The tower

Algebra Level 4

Given the following function (also called infinite exponential tower):

f ( a ) = a a a a a a a a . . . { { { { { { { { f(a)=a }^{ a } }^{ a } }^{ a } }^{ a } }^{ a } }^{ a } }^{ a } }^{ ... }

Where a a is a positive real number

Let a 0 { a }_{ 0 } be the maximum value of a a so that the function has a definite value and f ( a 0 ) = b f({ a }_{ 0 })=b

What is a 0 + b { a }_{ 0 }+b

Submit your answer as 10000 ( a 0 + b ) \left\lfloor 10000({ a }_{ 0 }+b) \right\rfloor

Note: . \left\lfloor . \right\rfloor is the floor function

*Note: though it may be possible to solve this problem manually all the way, my solution involves a little bit of programming. If anyone has a solution that is entirely manual, please tell me in the solution section. Thank you. *


The answer is 41629.

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3 solutions

Zhang Xiaokang
Jun 28, 2019

I'm not too familiar with latex, so here is a written 'manual solution' :)

Viet Tran
Jun 26, 2019

Let f ( a ) = x f(a)=x . It's easy to see that x = a x x={ a }^{ x }

Now we find the maximum value a 0 { a }_{ 0 } of a a so that the equation x = a x x={ a }^{ x } has a real solution x 0 { x }_{ 0 }

The solution is the intersection of two graphs: y = x y=x and y = a x y={ a }^{ x }

The maximum value of a is when the line y = x y=x is tangent to the curve y = a x y={ a }^{ x }

The tangent line equation y = y ( x x 0 ) + y 0 y = x y=y'(x-{ x }_{ 0 })+{ y }_{ 0 }\equiv y=x where ( x 0 , y 0 ) \left( { { x }_{ 0 },{ y }_{ 0 } } \right) is the point of intersection

y = a x 0 ln a ( x x 0 ) + a x 0 y={ a }^{ { x }_{ 0 } }\ln { a } (x-{ x }_{ 0 })+{ a }^{ { x }_{ 0 } }

We know y ( x 0 ) = 1 y'({ x }_{ 0 })=1 and the tangent line goes through ( 0 , 0 ) (0,0) so:

a x 0 ln a = 1 x 0 a x 0 = 0 \begin{matrix} { a }^{ { x }_{ 0 } }\ln { a } =1 \\ { x }_{ 0 }-{ a }^{ { x }_{ 0 } }=0 \end{matrix}

x 0 ln a = 1 x 0 a x 0 = 0 \equiv \begin{matrix} { x }_{ 0 }\ln { a } =1 \\ { x }_{ 0 }-{ a }^{ { x }_{ 0 } }=0 \end{matrix}

1 ln a a 1 ln a = 0 x 0 a x 0 = 0 \equiv \begin{matrix} \frac { 1 }{ \ln { a } } -{ a }^{ \frac { 1 }{ \ln { a } } }=0 \\ { x }_{ 0 }-{ a }^{ { x }_{ 0 } }=0 \end{matrix}

Now we use Newton's method to solve the first equation than substitute it into the second one. This is when programming comes in handy

Newton's method: x n = x n 1 f ( x n 1 ) f ( x n 1 ) { x }_{ n }={ x }_{ n-1 }-\frac { f({ x }_{ n-1 }) }{ f'({ x }_{ n-1 }) }

Now copy and paste all of these code lines into python (I'm sorry I don't know how to put a python window here):

The program gives:

a = 1.444667861

b = 2.718281802

a + b = 4.162949663

The "entirely manual" solution works like yours - once you get to the equation 1 log a a 1 log a = 0 \frac{1}{\log{a}}-a^\frac{1}{\log{a}}=0 , just substitute t = log a t=\log{a} . The equation becomes 1 t ( e t ) 1 t = 0 \frac{1}{t} - \left( e^t \right)^{\frac{1}{t}}=0 . But this is just 1 t e = 0 \frac{1}{t}-e=0 , and so we find a = e 1 e a=e^{\frac{1}{e}} .

Chris Lewis - 1 year, 11 months ago

As mentioned here , Euler showed that f ( a ) f\left(a\right) converges only when 1 e e a e e \displaystyle \frac{1}{e^e}\le a\le\sqrt[e]{e}

Aaghaz Mahajan - 1 year, 11 months ago

The maximum value of the given expression is b=e, when a=e^(1/e)

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