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Looking for an expression for radicals

  1. Aug 9, 2014 #1
    I am trying to find a general expression for radicals, for example,

    √k=f(k)

    Does anyone know of any?

    Thanks!
     
  2. jcsd
  3. Aug 9, 2014 #2

    HallsofIvy

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    I don't know what you mean by a "general expression". The square root, that you have there, can be written [itex]\sqrt{k}= k^{1/2}[/itex]. A general radical, the "nth root", can be written [tex]\sqrt[n]{k}= k^{1/n}[/tex]. Is that what you mean?
     
  4. Aug 9, 2014 #3
    No problem, let me clear it up
    Something along the lines of,

    √5 = 2(phi)-1

    Except a general form where,

    √k = some function of 'phi', or 'k' or both (or perhaps something completely different.)
     
  5. Aug 9, 2014 #4

    disregardthat

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    By phi, I assume you mean [itex]\frac{1+\sqrt{5}}{2}[/itex], so what you've done is just rewriting [itex]\sqrt{5}[/itex] in terms of this. I don't see any general about this. What exactly is it what you want, when you say you want a general expression?
     
  6. Aug 9, 2014 #5
    Okay, there is still some confusion, let me put it this way,

    Are there expressions for say, √3=? or √7=? or √11=? etc. etc.
    And if so do we have a general form where √x=some expression of 'x'

    I was using √5=2phi-1 simply as an example (and the only radical with an equivalent expression I am aware of that is known).
     
  7. Aug 9, 2014 #6

    disregardthat

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    [itex]\sqrt{x}[/itex] is an expression of x. It is also unclear why you would accept [itex]\sqrt{5} = 2 \cdot \frac{1+\sqrt{5}}{2} -1[/itex] as an 'expression' of [itex]\sqrt{5}[/itex]. Why not accept [itex]\sqrt{x} = 2 \cdot \frac{1+\sqrt{x}}{2} -1[/itex] as an 'expression' of x?
     
  8. Aug 9, 2014 #7
    It is correct but [itex]\sqrt{x} = 2 \cdot \frac{1+\sqrt{x}}{2} -1[/itex] is nothing more than √x=√x.

    Do you know of a form like √5=2phi-1 for say √3=?
     
  9. Aug 9, 2014 #8

    disregardthat

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    :confused: √5=2phi-1 is also nothing but √5=√5.

    I think you need to explain more clearly what you mean by a 'general expression'.
     
  10. Aug 9, 2014 #9
    Sorry about that, let's try it this way, can you write √3 as an expression with phi?
     
    Last edited: Aug 9, 2014
  11. Aug 9, 2014 #10

    disregardthat

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    Without using √ you mean? And why would you want that? And what is an expression with phi? Is √3 = √3 + phi - phi an expression with phi? Just explain what you are after in a clear and precise manner, it's no fun guessing.
     
  12. Aug 9, 2014 #11
    √3 + phi - phi is certainly an expression for √3 but not what I was looking for

    I apologize for not being more clear, please forgive my inability to put together the proper mathematical vernacular. It's okay though, it appears my question has been answered. Thank you for helping out!
     
  13. Aug 9, 2014 #12
    Hi mesa,

    You can approximate the square root of a number using this formula:
    $$
    f(x)=\sqrt{x}\approx1+\frac{1}{2}(x-1)+(-\frac{1}{4})(x-1)^2+\frac{3}{8}(x-1)^3+...=\sum_{n=0}^{...}f_n(x-1)^n
    $$
    where the coefficients ##f_n## can be computed by
    $$
    f_n=\frac{1}{n!}\prod_{m=0}^{n-1}(\frac{1}{2}-m)
    $$
    I haven't checked what values of x this is valid for though
     
  14. Aug 9, 2014 #13
    I stand corrected, that is a very nice approximation! I wrote a continued fraction but it is ugly :P
    Any chance you know who put this together?
     
  15. Aug 9, 2014 #14
    Yes. A guy named Taylor.

    Note that the coefficients are a bit off. It should be something like $$1+\frac{1}{2} (x-1)-\frac{1}{8} (x-1)^2+\frac{1}{16} (x-1)^3+\cdots.$$
     
  16. Aug 9, 2014 #15
    It's a Taylor Series expansion of ##f(x)=x^{1/2}## around the point ##x=1##. Originally I did it for ##f(x)=x^m## for arbitrary m. For that one, the expansion is this: $$
    f(x)=x^m=\sum_{n=0}^{\infty}f_n(x-1)^n$$ where ##f_0=1## and for ##n\neq 0## $$f_n=\frac{1}{n!}\prod_{k=0}^{n-1}(m-k)$$ So
    $$f(x)=1+\sum_{n=1}^{\infty}\frac{1}{n!}\Bigg(\prod_{k=0}^{n-1}(m-k)\Bigg)(x-1)^n$$
    Kind of a messy formula lol
     
  17. Aug 9, 2014 #16
    I should really look into Taylor series more... The schools engineering program hardly touches on it :P

    Anyway, I don't think it is all that messy, thanks for sharing! (it is always good to know the 'gaps' in order to correct them)
     
  18. Aug 10, 2014 #17

    Mark44

    Staff: Mentor

    If the engineering curriculum includes calculus, it's a safe bet that Taylor's series (and Maclaurin series) will be covered pretty extensively, usually after the course where integration is taught.
     
  19. Aug 10, 2014 #18
    We did cover Taylor series but it was very brief (not even on an exam).
     
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