Numerical analysis condition number

I have the following problem. Let $$p = p(a)$$ be the positive unique root of the equation $$x^n − a*e^{−x} = 0$$ where $$n$$ is a natural number and $$a > 0$$. Show that the condition number $$κ_{p}(a) <1/n$$.

Attempted to start the Proof and I realized I probably do not understand this.

By definition $$k_p(x^*)=|\frac{xf'(x^*)}{f(x)}|$$. Then, $$k_p(a)=|\frac{a(na^{n-1}+ae^{-a})}{a^n-ae^{-a}}|=|\frac{na^{n}+a^2e^{-a}}{a^n-ae^{-a}}| = |\frac{n+a^{2-n}e^{-a}}{1-a^{1-n}e^{-a}}|$$. This is bounded by $$n$$ if $$a$$ goes to infinity. But not by $$1/n$$.

I am not sure I understand what is $$p(a)=p$$. Does it mean $$a$$ is the root of $$x^n − a*e^{−x}$$ and $$x^n − a*e^{−x}$$ is $$p(x)$$? In this case I have a problem since the denominator of $$k_p(a)$$ is $$0$$. Also I do not see how having a unique root plays a role here. Any suggestions? Thanks and regards,

• The characterization of $p(a)$ here is $p(a)^n-ae^{-p(a)}=0$. So $\kappa_p(a)=\left | \frac{a p'(a)}{p(a)} \right |$ by your definition. Obviously $p(a)$ is not zero if $a>0$.
– Ian
Oct 10 at 23:36
• Taking the log on $x^n − a*e^{−x} =0$: $nln(x)-ln(a)+x=0$. Then $nln(p(a))+p(a)=ln(a)$. And the derivative is $n\frac{p'(a)}{p(a)}+p'(a)=0\Longrightarraow \frac{ap'(a)}{p(a)}=-\frac{p'(a)}{n}$. Am i on the right track? Oct 10 at 23:54
• You need to move $ae^{-x}$ to the other side before you take the log (there is no way to simplify the log with the subtraction in there). Doing that you get $n \ln(x)=\ln(a)-x$ which you can analyze the way you said. Note though that what you're differentiating with respect to here is $a$.
– Ian
Oct 11 at 0:02
• Yes, this is what I did but skipped to show that step. Oct 11 at 0:03
• There is a term with no $p'(a)$ namely the contribution to the derivative of $ae^{-p(a)}$ from differentiating the first factor of $a$. In my notation above this is the $\frac{\partial F}{\partial a}$ term.
– Ian
Oct 11 at 0:24

First we note that by definition, $$\kappa_{p}(a)=|\frac{ap'(a)}{p(a)}|$$. Where by hypothesis, $$p(a)$$ is a solution to $$x^n-ae^{-x}=0$$, and thus we can write $$p^n(a)-ae^{-p(a)}=0$$. By derivation, $$np^{n-1}(a)p'(a)+ae^{-p(a)}p'(a)-e^{-p(a)}=0$$ which implies $$p'(a) = \frac{e^{-p(a)}}{np^{n-1}(a)+ae^{-p(a)}}$$.
Therefore $$\kappa_{p}(a)=|\frac{ap'(a)}{p(a)}| = |\frac{ae^{-p(a)}}{p(a)[np^{n-1}(a)+ae^{-p(a)}]}| = |\frac{ae^{-p(a)}}{np^{n}(a)+ ae^{-p(a)}}|<\frac{1}{n}|\frac{ae^{-p(a)}}{np^{n}(a)}|=\frac{1}{n}$$ since $$p^n(a)-ae^{-p(a)}=0$$.
• Your instructor may wish to know why $a \rightarrow p(a)$ is a well-defined and differentiable function. The implicit function theorem can help settle this issue. Oct 11 at 11:49