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Solving y=x^x for x

  • Thread starter Izzhov
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Problem:
I'm trying to figure out how to solve [tex]y=x^x[/tex] for x as a function of y.

Related Equations:
[tex]\log_n a^b=b \ast \log_n a[/tex]

Attempt at Solution:
I took the natural logarithm of both sides and got: [tex]\ln y=x \ast \ln x[/tex] I don't really have any idea where to go from here.
 

Answers and Replies

Dr Transport
Science Advisor
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How about plotting both sides.....

Remember that the [tex] \ln(x) [/tex] is nonsense for [tex] x \leq 0 [/tex].
 
117
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How about plotting both sides.....

Remember that the [tex] \ln(x) [/tex] is nonsense for [tex] x \leq 0 [/tex].
Uh... by plot, do you mean graph? I'm not trying to find where the graphs intersect or anything... I'm just trying to change the equation from [tex]y=f(x)[/tex] to [tex]x=f(y)[/tex]
 
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Dr Transport
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You can't......
 
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You mean to say that [tex]x^x[/tex] has no inverse that can be defined in terms of elementary functions? Then what about using calculus? Can it be defined then?
 
HallsofIvy
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Depends upon what you mean by "using Calculus". The inverse of y= xx is the "Lambert W function".
 
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By "using calculus," I mean, for example, defining the function with sigma notation, the derivative of something, etc. Here's a good example: the indefinite integral [tex] \int x^x dx [/tex] can not be defined in terms of elementary functions, but can be defined as the indefinite integral of [tex] x^x [/tex] So, what I'm asking is, can the inverse of [tex] x^x [/tex] be defined in terms of elementary functions as well as indefinite integrals, series, sums (i.e. sigma notation), derivatives, and so on?
 
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I did some research on the "Lambert W function," and it's not the inverse of [tex]x^x[/tex]. It's the inverse of [tex]x \ast e^x[/tex].
 
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Nchimy

Solving y=x^x for x does have a solution in terms of Lambert W Function. Although this reply may have come late, one could still make use of it later.

To make the problem easy, lets assume that x and y are real. Then we can proceed as follows:

If y = x^x, .........(1)

then since we also have

x = exp[ln[x]], .........(2)

we may conviniently express (1) as

y=x^exp[ln[x]]. .........(3)

Now, taking logs on both sides of (3) (and noting that ln[d^c]=cln[d]) gives

ln[y] = ln[x] * exp[ln[x]] ..... (4).

Imediately, we see that (4) can be solved for ln[x], using Lambert W Function, as

ln[x] = W[ln[y]] ..... (5)

so that

x = exp[W[ln[y]]] .... (6) .

The solution in (6) is valid for y > 0 because ln[y] for real values of y makes sense only within this range. From (6) (and taking note that W[0]=1while W[e] =1) we see that

(a) x = 0 when y = 1,
(b) x = 1 when y = e
(c) x is only real when ln[y] >= -(1/e) (or equivalently x is real for y >= exp[-(1/e)) but complex and multivalued otherwise.
(d) x increases monotonically with increasing value of y.
(e) The exists a taylor series expansion of x, about ln[y]=0, with a radius of converges equal to -(1/e).

For more details on th Lambert W function, please refer to Corless R M et al, “On the Lambert W function”, Adv. Comput. Math, Vol. 5, pp.329-359.


Cheers.....
 
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