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Couple of Questions on proof

  1. Aug 3, 2004 #1


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    I'm writing up my maths notes to the computer, while looking over maths notes for series I spotted another way of proving the sum of natural numbers to n (to what I've been taught).

    By definition:

    [tex]\sum_{k=1}^n k = n + (n-1) + (n-2) + \ldots + (n - [n-2]) + (n - [n-1])[/tex]

    Separating terms we get:

    [tex]\sum_{k=1}^n k = (n + n + \ldots + n) - [1 + 2 + 3 + \ldots + (n-2) + (n-1)][/tex]

    [tex]\sum_{k=1}^n k = n^2 - \sum_{k=1}^{n-1} k[/tex]

    [tex]\sum_{k=1}^n k = n^2 - \left( \left[ \sum_{k=1}^{n} k \right] - n \right)[/tex]

    [tex]2\sum_{k=1}^n k = n^2 + n[/tex]

    [tex]\sum_{k=1}^n k = \frac{n(n+1)}{2}[/tex]

    So is this sufficient proof please?

    Furthermore I was trying to prove from first principles that:

    [tex]\frac{d}{dx} \left( x^n \right) = nx^{n-1}[/tex]

    For all real values of n, but so far I've only been able to prove it for all positive integers using binomial theorem. Could someone give me a clue at least please?
    Last edited: Aug 3, 2004
  2. jcsd
  3. Aug 3, 2004 #2
    1) proof looks surprisingly complicated (not that I knew a better one, though) but I can´t see a problem with it (except the "by definition: ..." because you already transformed the "real" definition 1+2+...+n).

    2) Had to dig out my old math book to see if it offers another way than the one I thought of spontaneously but sadly it didn´t:
    x^n = exp(n * ln(x))
    so with this apprach you need to have the chain-rule and the derivative of the logarithm. I wouldn´t even count on another method since spontaneously I couldn´t think of any other definition of x^n for n being real than above.
  4. Aug 3, 2004 #3
    This is how I was always taught that the deriv. of x^n=nx^(n-1) for n an element of R. Let y=x^n. ln|y|=ln|x^n|=ln(|x|^n)=nln|x|. Differentiating:

    y'/y=n/x so y'=ny/x=n(x^n)/x=nx^(n-1).
  5. Aug 3, 2004 #4


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    Thanks very much :biggrin:

    I suppose then the proof for the chain rule or the product rule is actually going to be very complex :confused: ?
  6. Aug 3, 2004 #5


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    Well, some tricks to keep in mind are that 1 = x^n * x^(-n) and x^r = (x^(r/s))^s
  7. Aug 3, 2004 #6
    That was a nice proof.
  8. Aug 3, 2004 #7


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    Proof of the product rule is not too difficult and just involves some simple algebra, you just have to add the right form of "0" and break the limit up accordingly. You'll also need the fact that a differentiable function is continuous.

    The chain rule is a similar idea, but you multiply by the right form of "1", and break up the limit accordingly. You'll also need to fall back on the continuity of both your functions.

    You should give them a try on your own, and ask for hints if you're stuck if you like. Or you can find them here:

  9. Aug 3, 2004 #8
    [tex]\sum_{k=1}^n k[/tex]

    Hmm... this might be the same. :eek:

    Gauss said (when he was 9 or 10) to:

    Write the sum S twice, once in the usual order and once in reverse order:

    [tex] S = 1 + 2 + 3 + \ldots + (n - 1) + n [/tex]
    [tex] S = n + (n - 1) + (n - 2) + \ldots + 2 + 1 [/tex]

    Then add columns vertically, getting:

    [tex] 2S = (n + 1) + (n + 1) + (n + 1) + \ldots + (n + 1) + (n + 1)[/tex]

    On the right side there are n terms, each of which is [tex]n + 1[/tex], so

    [tex]2S = n(n + 1)[/tex] or [tex]S = \frac{n(n+1)}{2}[/tex]

    hmm... (from Stewart's calculus book)
    Last edited: Aug 3, 2004
  10. Aug 4, 2004 #9


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    The chain rule or product rule isn't difficult to prove, but proving that the derivative of ln|x| is 1/x is.

    If n is a rational number, you can write it as [itex]\frac{p}{q}[/itex], so
    you can differentiate [itex]x^n[/itex] by using the product/chain rule on
    You`ll have to prove
    [tex]\frac{d}{dx} \left( x^n \right) = nx^{n-1}[/tex]
    holds for n=1/q though.
    When n is irrational you'll have to use a limiting process, but that's okay, because that's the way x^n is defined when n is irrational.
  11. Aug 4, 2004 #10


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    Thanks all :smile:

    Oh and gazzo I knew that proof thanks but I just prefered the one I posted so I just wanted to check it was o.k.
  12. Aug 4, 2004 #11
    hehe :blush:
  13. Aug 4, 2004 #12
    for lnx that's easiest of them all

    let y = ln x

    then e^y = x and differentiate implicitly

    dy/dx E^y = 1

    so dy/dx = 1/E^y

    but E^y = x

    so therefore dy/dx of ln x = 1/x

    looks long on here cause i don't know latex but from calc I remember this was the easiest thing i had ever seen.
  14. Aug 4, 2004 #13


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    Yeah that's possible, but I was thinking of proving it without resort to the derivative of e^x.

    The derivative of [itex]a^x[/itex] is
    [tex]a^x\lim_{h\rightarrow 0} \frac{a^h-1}{h}[/tex]
    and then you could define
    [tex]\ln(a)=\lim_{h\rightarrow 0} \frac{a^h-1}{h}[/tex]
    and define e by:
    [tex]\lim_{h\rightarrow 0} \frac{e^h-1}{h}=1[/tex]

    But then you wouldn't be introducing the logarithm as the inverse of the exponential.
    Last edited: Aug 4, 2004
  15. Aug 4, 2004 #14
    true , how do you prove without resorting to e^x i haven't seen it done.
  16. Aug 4, 2004 #15


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    If you want to differentiate ln(x) without using ex, the simplest way is to go back and define ln(x) as "integral from 1 to x of (1/t)dt". That makes derivative (including the fact that ln(x) is differentiable) obvious. Also all properties of ln(x) can be derived from that.
  17. Aug 4, 2004 #16
    true that makes sense now, but didin't it take years for anyone to see that when calc was being assembled shall I say.
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