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    The elliptic integrals of the first and second type are defined as:

    \displaystyle K(k) = \int_0^{\pi/2} \frac{1}{\sqrt{1-k^2 \sin^2 \theta}}\,d\theta

    \displaystyle E(k) = \int_0^{\pi/2} \sqrt{1-k^2 \sin^2 \theta}\,d\theta

    (For some reason, E(k) is called an integral of the 2nd type, even though it looks the more "fundamental" integral).

    The problem is to show (by differentiating under the integral sign) that

    \displaystyle k\frac{dE}{dk} = E(k)-K(k)

    and

    \displaystyle k\frac{dK}{dk} = \frac{E(k)}{1-k^2} -K(k)

    The first relation isn't too bad, but the second was a lot tougher and I found I needed a little help from Mathematica en route.

    The only web references I've found seem to imply these relations aren't too tricky (or at least they seem to think they follow easily from the definitions), but I'm not seeing it.

    This is part of proving a particularlt efficient formula for calculating \pi, in case you're wondering. (There results are needed to prove the Legendre relation for Elliptic Integrals, if that means anything to anyone).
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    huh??? where's the difficulty?
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    (Original post by chewwy)
    huh??? where's the difficulty?
    Don't suppose you'd put me out of my misery and post your working then?
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    (Original post by DFranklin)
    Don't suppose you'd put me out of my misery and post your working then?
    I'm curious too, the first one falls out quickly, but the second one seems to cause me nasty algebra
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    (Original post by nota bene)
    I'm curious too, the first one falls out quickly, but the second one seems to cause me nasty algebra
    If it helps, I think the 2nd result does actually depend on the limits (i.e. it wouldn't be true if they weren't 0 and pi/2), unlike the 1st one where I didn't actually use the limits at all.
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    In the absence of any suggestions, here's what I did for the 2nd integral:

    Write s=\sin \theta, c = \cos \theta.

    \displaystyle K(k) = \int_0^{\pi/2} \frac{1}{\sqrt{1-k^2s^2}}\,d\theta

    \displaystyle \frac{dK}{dk} = \int_0^{\pi/2} \frac{ks^2}{(1-k^2s^2)^{3/2}}\,d\theta

    \displaystyle k\frac{dK}{dk} = \int_0^{\pi/2} \frac{k^2s^2}{(1-k^2s^2)^{3/2}}\,d\theta

    \displaystyle k\frac{dK}{dk} = \int_0^{\pi/2} \frac{(k^2s^2-1)+1}{(1-k^2s^2)^{3/2}}\,d\theta

    \displaystyle k\frac{dK}{dk} = -\int_0^{\pi/2} \frac{1-k^2s^2}{(1-k^2s^2)^{3/2}} + \int_0^{\pi/2} \frac{1}{(1-k^2s^2)^{3/2}}\,d\theta

    \displaystyle k\frac{dK}{dk} = -K + \int_0^{\pi/2} \frac{1}{(1-k^2s^2)^{3/2}}\,d\theta

    So it is sufficient to show \displaystyle \int_0^{\pi/2} \frac{1}{(1-k^2s^2)^{3/2}}\,d\theta = \frac{E(k)}{1-k^2}, or equivalently,

    \displaystyle \int_0^{\pi/2} \frac{1-k^2}{(1-k^2s^2)^{3/2}}\,d\theta = \int_0^{\pi/2} \sqrt{1-k^2s^2} \,d\theta

    Now \displaystyle \frac{1-k^2}{(1-k^2s^2)^{3/2}} = \frac{(1-k^2)\sqrt{1-k^2s^2}}{(1-k^2s^2)^2} = \frac{1-k^2}{1-2k^2s^2+k^4s^4}\sqrt{1-k^2s^2}

    \displaystyle = \frac{1-k^2  -2k^2s^2+k^4s^4+2k^2s^2-k^4s^4}{1-2k^2s^2+k^4s^4}\sqrt{1-k^2s^2}

    \displaystyle = \frac{1-2k^2s^2+k^4s^4+k^2(2s^2-1)-k^4s^4}{1-2k^2s^2+k^4s^4}\sqrt{1-k^2s^2}

