True? $(1-x ) E\left(\frac{-4x }{ (1 - x)^2 }\right) = 2E(x^2)+(x^2-1)K(x^2)=(1+x )E\left(\frac{4x}{(1 + x)^2 }\right)$

Question

Note on Terminology

$\operatorname{K}(m)$ and $\operatorname{E}(m)$ are Complete Elliptic Integral functions of, respectively, the 1st and 2nd kind. Here I use the same convention as Wolfram Alpha in defining the parameter $m = k^2$ such that... $$\operatorname{K}(m) =\int_0^{\pi/2} \frac{\text{d}\theta}{\sqrt{1-m\sin^2\theta}} =\frac{\pi}{2}\sum_{n=0}^\infty \left( \frac{(2n)!}{2^{2n}(n!)^2}\right)^2m^n, $$ and $$\operatorname{E}(m) =\int_0^{\pi/2} \sqrt{1-m\sin^2\theta}~\text{d}\theta =\frac{\pi}{2}\sum_{n=0}^\infty \left( \frac{(2n)!}{2^{2n}(n!)^2}\right)^2\frac{m^n}{1-2n}. $$

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Context

In a recent question about the force exerted by a circular ring on an external co-planar point three solutions were presented. All three solutions appear to give the same result (checked by numerical calculations over a range of $0<=x<=1$).

The Gauss Transformation, applied to $\operatorname{K}()$ (Complete Elliptic Integrals of the 1st kind) functions, goes some way to explaining the equivalence of the three solutions.

$$ \frac{1}{\left(1-x\right)}\operatorname{K}\left(\frac{-4x }{ \left(1 - x\right)^2 }\right) = \operatorname{K}\left( \left(-x\right)^2 \right) = \operatorname{K}\left( \left(+x\right)^2 \right) = \frac{1}{\left(1+x\right)}\operatorname{K}\left(\frac{+4x }{ \left(1 + x\right)^2 }\right). $$

But it also seems that there is a similar identity involving $\operatorname{E}()$ (Complete Elliptic Integrals of the 2nd kind) functions. This identity also involves the $\operatorname{K}()$ function.

$$ {f(x):~} (1-x ) ~\operatorname{E}~\left(\frac{-4x }{ \left(1 - x\right)^2 }\right)$$ $$= {g(x):~} 2\operatorname{E}\left(x^2\right)+(x^2-1)\operatorname{K}\left(x^2\right) $$ $$= {h(x):~} (1+x ) ~\operatorname{E}~\left(\frac{+4x }{ \left(1 + x\right)^2 }\right). $$ [Equation $g(x)$ was corrected to replace $-(1+x)$ by $+(x^2-1)$ as multiplier of $K(x^2)$].

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Update 1

Using WolframAlpha gives for the Taylor Series expansion (at $x=0$) of the middle expression (denoted $g$ by /u/Claude Leibovici) :-

$$ g = 2\operatorname{E}(x^2)-(1-x^2)\operatorname{K}(x^2) = \frac{\pi }{2}+\frac{\pi x^2}{8}+\frac{\pi x^4}{128} +\frac{\pi x^6}{512} +\frac{25\pi x^8}{32678} +O\left(x^9\right) $$

which agrees (as far as the actual terms displayed) with the Taylor Expansions of the other two expression ($f$ and $h$) presented in the answer by /u/Claude Leibovici.

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Update 2

The following SEMaths question (by /u/jnm2) and answer (by /u/J.M.is a poor mathematician) Identity for complete elliptic integral of the second kind addresses one of the identities which I have asked about ($f$ = $h$):- $$ |1-x | ~\operatorname{E}~\left(\frac{-4x }{ \left(1 - x\right)^2 }\right) = 2\operatorname{E}\left(x^2\right)+(x^2-1)\operatorname{K}\left(x^2\right) = |1+x | ~\operatorname{E}~\left(\frac{+4x }{ \left(1 + x\right)^2 }\right). $$ Note the terms $|1+x|$ and $|1-x|$ are absolutes.

Quote from that answer:

"...the complete elliptic integral of the second kind satisfies the imaginary modulus identity ... specialized here to the complete case, $\phi=\pi/2$:

$$E(-m)=\sqrt{1+m}\,E\left(\frac{m}{1+m}\right)"$$

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Oustanding Question

[Upated following learning that $f(x)=h(x)$ is demonstrated elsewhere].

