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Thread: Help with Fractional Powers

  1. #1 Help with Fractional Powers 
    Forum Sophomore ChaosD.Ace's Avatar
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    Hi, How would I solve to find x in 967.261271503x5.45 + 0.0000403225x = 0.00000077575?


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  3. #2  
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    I could probably write a computer program to find the answer, or at least as close as need be. I think the program could also quickly determine, up front, if there is a solution. College math is behind me many, many years. I'm am very curious if and how this would be solved more "mathematically".


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  4. #3  
    Forum Sophomore ChaosD.Ace's Avatar
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    Oh you mean it's degree maths, and no quick way to find out?
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  5. #4  
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    Quote Originally Posted by ChaosD.Ace View Post
    Oh you mean it's degree maths, and no quick way to find out?
    No. Sorry to confuse you. I don't know how to solve it mathematically. It may be simple. It may be difficult.

    You can use this scientific calculator to plug in different values for 'x' and see where it takes you: Online Scientific Calculator | EEWeb
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  6. #5  
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    I am unclear if the power 5.45 is just on the x or on the complete 967.261271503x

    So does the distributive law of exponents apply here?

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    I assumed it was just on the 'x'. It has the "look" of a quadratic equation, except for the fractional power of 5.45 instead of 2.
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  8. #7  
    Forum Professor Daecon's Avatar
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    Is it possible to first multiply both numbers so that the power is a whole, and then re-divide the result after you've done the calculation?

    Or does that not work with this kind of equation?
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  9. #8  
    Forum Radioactive Isotope MagiMaster's Avatar
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    If you want to solve it numerically, just apply Newton's method and you should be able to find the answers to a reasonable degree of accuracy pretty quickly. High degree polynomials don't generally have nice solutions and noninteger powers are generally less nice.
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  10. #9  
    Forum Junior anticorncob28's Avatar
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    Let there be a primitive algorithm to solve any polynomial equation. Then
    967.261271503x^5.45 + 0.0000403225x = 0.00000077575
    967.261271503x^5.45 = 0.00000077575 - 0.0000403225x
    Then, raise both sides to the power of 20. The equation turns to something of the form:
    A(x^109) = P(x)
    P(x) is a polynomial of degree 20 and A is a constant (967.261271503^109, to be precise), multiplied by x^109. For some unknown reason it looks like a function A(x) on A(x^109). But it is actually a constant multiplied by x^109.
    A(x^109) - P(x) = 0
    This is a polynomial equation of degree 109. With a primitive algorithm to solve any polynomial equation, it can be solved. However with the primitive algorithms only of the axioms of the real numbers, it probably can't be solved. There is no general solution to polynomial equations of degree higher than 4, let alone degree 109, so unless this is a special case, it can't be solved using the facts we currently know about the real numbers.
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  11. #10  
    Forum Junior anticorncob28's Avatar
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    Let there be a primitive algorithm to solve any polynomial equation. Then
    967.261271503x^5.45 + 0.0000403225x = 0.00000077575
    967.261271503x^5.45 = 0.00000077575 - 0.0000403225x
    Then, raise both sides to the power of 20. The equation turns to something of the form:
    A(x^109) = P(x)
    P(x) is a polynomial of degree 20 and A is a constant (967.261271503^109, to be precise), multiplied by x^109. For some unknown reason it looks like a function A(x) on A(x^109). But it is actually a constant multiplied by x^109.
    A(x^109) - P(x) = 0
    This is a polynomial equation of degree 109. With a primitive algorithm to solve any polynomial equation, it can be solved. However with the primitive algorithms only of the axioms of the real numbers, it probably can't be solved. There is no general solution to polynomial equations of degree higher than 4, let alone degree 109, so unless this is a special case, it can't be solved using the facts we currently know about the real numbers.
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  12. #11  
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    Quote Originally Posted by ChaosD.Ace View Post
    Hi, How would I solve to find x in 967.261271503x5.45 + 0.0000403225x = 0.00000077575?
    Are you limiting yourself only to real values of x? The general collection of solutions is quite large...(although you are guaranteed at least one real solution here).
    Last edited by tk421; March 4th, 2014 at 08:54 PM.
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  13. #12  
    Forum Sophomore ChaosD.Ace's Avatar
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    Wow, thanks a lot guys, so many repliers.

    I am unclear if the power 5.45 is just on the x or on the complete 967.261271503x
    the power 5.45 is indeed on the x.


    If you want to solve it numerically, just apply Newton's method and you should be able to find the answers to a reasonable degree of accuracy pretty quickly. High degree polynomials don't generally have nice solutions and noninteger powers are generally less nice.
    I tried getting my A-Level maths teacher to solve this but he said he couldn't algebraicly, he told me to write a program to work it out numerically as well.

