Why can't a^x = x^b be solved analytically?!

Hmm. It feels a bit like integrating e^(x^2), doesn't it...

my maths teacher mentioned that, is it now defined to be the error function or something? which, if it is, is that the same as the cumulative normal distribution? [with 0 mean and a SD of something or other, of course]

Hmm. It feels a bit like integrating e^(x^2), doesn't it...

my maths teacher mentioned that, is it now defined to be the error function or something? which, if it is, is that the same as the cumulative normal distribution? [with 0 mean and a SD of something or other, of course]

Not quite, we have:

$\displaystyle \int_0^x{\mathrm e}^{-t^2}{\mathrm d}t = \sqrt{\frac{\pi}2}\operatorname{erf}(x)[br]$

and

$\displaystyle \int_0^x{\mathrm e}^{t^2}{\mathrm d}t = -i\sqrt{\frac{\pi}2}\operatorname{erf}(ix)[br]$

Obviously, it can be integrated.

But it doesn't have a very tidy/neat form that involves very common functions as OP was hinting at. I hadn't heard of the error function until I read about this particular integral, and yet the integral of e^x is just itself, which is a lot neater. xD

Oh, no, I wasn't saying that you'd said anything wrong, it was just that the OP misinterpreted e^x^2 to be the normal dist. which it isn't, that would be e^(-x^2) so I was just highlighting the difference for him.

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I'm sorry but I've attempted to answer a question in this form and I mean, wtf.. what is wrong with maths. I read some guys 'answer' and he said you'd have to use numerical methods to solve it.

Obviously plotting x^b gets a bit hairy when x is negative and b isn't an integer and a^x gets hairy when a is negative etc., but surely there's a neat complex solution to this. I mean... please guys, tell me someone's got a neat solution to this

Please.

Don't do this to me.




Not like this, maths.













Not like this.

I used to think like you. Unfortunately, not all things have a closed-form solution. It's just something I came to accept.. eventually...

The fact that 2 graphs can intercept at a point corresponding to an infinite sum (e.g. a taylor/power series) is interesting to think about though. I would say that such solutions are just another step up from irrational numbers, in that they both require infinite "amounts" to express properly without written abbreviations.

What interests me is that, if we came up with another angle and somehow saw things from another perspective, maybe everything would flip around. E.e. the things that aren't tidy become simple and tidy, but the currently tidy things become the complicated ones...

There is an 'analytical' solution, if you're prepared to deal with the Lambert W Function.

Then the solution is $\displaystyle x = - \frac{b W\left(-\frac{\log a}{b}\right)}{\ln a}$

I'm sorry but I've attempted to answer a question in this form and I mean, wtf.. what is wrong with maths. I read some guys 'answer' and he said you'd have to use numerical methods to solve it.

Obviously plotting x^b gets a bit hairy when x is negative and b isn't an integer and a^x gets hairy when a is negative etc., but surely there's a neat complex solution to this. I mean... please guys, tell me someone's got a neat solution to this

Please.

Don't do this to me.




Not like this, maths.













Not like this.

There is an 'analytical' solution, if you're prepared to deal with the Lambert W Function.

Then the solution is $\displaystyle x = - \frac{b W\left(-\frac{\log a}{b}\right)}{\ln a}$

After reading a little bit about it; how can you still have a useful function that isn't expressed in elementary functions? Is it defined in terms of derivatives/ integrals or something? How can someone put a value into it and get a value out?

Wait Zacken, I thought you are in Year 13. Are you an undergrad?

Well, it's a bit of a complicated situation, I've finished my GCSE's last year november, basically...

So you are applying to uni this year? I guess I've not done enough further reading if I'm up agaist people who have studied things like this in their own time!