    \displaystyle = \sqrt{1-k^2s^2} +k^2\frac{2s^2-1-k^2s^4}{1-2k^2s^2+k^4s^4}\sqrt{1-k^2s^2}

    \displaystyle = \sqrt{1-k^2s^2} +k^2\frac{2s^2-1-k^2s^4}{(1-k^2s^2)^{3/2}}

    So \displaystyle \int_0^{\pi/2} \frac{1-k^2}{(1-k^2s^2)^{3/2}}\,d\theta = \int_0^{\pi/2} \sqrt{1-k^2s^2} \,d\theta + k^2 \int_0^{\pi/2}\frac{2s^2-1-k^2s^4}{(1-k^2s^2)^{3/2}} \,d\theta

    So we now only(!) need to show that \displaystyle \int_0^{\pi/2}\frac{2s^2-1-k^2s^4}{(1-k^2s^2)^{3/2}} \,d\theta = 0

    This was where a little experimenting with Mathematica lead me to look at

    \displaystyle \frac{d}{d\theta} \frac{sc}{\sqrt{1-k^2s^2}} = \frac{k^2s^2c^2}{(1-k^2s^2)^{3/2}} + \frac{c^2-s^2}{\sqrt{1-k^2s^2}}

    \displaystyle =\frac{k^2s^2c^2+(1-k^2s^2)(c^2-s^2)}{(1-k^2s^2)^{3/2}}

    \displaystyle = \frac{k^2s^2c^2+c^2-s^2-k^2s^2c^2+k^2s^4}{(1-k^2s^2)^{3/2}}

    \displaystyle = \frac{c^2-s^2+k^2s^4}{(1-k^2s^2)^{3/2}}

    \displaystyle = \frac{1-2s^2+k^2s^4}{(1-k^2s^2)^{3/2}}

    Thus \displaystyle \int_0^{\pi/2}\frac{2s^2-1-k^2s^4}{(1-k^2s^2)^{3/2}} \,d\theta = \left[ -\frac{sc}{\sqrt{1-k^2s^2}}\right]_0^{\pi/2}\right] = 0.

    So, that proves the result, but it's hardly pretty, and without the Mathematica inspired guess I don't think I'd have got that integral out.

    [Waits for someone to point out some obvious substitution that makes it all come out in 2 lines...]
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    In the absence of other posters I think you can take it that this is a pretty non-trivial exercise

    I have tucked away at home a copy of Whitaker & Watson that I bought some years ago as an investment (although I'll probably never get chance to read it until I'm retired, I like to take it out occasionally and admire the absurd level of difficulty of some of the early 1900s Tripos questions )
    Anyway, their "proof" (by which I mean a 2-line statement along the lines of here's a relation, here's another relation, hence the result) involves introducing the functions dn and sd within the integral and then referring to previously quoted properties of these elliptic functions!

    I also got chance to flick through an old Arfken and Weber Methods book last night. They have this problem set as an exercise, but after reducing the problem to showing that
    E(k)/(1-k^2) = the horrible integral with a 3/2 power in the denominator, their "hint" is to expand both sides as power series and compare coefficients

    I haven't tried to follow through your integration fully, but you've certainly produced a more explicit demonstration than I've been able to find either in the literature or on the web (* waits for a Greek mathematician to turn up to point out how easy it really is *)
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    (Original post by davros)
    In the absence of other posters I think you can take it that this is a pretty non-trivial exercise
    I found one paper on the web that purports to "prove" the Legendre relation. For these two relations (which are needed to prove the Legendre one), it just says "Proof: This follows directly from the definitions". :mad:

    I have tucked away at home a copy of Whitaker & Watson that I bought some years ago as an investment (although I'll probably never get chance to read it until I'm retired, I like to take it out occasionally and admire the absurd level of difficulty of some of the early 1900s Tripos questions )
    Ah, but that was in the old "Classical" Tripos days, when there was still a Greek influence on the questions...

    It would actually be really interesting to see how tripos questions have changed over the years - I recall seeing some quite old questions and thinking they weren't too bad, but perhaps my memory is playing tricks (i.e. 1970 seemed old at the time...)
 
 
 
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