Show that the conjectured identity $ g(x) = h(x) $ ... $$ 2\operatorname{E}\left(x^2\right)+(x^2-1)\operatorname{K}\left(x^2\right) = {h:~} (1+x ) ~\operatorname{E}~\left(\frac{+4x }{ \left(1 + x\right)^2 }\right), $$

( or its equivalent $g(x) = f(x)$) $$ 2\operatorname{E}\left(x^2\right)+(x^2-1)\operatorname{K}\left(x^2\right) = {h:~} (1-x ) ~\operatorname{E}~\left(\frac{-4x }{ \left(1 - x\right)^2 }\right). $$ is true.

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Update 3

User /u/user reports that the $g(x)$ series in powers of $x$ is also given by the formula... $$g(x) = \frac{\pi}{2}\sum_{n=0}^\infty \left( \frac{(2n)!}{2^{2n}(n!)^2}\frac{}{2n-1}\right)^2x^{2n}. $$

Update 4

Below is a table of some initial power series terms for each term in $h(x)$ and $g(x)$. (NB The arguments in $E(k)$ and $K(k)$ conform to "wikipedia convention" rather than "WolframAlpha convention" which would be $E(m)$ and $K(m)$ where $m=k^2$).

Note how the odd-power terms in $h(x)$ cancel to zero and how $g(x)$ does not have any odd-power terms. For every power term, the sums of $h(x)$ and $g(x)$ are equal.

(Links to Wolfram Alpha expansions: $h(x)_1$, $h(x)_2$, $g(x)_1$, $g(x)_2$, $g(x)_3$.)

\begin{array}{|Term|c|c|c|c|c|c|c|c|c|c|c|c|} \hline h(x)_1 & +1E(\gamma_x) & \frac{1\pi x^0}{2}& \frac{-1\pi x^1}{2}& \frac{+5\pi x^2}{8}& \frac{-5\pi x^3}{8}& \frac{+81\pi x^4}{128}& \frac{-81\pi x^5}{128}& \frac{+325\pi x^6}{512}& \frac{-325\pi x^7}{512} & \frac{+20825\pi x^8}{32768} & \frac{-20825\pi x^9}{32768} & \frac{+83349\pi x^{10}}{131072} & \\ \hline h(x)_2 & +xE(\gamma_x) & & \frac{+1\pi x^1}{2}& \frac{-1\pi x^2}{2}& \frac{+5\pi x^3}{8}& \frac{-5\pi x^4}{8}& \frac{+81\pi x^5}{128}& \frac{-81\pi x^6}{128}& \frac{+325\pi x^7}{512} & \frac{-325\pi x^8}{512} & \frac{+20825\pi x^9}{32768} & \frac{-20825\pi x^{10}}{32768} & \\ \hline h(x) & SUM & \frac{1\pi x^0}{2}& 0 & \frac{+1\pi x^2}{8}& 0 & \frac{+1\pi x^4}{128}& 0 & \frac{+1\pi x^6}{512}& 0 & \frac{+25\pi x^8}{32768} & 0 & \frac{+49\pi x^{10}}{131072} & \\ \hline g(x)_1 & -1K(x) & \frac{-1\pi x^0}{2}& 0 & \frac{-1\pi x^2}{8}& 0 & \frac{-9\pi x^4}{128}& 0 & \frac{-25\pi x^6}{512}& 0 & \frac{-1225\pi x^8}{32768}& 0 & \frac{-3969\pi x^{10}}{131072}& \\ \hline g(x)_2 & +x^2 K(x) & 0 & 0 & \frac{+1\pi x^2}{2}& 0 & \frac{+1\pi x^4}{8}& 0 & \frac{+9\pi x^6}{128}& 0 & \frac{+25\pi x^8}{512}& 0 & \frac{+1225\pi x^{10}}{32768}& \\ \hline g(x)_3 & +2 E(x) & \frac{2\pi x^0}{2}& 0 & \frac{-2\pi x^2}{8}& 0 & \frac{-6\pi x^4}{128}& 0 & \frac{-10\pi x^6}{512}& 0 & \frac{-350\pi x^8}{32768}& 0 & \frac{-882\pi x^{10}}{131072}& \\ \hline g(x) & SUM & \frac{1\pi x^0}{2}& 0 & \frac{+1\pi x^2}{8}& 0 & \frac{+1\pi x^4}{128}& 0 & \frac{+1\pi x^6}{512}& 0 & \frac{+25\pi x^8}{32768}& 0 & \frac{+49\pi x^{10}}{131072}& \\ \hline \end{array}

Answer 1

To me, there is a problem somewhere with the term in the middle.

