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

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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 :cry2:

    Please.

    Don't do this to me.




    Not like this, maths.













    Not like this.
    :emo:
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    I haven't actually studied it before, but would Taylor series or something of the sort allow you to solve this?
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    And if you know any more equations that have no analytical solutions, feel free to post them here, because I'm quite interested...

    interested in running my previous image of maths further into the ground, into the abyss in my cold, empty heart where it now resides. pfft. No neat solution. What do you take me for. You don't know me. There's a neat solution, I assure you. You're just too stupid to find it. I know there's a solution, DONT TELL ME THERES NOT.
    *sobs quietly*



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

    :teehee:
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    (Original post by Maria G Agnesi)
    Hmm. It feels a bit like integrating e^(x^2), doesn't it...

    :teehee:
    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]
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    (Original post by Callum Scott)
    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]
    Yes, something about the error function. :mmm:

    But the integral doesn't have a 'nice' general solution as many others do.

    I don't know what useful applications the integral could have. Any ideas?
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    (Original post by Maria G Agnesi)
    Hmm. It feels a bit like integrating e^(x^2), doesn't it...

    :teehee:
    (Original post by Callum Scott)
    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)

    and

    \displaystyle \int_0^x{\mathrm e}^{t^2}{\mathrm d}t = -i\sqrt{\frac{\pi}2}\operatorname  {erf}(ix)
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    (Original post by Zacken)
    Not quite, we have:

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

    and

    \displaystyle \int_0^x{\mathrm e}^{t^2}{\mathrm d}t = -i\sqrt{\frac{\pi}2}\operatorname  {erf}(ix)
    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
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    (Original post by Maria G Agnesi)
    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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    (Original post by Zacken)
    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.
    Ah, right, thank you for clearing that up.
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    (Original post by Callum Scott)
    x
    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}
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    (Original post by Callum Scott)
    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 :cry2:

    Please.

    Don't do this to me.




    Not like this, maths.













    Not like this.
    :emo:
    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...
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    (Original post by TooEasy123)
    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...
    I always hope for things to turn out like that lol. When I was first confronted with the fact that x^2 + 1 = 0 has no solutions, somewhere deep down, it just didn't settle. Then complex numbers come into play and I'm like woahoahwoah, THEEES; I LIKE. I'm always a hater of \frac{1}{0} as well. There just can't be a 'gap' in how we understand things. Some video on youtube mentioned that if we consider \frac{1}{0} to be an actual point, we could do fancy things with cylinders and complex numbers and yada yada yada... idk what it was, but sweet jesus did it turn me on.
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    (Original post by Zacken)
    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?
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    (Original post by Callum Scott)
    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 :cry2:

    Please.

    Don't do this to me.




    Not like this, maths.













    Not like this.
    :emo:
    https://www.youtube.com/watch?v=X02EwkGqyto
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    (Original post by Zacken)
    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}
    Wait Zacken, I thought you are in Year 13. Are you an undergrad?
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    (Original post by Callum Scott)
    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?
    It's defined as the solution to a certain equation, or satisfying a particular equation, I can't quite recall it.

    But yes, you can put values into it and get other valuesout. For example: w(0) = 0, w(e)=1, etc...
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    (Original post by Louisb19)
    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...
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    (Original post by Zacken)
    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!
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    (Original post by Louisb19)
    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!
    Oh, trust me, I'm just as scared if not more about going up against people, I need to work on my problem solving ability big time.

    Yeah, I'm applying a year early.
 
 
 
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