[unreality]

Today on my first day of break I found myself bored already and messing around with exponential equations.

A lot of very simple equations, such as a^b = b^a, are not solvable by typical algebra, although sometimes variants are. For example, it's possible if b is a known multiple of a... ie, x^(kx) = (kx)^x, solving for x in terms of k, you get x = k^(1/(k-1)). But most equations with exponentials, including things as simple as e^x = x (which has no real solution and only one complex solution, x = 0.31813150520476413531 - 1.3372357014306894*i) are just impossible with standard algebra unless you're clever or have a computer.

The point here is to not google/wolframalpha/maple/matlab these but instead try to figure them out yourselves, without using the kinds of non-elementary functions people have devised (ie, LambertW). For an example of what I mean, see the hint I posted for #1.

Roughly in order of difficulty...

[1]

e^y = x*y

Find y in terms of x, ie, find the function y(x)

All you need is one function

[2]

For a function f(x) = x^(x^(x^(x^(x^(...

which goes on to infinity,

find the derivative f ' (x)

And also find the values of x such that f(x) is finite.

[3]

Now introducing the "W" function, which solves this equation: z = W*e^W. So W(0) is the solution for W of 0 = W*e^W, i.e. the value is W(0) = 0. Other values are W(-pi/2) = i*pi/2, and W(-1/e) = -1, and W(1) is the omega constant (.56714).

Armed with this "non elementary" function, solve this analog of the quadratic equation for x:

a*ln(ln(x)) + b*ln(x) + c*x = 0

(hint: try it with blnx + cx = 0 first)

I haven't gotten this one yet :lol:

[4]

Use W to invert these functions:

f(x) = x^x

f(x) = x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(...

f(x) = x*ln(x)

[Guest]

I'm not sure if I understand this correctly but it seems like f(x) = lim as n-> infinity of the sequence of functions defined by f1= x, fn+1= x^(fn). If that is the case then f(x) is undefined for x<0, f(x)=0 for x=0, f(x) = 1 for 0<x<=1, and f(x) diverges to positive infinity for x>1. This would make f'(x)=1 where it exists. But this seems a lot simpler than problem 1 so it's probably wrong to some extent.

[unreality]

I'm not sure if I understand this correctly but it seems like f(x) = lim as n-> infinity of the sequence of functions defined by f1= x, fn+1= x^(fn). If that is the case then f(x) is undefined for x<0, f(x)=0 for x=0, f(x) = 1 for 0<x<=1, and f(x) diverges to positive infinity for x>1. This would make f'(x)=1 where it exists. But this seems a lot simpler than problem 1 so it's probably wrong to some extent.

yeah it's a bit more complex than that

[k-man]

I came up with f'(x) = 1 / x^(f(x)-1)

The function f(x) is finite for all 0 <= x <= sqrt(2) and infinite for all x>sqrt(2)

[unreality]

I came up with f'(x) = 1 / x^(f(x)-1)

The function f(x) is finite for all 0 <= x <= sqrt(2) and infinite for all x>sqrt(2)

sqrt(2) is 1.41421356

However,

1.44466786

And regarding the derivative, that may be another valid way of expressing it, I don't know without your steps, but if you:

you might get something like

f'(x) = f(x)^2 / (x - x*f(x)*ln(x))

[Guest]

e^y = x*y

=> y = log (x*y)

=> y = log x + log y

=> y = log x + log (log x + log y)

=> y = log x + log (log x + log (log x + log (log x + log (log x + ....

now, if x = e^z then

y = z + log (z + log (z + log ( z + ...

[Guest]

since it's powers, so any value > 1 should reach to countable infinity with sufficiently large iterations...

Also, being infinite

(x^x)^((x^x)^((x^x)^...

[k-man]

sqrt(2) is 1.41421356

However,

1.44466786

I stand corrected.

... the answer is sqrt(e)

And regarding the derivative, that may be another valid way of expressing it, I don't know without your steps, but if you:

you might get something like

f'(x) = f(x)^2 / (x - x*f(x)*ln(x))

No, mine is definitely wrong. I haven't done any advanced calculus in a long time, so I tried to retrace my own steps and couldn't repeat it. Found a mistake. However, I couldn't get your answer either. Care to post how did you get your answer?

[unreality]

since it's powers, so any value > 1 should reach to countable infinity with sufficiently large iterations...

That's what intuition may tell you, but not true. Try it with sqrt(2).

sqrt(2) = 1.414214

sqrt(2)^sqrt(2) = 1.632527

sqrt(2)^ 1.632527 = 1.76084

sqrt(2)^1.76084 = 1.840911

keep going and it converges on 2.

However sqrt(2) isn't quite the limit, you can go higher yet. (But not so high as sqrt(e) which was k-man's last guess)

I stand corrected.

... the answer is sqrt(e)

sqrt(e) = e^(1/2) is too high, it doesn't work. You're getting warmer though

No, mine is definitely wrong. I haven't done any advanced calculus in a long time, so I tried to retrace my own steps and couldn't repeat it. Found a mistake. However, I couldn't get your answer either. Care to post how did you get your answer?

y = x^(x^(x^(...

ln(y) = x^(x^(x..)))))..) * ln(x) = y * ln(x)

so we have

ln(y) = y*ln(x)

now take the derivative with respect to x:

(dy/dx) / y = (dy/dx)*ln(x) + y/x

from here it's just algebra:

-y/x = (dy/dx) * (ln(x) - 1/y)

(dy/dx) = -y / ( x*(ln(x) - 1/y) )

f ' (x) = -f(x) / (xlnx - x/f(x))

which is the same as:

f ' (x) = f(x)^2 / (x - x*ln(x)*f(x) )

[unreality]

e^y = x*y

=> y = log (x*y)

=> y = log x + log y

=> y = log x + log (log x + log y)

=> y = log x + log (log x + log (log x + log (log x + log (log x + ....

now, if x = e^z then

y = z + log (z + log (z + log ( z + ...

yep! kudos

[Guest]

That's what intuition may tell you, but not true. Try it with sqrt(2).

sqrt(2) = 1.414214

sqrt(2)^sqrt(2) = 1.632527

sqrt(2)^ 1.632527 = 1.76084

sqrt(2)^1.76084 = 1.840911

keep going and it converges on 2.