Considering $$f=(1-x) E\left(-\frac{4 x}{(1-x)^2}\right)$$ $$g=2 E\left(x^2\right)-(x+1) K\left(x^2\right)$$ $$h=(1+x) E\left(\frac{4 x}{(1+x)^2}\right)$$

Being lazy, I just computed the Taylor series $$f=\frac{\pi }{2}+\frac{\pi x^2}{8}+\frac{\pi x^4}{128}+\frac{\pi x^6}{512}+O\left(x^7\right)$$ $$g=\frac{\pi }{2}-\frac{\pi x}{2}-\frac{3 \pi x^2}{8}-\frac{\pi x^3}{8}-\frac{15 \pi x^4}{128}-\frac{9 \pi x^5}{128}-\frac{35 \pi x^6}{512}+O\left(x^7\right)$$ $$h=\frac{\pi }{2}+\frac{\pi x^2}{8}+\frac{\pi x^4}{128}+\frac{\pi x^6}{512}+O\left(x^7\right)$$ So $f=h\neq g$.

Edit

The above was written before the edit of the post.

If $$g=2E(x^2)-(1-x^2)K(x^2)$$ the story is totally different since the expansion gives $$g=\frac{\pi }{2}+\frac{\pi x^2}{8}+\frac{\pi x^4}{128}+\frac{\pi x^6}{512}+O\left(x^7\right)$$ making the statement to be true.

Answer 2

Notes

1. This is a slowly-developing answer, still incomplete.

2. For this answer I will use the "wikipedia" argument convention $\operatorname{K}(k)$ rather than the "Wolfram Alpha" convention $\operatorname{K}(m=k^2).$

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Approach

The expressions $f(x)$, $g(x)$ and $h(x)$ all output an infinite series of additive terms in powers of $x$.

For each expression I will attempt to define a formula for the numeric coefficient $C_n$ of the term containing $x^n$. Then if all formulae give the same value for $C_n$ the identity: $f(x)=g(x)=h(x)$ will be proven.

For expression $g(x)$ all the arguments are in terms of $x$, whereas $f(x)$ and $h(x)$ both have argument $\gamma= \sqrt\frac{4x}{((1+x)^2)}$ .

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Coefficients of Expression $g(x)$

For $g(x)$, in order to obtain a formula for the numerical coefficient of each $x$-power term, it is relatively straightforward. We use the power series formulae for $\operatorname{K}$ and $\operatorname{E}$.

Consider the following expression of $g(x)$ :- $$ g(x) = -\operatorname{K}(x) ~+ x^2* \operatorname{K}(x) ~+2\operatorname{E}(x) $$ and defining $P_n$ as the Legendre Polynomial $P_{2n}(0)$ expressed as ... $$ P_n =\left( \frac{(2n)!} {4^n ~n!~n!} \right). $$ We then have $$ \frac{2}{\pi}g(x) = -\sum_{n=0}^{\infty} P^2_n x^{2n} +x^2\sum_{n=0}^{\infty} P^2_n x^{2n} +2\sum_{n=0}^{\infty} P^2_n \frac{1}{1-2n}x^{2n} $$

$$ \frac{2}{\pi}g(x) = -\sum_{n=0}^{\infty} P^2_n x^{2n} +\sum_{n=0}^{\infty} P^2_n x^{2n+2} +\sum_{n=0}^{\infty} 2 P^2_n \frac{1}{1-2n}x^{2n} $$ It can be seen that the power series will only contain even powers of $x$:$(0,2,4,6...)$.

We can deduce that for a given term $C_r.x^r$ with $r$ being the particular power of $x$, the coefficient $C_r$ from $g(x)$, symbolized as $C_{r(g)}$ will be given by the sum of three contributions:-

$$C_{r(g)} = \frac{\pi}{2} \left[ - P^2_{(n=r/2)} + P^2_{(n=(r-2)/2)} + 2\frac{P^2_{(n=r/2)}}{1-2{(n=r/2)}}\right]$$

$$C_{r(g)} = \frac{\pi}{2} \left[ - P^2_{(r/2)} + P^2_{(r/2 -1)} + 2\frac{P^2_{(r/2)}}{1-r}\right]$$

This applies for $r=2,4,6,8,...$. For $r=0$ there is no contribution from the middle term:- $x^2K(x)$.

