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# $$\alpha^ \alpha = \beta ^ \beta$$

Find all pairs of positive real numbers $$\alpha$$ and $$\beta$$ such that $$\alpha \neq \beta$$ but $$\alpha^ \alpha = \beta ^ \beta$$.

This is a list of Calculus proof based problems that I like. Please avoid posting complete solutions, so that others can work on it.

Note by Calvin Lin
3 years, 10 months ago

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Just a couple of hints from someone who has been defeated by this problem :P

Where is the symmetry? Let $$f(x) = x^x$$. We have $$f'(x) = x^x(\ln x + 1)$$, noticing $$f'(\tfrac{1}{e}) = 0$$. It means $$x = \tfrac{1}{e}$$ is an inflection point of $$f(x)$$. As $$x \rightarrow 0$$, we have that $$x^x \rightarrow 1$$, thus for $$x > 1, f(x)$$ is injective, and no value in this interval will satisfy our desired condition.

Can we assume a simple or proportional symmetry? Because $$|\tfrac{1}{e} - 0| \neq |1 - \tfrac{1}{e}|$$, the symmetry of pairs is not of the type $$\tfrac{1}{e} \pm \delta$$. A graphical analysis shows us that the symmetry is also not proportional: the left side is $$(e-1)$$ times smaller than right side, but still $$f(\tfrac{1}{e}-\delta) \neq f(\tfrac{1}{e}+(e-1)\delta)$$.

- 3 years, 10 months ago

We are seeking REAL numbers $$\alpha$$ and $$\beta$$ such that $$\alpha^{\alpha}=\beta^{\beta}$$.

Let $$\beta=k \alpha$$ for a real number $$k$$ . Substituting this in our condition and taking the natural logarithm of both sides, we get:

$$(k\alpha)^{k\alpha}=\alpha^{\alpha}$$
$$k\alpha\ln (k\alpha)=\alpha\ln (\alpha)$$

Rearranging, reducing, and using log properties, we get:

$$\ln \alpha=\frac{k\ln k}{1-k}$$ . Then,using exponentials : $$e^{\ln \alpha}=e^{\frac{k\ln k}{1-k}}$$ and thus $$e^{\ln \alpha}=(e^{\ln k})^{\frac{k}{1-k}}$$ which is equivalent to $$\boxed{\alpha=k^{\frac{k}{1-k}}}$$ . Consequently, $$\beta=k\alpha=k \times k^{\frac{k}{1-k}}= k^{1+\frac{k}{1-k}}=k^{\frac{1}{1-k}}$$ . $$\boxed{\rightarrow \beta=k^{\frac{1}{1-k}}}$$

Checking the values of k:

For k=1, we have $$\alpha=\beta$$ and this contradicts a condition in the question.

For k=0, we have $$\beta=0$$ which is rejected since $$\alpha,\beta$$ are positive.

for k<0, we have a negative $$\alpha$$ or $$\beta$$, so it is also rejected. Therefore :

The solution of the equation $$\alpha^{\alpha}=\beta^{\beta}$$ for $$\alpha,\beta \in R$$ is the set $$\{(\alpha,\beta)|(\alpha,\beta)=(k^{\frac{k}{1-k}},k^{\frac{1}{1-k}}) , \forall k\neq1 \land k \in R^+\}$$

- 3 years, 6 months ago

Because $$f(x) = x^x$$ is injective for $$x > 1$$, we have $$\alpha , \beta < 1$$. Let $$\alpha \cdot a = 1, \beta \cdot b = 1$$. We now have that $$\left ( \tfrac{1}{a} \right )^{\tfrac{1}{a}} = \left ( \tfrac{1}{b} \right )^{\tfrac{1}{b}} \Leftrightarrow \boxed{a^b = b^a}$$.

The only $$\mathbb{Z} - \left \{ 0, 1 \right \}$$ solutions for $$(a,b)$$ are permutations of $$(2,4)$$ (trial and error!). Thus, we have found the only two rational pairs for $$(\alpha , \beta)$$, which are permutations of $$(\frac{1}{2}, \frac{1}{4})$$.

PS: A cheating at WolframAlpha tells me that the pairs are of the form

$\left (x, e^{W(b \log b)} \right )$

where the function $$W(z)$$ is the Lambert W function, defined one way by $$z = W(z) \cdot e^{W(z)}$$.

I have no idea what I'm talking about.

- 3 years, 9 months ago