X^X == 10^N

[Zaphod Beeblebrox]

Write a program that given a value N, will return x, such that X^X=10^N.

My best program so far is 45 bytes, after subtracting the name.

EDIT: Find the closest approximation of x, not necessarily exactly x. And also, it should support values of N > 100.

[thornahawk]

This is all I have:


Nln(10
solve(Ans-Yln(Y),Y,1+Ans


which is fine unless you wanted the *other branch* of the inverse of x^x, or something that works for complex number inputs as well.

(edit)

...or it turns out that this is also OK:


solve(N-Ylog(Y),Y,1+N


thornahawk

[Zaphod Beeblebrox]

Well, I was thinking you do the algorithm yourself, rather than having the calculator do it for you. This was (supposed to be) an exercise in creating your own algorithm and optimizing it. Oh well. I guess I didn't say that.

For example:

[elfprince13]

[DarkerLine]

You should probably be aware that EN causes an error message.

[Ed H]

Edit: Check my later post for a program that works way better.

32+2=34 bytes:

[elfprince13]

urgh, that sucks, so we have to do 10^N instead of EN? I haven't programmed Basic in too long.

Ed, that mean's yours is broken too

[Ed H]

elfprince, that's the base of the natural log, not the scientific notation exponent operation.

[simplethinker]

[DarkerLine]

It does work, actually: log is what you want in this case. Except it crashes if Ans doesn't happen to be a positive number.

[Ed H]

Edit: Hmm, my program doesn't converge or work on a lot of inputs. I'm not quite sure how to fix this. It seems to work alright if I put abs( in all my logs, but then I get overflow errors for large N, such as 9000.

@simplethinker:
Try it first, simplethinker. I do take into account the fact that log is base-10 and ln is base-e. I don't remember exactly the steps I followed in getting to that expression, but I plan on posting them later.

@Darkerline:
Ah yes, that is a rather serious error... I wonder if it will always converge if Ans starts at 1?

[simplethinker]

I have 42 bytes.

[Ed H]

I found what was wrong with my program. I entered in an X instead of an Ans - funny thing, then, that it still worked for N up to like, 500! But now it converges almost instantly, and works for up to 1E98. Unfortunately, still no negatives.

Here is the actual program, at 34 bytes.

[simplethinker]

I just realized that using log(log( is kind of pointless. The problem with using log( 10^N=x^x) is with |N|<1.

[Ed H]

Here's a version that converges for everything, in 48 bytes:

[thornahawk]

Without using solve( , I couldn't do better than this:


N+1
Repeat not(abs(Ans−F \\ this complication is needed so it does not break on negative/complex N
Ans→F
log(℮Ans)ֿ¹(N+Anslog(℮
End
Ans


Again, this is only the main branch of the inverse of log₁₀(x^x). Something more elaborate would be needed if you wish to see the other branches.

thornahawk