However sqrt(2) isn't quite the limit, you can go higher yet. (But not so high as sqrt(e) which was k-man's last guess)

hmm, in that case...

y = x ^ (x ^ (x ^ ( x ^ ...

=> y = x ^ y

=> x = y ^ 1/y

at y = (inf), you are asking me to define 1/(inf) and (inf) ^ (1/(inf))

No problem, let me try...

[k-man]

However sqrt(2) isn't quite the limit, you can go higher yet. (But not so high as sqrt(e) which was k-man's last guess)

sqrt(e) = e^(1/2) is too high, it doesn't work. You're getting warmer though

I saw that the largest finite number that this sequence converges to is e. So I just wrote sqrt(e) since I saw that sqrt(2) converges to 2. The correct answer (and I verified it) is e^1/e

[unreality]

yep good job There hasn't been much progress so I'll go ahead and post the answers

[1]

e^y = x*y

Find y in terms of x, ie, find the function y(x)

y = log x + log (log x + log (log x + log (log x + log (log x + ....

KlueMaster got it!

[2]

For a function f(x) = x^(x^(x^(x^(x^(...

which goes on to infinity,

find the derivative f ' (x)

And also find the values of x such that f(x) is finite.

f'(x) = f(x)^2 / (x - x*f(x)*ln(x))

the bounds of convergence are e^(-e) and e^(1/e)

[3]

Now introducing the "W" function, which solves this equation: z = W*e^W. So W(0) is the solution for W of 0 = W*e^W, i.e. the value is W(0) = 0. Other values are W(-pi/2) = i*pi/2, and W(-1/e) = -1, and W(1) is the omega constant (.56714).

Armed with this "non elementary" function, solve this:

(hint: try it with blnx + cx = 0 first)

x = (b/c) * W(c/b)

[4]

Use W to invert these functions:

f(x) = x^x

f(x) = x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(...

f(x) = x*ln(x)

If g(x) is the inverse function of f(x) then:

f(x) = x^x

g(x) = ln(x) / W(ln(x))

f(x) = x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(x^(...

You don't even need the W function for this one:

g(x) = e^( ln(x) / x) = x^(1/x)

f(x) = x*ln(x)

g(x) = x / W(x)

[dark_magician_92]

That's what intuition may tell you, but not true. Try it with sqrt(2).

sqrt(2) = 1.414214

sqrt(2)^sqrt(2) = 1.632527

sqrt(2)^ 1.632527 = 1.76084

sqrt(2)^1.76084 = 1.840911

keep going and it converges on 2.

However sqrt(2) isn't quite the limit, you can go higher yet. (But not so high as sqrt(e) which was k-man's last guess)

sqrt(e) = e^(1/2) is too high, it doesn't work. You're getting warmer though

y = x^(x^(x^(...

ln(y) = x^(x^(x..)))))..) * ln(x) = y * ln(x)

so we have

ln(y) = y*ln(x)

now take the derivative with respect to x:

(dy/dx) / y = (dy/dx)*ln(x) + y/x

from here it's just algebra:

-y/x = (dy/dx) * (ln(x) - 1/y)

(dy/dx) = -y / ( x*(ln(x) - 1/y) )

f ' (x) = -f(x) / (xlnx - x/f(x))

which is the same as:

f ' (x) = f(x)^2 / (x - x*ln(x)*f(x) )

ya i got f'(x) by doing it implicitly but like k-man, my intuition tells me sqrt(2) might work and tend to be finite, are u sure about d answer?

Edited December 30, 2010 by dark_magician_92

[dark_magician_92]

yep! kudos

cool buddy, well done that replacement of x by z was very well thought

[unreality]

ya i got f'(x) by doing it implicitly but like k-man, my intuition tells me sqrt(2) might work and tend to be finite, are u sure about d answer?

One way to show the e^(1/e) limit is this...

so we have y' = (y^2) / (x - x y ln x)

we'll have critical points where y' is 0 or undefined.

0 when y^2 = 0, so y = 0. We find the inverse function x = y^(1/y) and y=0, so 0^(1/0) isn't really defined. So look just at the bottom:

x(1 - ylnx) = 0

x = 0 and 1 = ylnx

1 = ylnx

raise e to both sides to get:

e = y

then use the inverse (x = y^(1/y)) to get x = e^(1/e)

That's one way to do it, kind of sloppy though

[dark_magician_92]

One way to show the e^(1/e) limit is this...

so we have y' = (y^2) / (x - x y ln x)

we'll have critical points where y' is 0 or undefined.

0 when y^2 = 0, so y = 0. We find the inverse function x = y^(1/y) and y=0, so 0^(1/0) isn't really defined. So look just at the bottom:

x(1 - ylnx) = 0

x = 0 and 1 = ylnx

1 = ylnx

raise e to both sides to get:

e = y

then use the inverse (x = y^(1/y)) to get x = e^(1/e)

That's one way to do it, kind of sloppy though

thanks a lot, i solved the problem upto that "ylnx=1", but couldnt solve further. Thanks n I feel happy now to have joined brainden