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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestJust 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 \(f(\tfrac{1}{e}-\delta) \neq f(\tfrac{1}{e}+(e-1)\delta)\).Log in to reply

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 \(\{(\alpha,\beta)|(\alpha,\beta)=(k^{\frac{k}{1-k}},k^{\frac{1}{1-k}}) , \forall k\neq1 \land k \in R^+\}\)

Log in to reply

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.

Log in to reply