Initial terms calculated from this formula are:- $$ \pi \left( 1+\frac{1x^2}{8} +\frac{1x^4}{128} +\frac{1x^6}{512} +\frac{25x^8}{32768} +\frac{49x^{10}}{131072} \text{...}\right) $$

This agrees with the initial terms calculated (by Wolfram Alpha) for $h(x)$ and $g(x)$.

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User's Hint

$$g(x)_{user} = \frac{\pi}{2} \sum^{\infty}_{n=0} \left( \binom{2n}{n} * \frac{1}{4^n} * \frac{1}{(2n-1)} \right)^2 x^{2n} $$

For a required power $R$ of $x$ the coefficient $C_{r(u)}$ can be calculated quite simply. First we note that there are no odd powers of $x$ since the power term is $2n$ and $n$ is an integer $(0,1,2,3...)$.

The value of $C_{r(u)}$ is given by:- $$ C_{r(u)} = \frac{\pi}{2} \left( \binom{r}{r/2} * \frac{1}{4^{r/2}} * \frac{1}{r-1} \right)^2 $$

The initial terms of $g(x)$ calculated (by excel) with this formula are $$ \pi \left( 1+\frac{1x^2}{8} +\frac{1x^4}{128} +\frac{1x^6}{512} +\frac{25x^8}{32768} +\frac{49x^{10}}{131072} \text{...}\right) $$

This agrees with the initial terms calculated (by Wolfram Alpha) for $h(x)$ and $g(x)$.

Note: I cannot apply the User hint formula in a proof because (so far) I have not been able to derive it from the initial formulae for $f(x)$, $g(x)$ or $h(x)$.

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Remaining tasks

(1) obtain a similar formula for the coefficients of $h(x)$ (or $f(x)$);

(2) show that for any value of $n>0$ (because these formulae dont work for $n=0$) the coefficient formulae give equal values of $C_n$;

(3) show that the coefficients for $n=0$ for expressions $g()$ and $h()$ (or $f()$) have the same values .

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NOTE Need to insert factor $\pi /2$ as appropriate below

Coefficients of Expression $h(x)$

This will be more difficult than for $g(x)$ because the argument ($\gamma=\sqrt{\frac{4x}{(1+x)^2}}$) of the complete elliptic integral ($\operatorname{E}$) expands to a more complicated expression...

$$ h(x) = (1+x) \operatorname{E}(\gamma) = (1+x) \sum_{n=0}^\infty \frac{{P_n}^2}{1-2n} \gamma^{2n} = (1+x) \sum_{n=0}^\infty \frac{{P_n}^2}{1-2n} \left( \frac{4x}{(1+x)^2}\right)^n $$

$$ h(x) = (1+x) \sum_{n=0}^\infty \frac{{P_n}^2}{1-2n} (4x)^n(1+x)^{-2n} $$

$$ h(x) = (1+x) \sum_{n=0}^\infty \frac{{P_n}^2}{1-2n} (4x)^n(1-2x+3x^2-4x^3 ...)^{n} $$

$$ h(x) = (1+x) \sum_{n=0}^\infty \frac{{P_n}^2}{1-2n} (4)^n(x-2x^2+3x^3-4x^4 ...)^{n} $$

$$ h(x) = \sum_{n=0}^\infty \frac{{P_n}^2}{1-2n} (4)^n(x-2x^2+3x^3-4x^4 ...)^{n} + \sum_{n=0}^\infty \frac{{P_n}^2}{1-2n} (4)^n(x-2x^2+3x^3-4x^4 ...)^{n} (x) $$

Actually this is not the best way to go about things.

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A formula is available - the Binomial Series Expansion which applies for all fixed real $a$ and for all $x$ in the interval $-1<x<+1$ (UEOM: Universal Encyclopedia of Mathematics, page 574) for obtaining the coefficients of the different $x$-powers for expressions of the form $(1+x)^{n/m}$ namely $$(1+x)^{n/m}= 1 + \frac{n}{m}x^1 - \frac{n(m-n)}{2!m^2} x^2+ \frac{n(m-n)(2m-n)}{3!m^3} x^3 - ... + (-1)^{k+1} \frac{n(m-n)(2m-n)...[(k-1)m - n]}{k!m^k}x^k + ...$$ In the present problem the fractional form can be replaced by the simpler integer form (UEOM, page 69): $$ (1+x)^{\alpha} = 1 + \binom{\alpha}{1}x+ \binom{\alpha}{2}x^2+ \binom{\alpha}{3}x^3 + ... = \sum_{k=0}^{\infty} \binom{\alpha}{k}x^k $$ Where the coefficients are calculated from this formula: $$ \binom{\alpha}{k} = \frac{\alpha(\alpha-1)(\alpha-2)...(\alpha-k+1)}{k!} $$

