Numerical analysis condition number

Question

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 '21 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? — Mihai.Mehe, Oct 10 '21 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 '21 at 00:02
Fair enough, then the mistake is that the derivative of $\ln(a)$ is $1/a$ not $0$. (You should have viewed your result with suspicion immediately since you could divide by $p'(a)$ and then explicitly solve for $p(a)$ which is obviously not possible.) — Ian, Oct 11 '21 at 00:04
Incidentally this step with the logarithm wasn't actually necessary anyway, the point here is that if $F(p,a)=p^n-ae^{-p}$ then $\frac{d}{da} F(p(a),a)=0=\frac{\partial F}{\partial p} p'(a) + \frac{\partial F}{\partial a}$, which you can solve for $p'(a)$. — Ian, Oct 11 '21 at 00:07
I tried that first but I am getting something odd by deriving $p^n(a)-ae^{-p(a)}=0$ $p'(a)(np^{n-1}(a)+ae^{-p(a)})=0$ — Mihai.Mehe, Oct 11 '21 at 00:20
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 '21 at 00:24
And $|\frac{ap'(a)}{p(a)}|<\frac{1}{n}|\frac{ae^{-p(a)}}{p^{n}(a)}|$ — Mihai.Mehe, Oct 11 '21 at 00:46
...and then finally you know what $ae^{-p(a)}/p^n(a)$ is from the definition of the problem. So now you should be able to write and accept an answer to your own question. — Ian, Oct 11 '21 at 00:48

score 0 · Accepted Answer · answered Oct 11 '21 at 01:30

I summarized the solution given in the comments above by Ian.

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. — Carl Christian, Oct 11 '21 at 11:49

Numerical analysis condition number

1 Answers1