is a fairly challenging argument to show such a thing.</p>