αα=ββ \alpha^ \alpha = \beta ^ \beta

Find all pairs of positive real numbers α \alpha and β \beta such that α≠β \alpha \neq \beta but αα=ββ \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.

#Calculus #Proofs

Note by Calvin Lin
7 years, 2 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list
  • bulleted
  • list

1. numbered
2. list
  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

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

Where is the symmetry? Let f(x)=xxf(x) = x^x. We have f′(x)=xx(ln⁡x+1)f'(x) = x^x(\ln x + 1), noticing f′(1e)=0f'(\tfrac{1}{e}) = 0. It means x=1ex = \tfrac{1}{e} is an inflection point of f(x)f(x). As x→0x \rightarrow 0, we have that xx→1x^x \rightarrow 1, thus for x>1,f(x)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 ∣1e−0∣≠∣1−1e∣|\tfrac{1}{e} - 0| \neq |1 - \tfrac{1}{e}|, the symmetry of pairs is not of the type 1e±δ\tfrac{1}{e} \pm \delta. A graphical analysis shows us that the symmetry is also not proportional: the left side is (e−1)(e-1) times smaller than right side, but still f(1e−δ)≠f(1e+(e−1)δ)f(\tfrac{1}{e}-\delta) \neq f(\tfrac{1}{e}+(e-1)\delta).

Guilherme Dela Corte - 7 years, 2 months ago

We are seeking REAL numbers α \alpha and β\beta such that αα=ββ\alpha^{\alpha}=\beta^{\beta}.

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

(kα)kα=αα(k\alpha)^{k\alpha}=\alpha^{\alpha}
kαln⁡(kα)=αln⁡(α)k\alpha\ln (k\alpha)=\alpha\ln (\alpha)

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

ln⁡α=kln⁡k1−k\ln \alpha=\frac{k\ln k}{1-k} . Then,using exponentials : eln⁡α=ekln⁡k1−ke^{\ln \alpha}=e^{\frac{k\ln k}{1-k}} and thus eln⁡α=(eln⁡k)k1−ke^{\ln \alpha}=(e^{\ln k})^{\frac{k}{1-k}} which is equivalent to α=kk1−k\boxed{\alpha=k^{\frac{k}{1-k}}} . Consequently, β=kα=k×kk1−k=k1+k1−k=k11−k\beta=k\alpha=k \times k^{\frac{k}{1-k}}= k^{1+\frac{k}{1-k}}=k^{\frac{1}{1-k}} . →β=k11−k\boxed{\rightarrow \beta=k^{\frac{1}{1-k}}}

Checking the values of k:

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

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

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

The solution of the equation αα=ββ\alpha^{\alpha}=\beta^{\beta} for α,β∈R\alpha,\beta \in R is the set {(α,β)∣(α,β)=(kk1−k,k11−k),∀k≠1∧k∈R+}\{(\alpha,\beta)|(\alpha,\beta)=(k^{\frac{k}{1-k}},k^{\frac{1}{1-k}}) , \forall k\neq1 \land k \in R^+\}

Hasan Kassim - 6 years, 10 months ago

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

The only Z−{0,1}\mathbb{Z} - \left \{ 0, 1 \right \} solutions for (a,b)(a,b) are permutations of (2,4)(2,4) (trial and error!). Thus, we have found the only two rational pairs for (α,β)(\alpha , \beta), which are permutations of (12,14) (\frac{1}{2}, \frac{1}{4}) .

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

(x,eW(blog⁡b))\left (x, e^{W(b \log b)} \right )

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

I have no idea what I'm talking about.

Guilherme Dela Corte - 7 years, 2 months ago
×

Problem Loading...

Note Loading...

Set Loading...