    Are you limiting yourself only to real values of x? The general collection of solutions is quite large...(although you are guaranteed at least one real solution here).
    I believe so. This equation results from the use of solid of rotation to find the volume of a solid object. Then I halved the volume and pluged that back into the integral in order to find the new boundary which corresponds to half the volume. The equation above is precisely that. In fact I made a mistake, it should be = 0.00000077575/pi (which is just a number.) Anyway this means that the answer will be bellow 0.015 (because that's the upper boundary for the whole volume.) Hope that makes sense.
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  14. #13  
    Brassica oleracea Strange's Avatar
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    If you don't need to know how to solve it, you can just look it up:
    http://www.wolframalpha.com/input/?i=solve+967.261271503*x^5.45+%2B+0.0000403225x+%3 D+0.00000077575
    Without wishing to overstate my case, everything in the observable universe definitely has its origins in Northamptonshire -- Alan Moore
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  15. #14  
    Forum Sophomore ChaosD.Ace's Avatar
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    I would like to know how. That stuff anticorncob was writting looks interesting. Also how reliable is wolfram?

    Lol, wolfram said standard computational time exceeded.
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  16. #15  
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    Quote Originally Posted by ChaosD.Ace View Post
    Also how reliable is wolfram?
    Wolfram is Mathematica.
    It is industrial strength mathematical programming and pretty much the gold standard in the field.
    It is a bit expensive for just hobby use at about $150.00 for the student edition or about $300.00 for the home edition.
    Industrial and commercial versions cost a lot more.

    Wolfram Mathematica: Technical Computing Software—Taking You from Idea to Solution
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  17. #16  
    Forum Sophomore ChaosD.Ace's Avatar
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    Quote Originally Posted by dan hunter View Post
    Quote Originally Posted by ChaosD.Ace View Post
    Also how reliable is wolfram?
    Wolfram is Mathematica.
    It is industrial strength mathematical programming and pretty much the gold standard in the field.
    It is a bit expensive for just hobby use at about $150.00 for the student edition or about $300.00 for the home edition.
    Industrial and commercial versions cost a lot more.

    Wolfram Mathematica: Technical Computing Software—Taking You from Idea to Solution
    Darn, That sucks. So, how about that whole programming myself solution, how would I start.
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  18. #17  
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    Quote Originally Posted by ChaosD.Ace View Post
    Quote Originally Posted by dan hunter View Post
    Quote Originally Posted by ChaosD.Ace View Post
    Also how reliable is wolfram?
    Wolfram is Mathematica.
    It is industrial strength mathematical programming and pretty much the gold standard in the field.
    It is a bit expensive for just hobby use at about $150.00 for the student edition or about $300.00 for the home edition.
    Industrial and commercial versions cost a lot more.



    Wolfram Mathematica: Technical Computing Software—Taking You from Idea to Solution
    Darn, That sucks. So, how about that whole programming myself solution, how would I start.
    Strange gave you a link to Wolfram's free calculator service, Wolfram Alpha.
    If you go there all you need to do is enter your equation and it will give you the answer.
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    43/2 = either 8 or -8. Which is, √(43) = (43)1/2 should help you understand it; 43 = 64, and 641/2 = √64 = 8 or -8.

    So raise the base to the power of the numerator of the fractional power, then take the denominator-th root of the result.

    5.45 is 545/100 and you may reduce the fraction before taking the powers and roots. It's 109/20. Thus you take the 109th power of the base, and then take the 20th root of the result. That's what raising to the 5.45 power means.

    Also its commutative; you can take the 20th root of the base and then the 109th power of the result, if that's more convenient. Same thing.
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  20. #19  
    Forum Radioactive Isotope MagiMaster's Avatar
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    If you want to program it, use Newton's method. There are other ways to do the same thing faster, but Newton's method is simple. Just rewrite your equation in the form f(x) = 0, then plug it in to the formula given on the Wiki site. Just be aware that fractional powers often give imaginary results, so you may need a complex number class.
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    Quote Originally Posted by MagiMaster View Post
    If you want to program it, use Newton's method. There are other ways to do the same thing faster, but Newton's method is simple. Just rewrite your equation in the form f(x) = 0, then plug it in to the formula given on the Wiki site. Just be aware that fractional powers often give imaginary results, so you may need a complex number class.
    Newton's Method is nonterminating.
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  22. #21  
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    Quote Originally Posted by Schneibster View Post
    Quote Originally Posted by MagiMaster View Post
    If you want to program it, use Newton's method. There are other ways to do the same thing faster, but Newton's method is simple. Just rewrite your equation in the form f(x) = 0, then plug it in to the formula given on the Wiki site. Just be aware that fractional powers often give imaginary results, so you may need a complex number class.
    Newton's Method is nonterminating.
    True, but that does not mean one can't get useful results from Newton's method. It does help if one has at least a rough idea of the value of the root(s) in question. Using that initial estimate as a seed frequently helps a great deal (although convergence is still not guaranteed).