Note that in the present problem the Binomial Theorem (UEOM, page 69) does NOT apply in general because the exponent $\alpha = -2n$ is usually $<-1$

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For any chosen power $i$ of $x$ we can apply the Binomial Series formula to our denominator factor $(1+x)^{-2}$ to obtain a coefficient which contributes to the coefficient $C_i$ in the term $C_ix^i$ in the expansion of $h(x)$. But we must also consider the other factors, namely $(4x)^n$, $(1+x)$ and $P_n$, which contribute to the value of $C_i$.

For example $$ \text{factor }(1+x)^{\alpha=-2} \text{ contributes the term } \binom{\alpha=-2}{k}x^{k} $$

Let us re-express the formula for $h(x)$:- $$ h(x) = (1+x) \sum_{n=0}^\infty \frac{{P_n}^2}{1-2n} \left( \frac{4x}{(1+x)^2}\right)^n $$ $$ h(x) = (1+x) \sum_{n=0}^\infty \frac{{P_n}^2~4^n}{1-2n} x^n~(1+x)^{-2n} $$ Let us split the RHS into two "streams", the "1-stream" and the "$X$-stream." $$ h(x) = (1) \sum_{n=0}^\infty \frac{{P_n}^2~4^n}{1-2n} x^n~(1+x)^{-2n} +(x) \sum_{n=0}^\infty \frac{{P_n}^2~4^n}{1-2n} x^n~(1+x)^{-2n} $$

$$ h(x) = \sum_{n=0}^\infty \frac{{P_n}^2~4^n}{1-2n} x^n~(1+x)^{-2n} + \sum_{n=0}^\infty \frac{{P_n}^2~4^n}{1-2n} x^{n+1}~(1+x)^{-2n} $$

$$ h(x) = \sum_{n=0}^\infty \frac{{P_n}^2~4^n}{1-2n} x^n~\sum_{k=0}^{\infty} \binom{-2n}{k}x^k + \sum_{n=0}^\infty \frac{{P_n}^2~4^n}{1-2n} x^{n+1}~\sum_{k=0}^{\infty} \binom{-2n}{k}x^k $$

For clarity let us replace $$ \frac{{P_n}^2~4^n}{1-2n} ~\text{ by }~Q_n $$ so now the RHS is expressed as the sum of two streams, with each stream now comprising a series in $n$ containing a "nested" series in $k$. $$ h(x) = \sum_{n=0}^\infty Q_n x^n~\sum_{k=0}^{\infty} \binom{-2n}{k}x^k + \sum_{n=0}^\infty Q_n x^{n+1}~\sum_{k=0}^{\infty} \binom{-2n}{k}x^k $$

It can be seen that a particular pair of $n,k$ values will output a value of $$ Q_n x^n.\binom{-2n}{k}x^{k}+ Q_n x^{n+1}.\binom{-2n}{k}x^{k} $$ $$ = Q_n x^{k+n}.\binom{-2n}{k} + Q_n x^{k+n+1}.\binom{-2n}{k} $$

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Nothing forces us to pair with the same $n$ across the two streams. We could arrange it so that like powers of $x$ are collected by pairing the $(n+1,k)$ and $(n,k)$ terms from respective streams, thus... $$ = Q_{n+1} x^{k+n+1}.\binom{-2(n+1)}{k} + Q_n x^{k+n+1}.\binom{-2n}{k}. $$ I will call such an arrangement a "staggered" arrangement.

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Now let us consider what it takes to collect, in either stream, all the contributions $x^r$ in a particular "required" power $r:r>0$, of $x$.

