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im trying to find the exact minimum of this function. Upon plotting, i see that the value is close to $0.104$. However, i am not sure as to the exact value of said minima. The function is as follows:

$$ f(x)=x^x+x $$

i know that to find the minimum and maximums of a function, you must evaluate where the first derivative is $0$. I took the first derivative, and ended up with this:

$$ f'(x)=x^x\left(\ln\left(x\right)+1\right)+1 $$

However, when i set this to zero i am unable to isolate x into finding an exact solution. If i were to take a guess, the exact solution would involve the Lambert W function, due to the $x^x$ term.

Jacob
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1 Answers1

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I assume you mean minimum? The function has a local maximum at x=0 and is unbounded as x increases without bound.

Even with the Lambert W, you'll probably need a numerical method to find the min.

Are you familiar with Newton's Method?

Pick a guess as your $x_0$

Then let $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$. It converges in most cases where f is twice differentiable, $|f''(x)f(x)|<f'(x)^2$ at root x.

So let $g(x)=x^x(\ln(x)+1)+1.$ Then $g'(x)=x^x(\ln(x)+1)^2+x^{x-1}$. Use Newton's method for $g(x)=0$.

$g'(x)=[g(x)-1](\ln(x)+1)+\frac{[g(x)-1]}{x(\ln(x)+1)}$ is a shortcut that will save you a few calculations per iteration.

TurlocTheRed
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