    In more difficult cases, I find that Laguerre's method does a remarkably good job. It seems (almost) always to converge, independent of the seed value, and converges relatively quickly.
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    Actually, Newton's Method is Chaotic.



    Which turns out to be very cool; it's a fractal function that creates four basins of attraction depending on your choice of initial estimate. ETA: Yes, yes, I know about the number of complex roots. I'm talking about the simplest case and we all know it spans the class.

    But it also makes it Turing- and Kolmogorov-undecidable.

    Not that that's ultimately all that freaky, what we're really saying is we can't differentiate between 8 and -8 for certain in reverse physics of that type; which is merely affirming the 2LOT.
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  24. #23  
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    Quote Originally Posted by ChaosD.Ace View Post
    Wow, thanks a lot guys, so many repliers.

    I am unclear if the power 5.45 is just on the x or on the complete 967.261271503x
    the power 5.45 is indeed on the x.


    If you want to solve it numerically, just apply Newton's method and you should be able to find the answers to a reasonable degree of accuracy pretty quickly. High degree polynomials don't generally have nice solutions and noninteger powers are generally less nice.
    I tried getting my A-Level maths teacher to solve this but he said he couldn't algebraicly, he told me to write a program to work it out numerically as well.

    Are you limiting yourself only to real values of x? The general collection of solutions is quite large...(although you are guaranteed at least one real solution here).
    I believe so. This equation results from the use of solid of rotation to find the volume of a solid object. Then I halved the volume and pluged that back into the integral in order to find the new boundary which corresponds to half the volume. The equation above is precisely that. In fact I made a mistake, it should be = 0.00000077575/pi (which is just a number.) Anyway this means that the answer will be bellow 0.015 (because that's the upper boundary for the whole volume.) Hope that makes sense.
    I don't know if you've got Matlab or have ever used it, but it has a decent root-finder. If you do not have Matlab, there's a free alternative called Octave. I strongly recommend it to any and all.
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    One does have to confess that raising to the 109th power is not trivial.
    tk421 likes this.
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  26. #25  
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    Quote Originally Posted by Schneibster View Post
    Quote Originally Posted by MagiMaster View Post
    If you want to program it, use Newton's method. There are other ways to do the same thing faster, but Newton's method is simple. Just rewrite your equation in the form f(x) = 0, then plug it in to the formula given on the Wiki site. Just be aware that fractional powers often give imaginary results, so you may need a complex number class.
    Newton's Method is nonterminating.
    Actually, in this case, where you have 5 significant figures in the least accurate input, there's no point in calculating the root out beyond 5 figures, which should take less than 10 iterations. It'd probably only take 2 or 3 more to hit the 12 figures of the most accurate input. After that, any more digits you get are worthless. And in general, you'll fill out the significant figures of a floating point value pretty quickly.

    As for it's chaotic nature, that's true in general, but not in every case.

    I'm not saying there's aren't better methods, but you should try Newton's method first just because it's so simple. (And any numerical root finder will be non-terminating.)
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    Quote Originally Posted by MagiMaster View Post
    Quote Originally Posted by Schneibster View Post
    Quote Originally Posted by MagiMaster View Post
    If you want to program it, use Newton's method. There are other ways to do the same thing faster, but Newton's method is simple. Just rewrite your equation in the form f(x) = 0, then plug it in to the formula given on the Wiki site. Just be aware that fractional powers often give imaginary results, so you may need a complex number class.
    Newton's Method is nonterminating.
    Actually, in this case, where you have 5 significant figures in the least accurate input, there's no point in calculating the root out beyond 5 figures, which should take less than 10 iterations. It'd probably only take 2 or 3 more to hit the 12 figures of the most accurate input. After that, any more digits you get are worthless. And in general, you'll fill out the significant figures of a floating point value pretty quickly.

    As for it's chaotic nature, that's true in general, but not in every case.

    I'm not saying there's aren't better methods, but you should try Newton's method first just because it's so simple. (And any numerical root finder will be non-terminating.)
    Heh, absolutely, you're correct, and that's close enough for most work. I once upon a time bought a calculator that could numerically calculate a definite integral; it was the newest hottest thing.

    I bought a graphing calculator six or eight years ago that blows it off the planet.

    I never bothered with any of the computer solutions; my TI does it fine. If I did the stuff for a living I might drill deeper.
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  28. #27  
    Forum Sophomore ChaosD.Ace's Avatar
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    Thanks a lot guys. I'll get cracking.
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