Help needed solving an inequality involving the natural log

• Jul 25th 2012, 03:56 PM
Giestforlife
Help needed solving an inequality involving the natural log
I'm trying to solve a simple alternative series question, but I cannot get past the following inequality to calculate my nth term for the required error:

e^(-n-1).(n+1) < 0.01

I solved it all the way to:

ln n < (n + 1) + ln 0.01

I can't figure out how to solve for n after this.

Any help will be greatly appreciated,

Giest
• Jul 25th 2012, 04:33 PM
HallsofIvy
Re: Help needed solving an inequality involving the natural log
There is no elementary way to solve that. What we can do is let u= n+ 1 so the function can be written \$\displaystyle e^{-u}u\$. We can further let v= -u so that \$\displaystyle e^{-u}u= e^v(-v)= -e^vv\$. Now, the best way to solve such an inequality is to first solve the associate equation- here, \$\displaystyle e^{-(n+1)}(n+1)= -ve^v= 0.01\$ which is, of course, the same as \$\displaystyle ve^v= -0.01\$. And the solution to that is x= W(-0.01) where W is the "Lambert W function" (also known as the "omega function" or "product log" function) which is defined as the inverse function to \$\displaystyle f(x)= xe^x\$. Looking at a graph of \$\displaystyle y= xe^x\$ we can see there are two roots for \$\displaystyle xe^x= -0.01\$, one at about -0.01 and the other at about -6.47. The inequality is satisfied for x below the root around -6 and above the root around -.01. (I used the "Newton-Raphson" iteration method to solve for those two values.)
• Jul 25th 2012, 04:40 PM
Giestforlife
Re: Help needed solving an inequality involving the natural log
Thanks for the help on this one.