Firstly note that, for any $n>=1$, expansion of the factor $(1+x)^{-2n}$ generates an infinite series of ascending positive $x$-power terms $x^i$: with indices $i = 0, 1,2,3,4,5,6...\infty$ for example:- $$\begin{align} {n=0}; (1+x)^{-2n} & = 1 \\ {n=1}; (1+x)^{-2n} & = 1 - 2x^1 + 3x^2 - 4x^3 + 5x^4 - 6x^5 ...\\ {n=2}; (1+x)^{-2n} & = 1 - 4x^1 + 10x^2 - 20x^3 + 35x^4 - 56x^5 ...\\ {n=3}; (1+x)^{-2n} & = 1 - 6x^1 + 21x^2 - 56x^3 + 126x^4 -252x^5 ...\\ {n=4}; (1+x)^{-2n} & = 1 - 8x^1 + 36x^2 -120x^3 + 330x^4 -792x^5 ...\\ \end{align} $$

Therefore, in either stream, for $(n>0)$, every value of $n$ will have an associated expansion of the $(1+x)^{-2n}$ factor which contains an $x$-power term that will contribute to $x^r$.

For example lets say we wish to determine the ultimate coefficient of the term $x^4$. In each stream (without staggering), the contributions will come from two sources, the primary source (located just after $Q_n$) of either $x^{n}$ (from 1-stream) or $x^{n+1}$ (from X-stream) and the secondary source (located within the $(1+x)^{-2n}$ expansion series) of $x^{i}$.

For the 1-stream we require the condition must be satisfied that $n+i=r$, so in this example $n+i=4$ and we can see that contributions to $x^4$ will come from the following $(n,i)$ pairs:- (1,3) (2,2) (3,1) and (4,0) only. Obviously as $r$ increases so the number of contributing $(n,i)$ pairs will increase also.

This allows us to write a deterministic formula for the pre-ultimate coefficient $C_{r(h)_1}$ of the required term $C_r.x^r$ coming from the 1-stream (primary source: $x^n$):- $$ C_{r(h)_1} = \sum_{n=1}^{r} Q_n * \binom{-2n}{r-n} $$

A similar formula can be written for the pre-ultimate coefficient $C_{r(h)_X}$ of the required term $C_r.x^r$ coming from the $X$-stream (primary source: $x^{n+1}$):- $$ C_{r(h)_X} = \sum_{n=1}^{r-1} Q_n * \binom{-2n}{r-1-n} $$

Therefore the coefficient $C_r$ of the required $x$-power term $C_r.x^r$ is given by

$$ C_{r(h)}=C_{r(h)_1}+C_{r(h)_X} = \sum_{n=1}^{r} Q_n * \binom{-2n}{r-n} + \sum_{n=1}^{r-1} Q_n * \binom{-2n}{r-1-n} $$

$$ C_{r(h)} = Q_r * \binom{-2r}{r-r=0} + \sum_{n=1}^{r-1} Q_n * \left( \binom{-2n}{r-n} + \binom{-2n}{r-n-1} \right ). $$

Now $\binom{A}{0}= 1$, for any $A$ so... $$ C_{r(h)} = Q_r * 1 + \sum_{n=1}^{r-1} Q_n * \left( \binom{-2n}{r-n} + \binom{-2n}{r-n-1} \right ) $$

Now we can apply the Addition(Induction) Rule of binomial coefficients, namely $\binom{N}{K} = \binom{N-1}{K} +\binom{N-1}{K-1}$ from which:- $\binom{N+1}{K} = \binom{N}{K} +\binom{N}{K-1}$, so... $$ C_{r(h)} = Q_r + \sum_{n=1}^{r-1} Q_n * \binom{-2n+1}{r-n} = Q_r + \sum_{n=1}^{r-1} Q_n * \binom{-(2n-1)}{r-n} $$

Upper Negation

To remove the negative $n$ terms we apply the Upper Negation Identity for binomial coefficients, namely $\binom{-N}{K} = (-1)^K * \binom{N+K-1}{K}$ $$ C_{r(h)} = Q_r + \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-1 + r-n -1}{r-n} $$

$$ C_{r(h)} = Q_r + \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{r-n} $$

Applying the Symmetry Rule:- $\binom{N}{K} = \binom{N}{N-K}$ gives...

$$ C_{r(h)} = Q_r + \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{(2n-2 + r-n )-(r-n)} $$

$$ C_{r(h)} = Q_r + \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{2n-2} $$

(This expression tested OK for $r>=2$).

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Unfortunately this expression for $C_{r(h)}$ is still not very similar in pattern to the previously-derived expressions of

(i) $C_{r(g)}$ from $g(x)$:-

$$C_{r(g)} = \frac{\pi}{2} \left[ - P^2_{(r/2)} + P^2_{(r/2 -1)} + 2\frac{P^2_{(r/2)}}{1-r}\right]$$

or (ii) $C_{r(u)}$ from $g(x)_{user}$:- $$ C_{r(u)} = \frac{\pi}{2} \left( \binom{r}{r/2} * \frac{1}{4^{r/2}} * \frac{1}{r-1} \right)^2. $$

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Proof by Induction - Pending

One possible line of investigation is to consider obtaining formulae for the change in coefficient value from $C_r$ to $C_{r+1}$ for any $r$. If (i) the formula is the same for $g(x)$ and $h(x)$; and (ii) the value of $C_r$ for some initial value of $r$ is the same for $g(x)$ and $h(x)$; then this would constitute a proof that $g(x)$ = $h(x)$.

Induction for $h(x)$

We have $$ C_{r(h)} = Q_r + \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{2n-2} $$ and so $$ C_{r+1(h)} = Q_{r+1} + \sum_{n=1}^{r+1-1} Q_n * (-1)^{r-n+1} \binom{2n-2 + r-n +1}{2n-2} $$

and so the change in $C_{r(h)}$ from $r$ to $r+1$ is given by:- $$\text{delta}~C_{r(h)} = Q_{r+1} -Q_{r} $$ $$+ \sum_{n=1}^{r+1-1} Q_n * (-1)^{r-n+1} \binom{2n-2 + r-n +1}{2n-2} $$ $$- \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{2n-2} $$

$$=Q_{r+1} -Q_{r} + Q_r * (-1)^{r-r+1} \binom{2r-2 + r-r +1}{2r-2} $$ $$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n+1} \binom{2n-2 + r-n +1}{2n-2} $$ $$-- \sum_{n=1}^{r-1} Q_n * (-1)^{r-n+1} \binom{2n-2 + r-n }{2n-2} $$

$$=Q_{r+1} -Q_{r} + Q_r * (-1) \binom{2r-2 +1}{2r-2} $$ $$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n+1} \left( \binom{2n-2 + r-n +1}{2n-2} + \binom{2n-2 + r-n }{2n-2} \right) .$$

Using the Addition/Induction Rule and the Absorption/Extraction Rule $\left(\binom{N}{K}=\frac{N}{K}\binom{N-1}{K-1}\right)$ we can show that $\binom{N}{K}+\binom{N+1}{K} = \left(2-\frac{K}{N+1}\right)\binom{N+1}{K}$, hence

$$\text{delta}~C_{r(h)} = Q_{r+1} -Q_{r} - Q_r * \binom{2r-2 +1}{2r-2} $$ $$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n+1} *\left( 2 - \frac{2n-2}{2n-2 + r-n +1}\right) *\binom{2n-2 + r-n +1}{2n-2} $$

$$\text{delta}~C_{r(h)} = Q_{r+1} -Q_{r} - Q_r * \binom{2r-2 +1}{2r-2} $$ $$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n+1} *\left( 2 - \frac{2n-2}{n-1 + r }\right) *\binom{n-1 + r}{2n-2} $$

$$\text{delta}~C_{r(h)} = Q_{r+1} -Q_{r} - Q_r * \binom{2r-2 +1}{2r-2} $$ $$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n+1} *\left( \frac{ 2n-2 + 2r + 2n-2}{n-1 + r }\right) *\binom{n-1 + r}{2n-2}$$

$$\text{delta}~C_{r(h)} = Q_{r+1} -Q_{r} - Q_r * \binom{2r-2 +1}{2r-2} $$ $$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n+1} *\left( \frac{2r}{n-1 + r }\right) *\binom{n-1 + r}{2n-2} $$

Unfortunately this does not look very tractable :-(.

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We might obtain more clarity if we treated $g(x)$ as two streams - even and odd powers of $x$. We would expect that $dC_{r(h,odd)}$ will equal $0$ which will allow us to ignore odd powers of $x$ (given the calculated seed that $C_{r(h)}x^1=0$). And we might get a simpler expression of $dC_{r(h,even)}$ which is easier to compare with that for $dC_{r(g)}$.

We have $$ C_{r(h)} = Q_r + \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{2n-2} $$ and so $$ C_{r+2(h)} = Q_{r+2} + \sum_{n=1}^{r+2-1} Q_n * (-1)^{r-n+2} \binom{2n-2 + r-n +2}{2n-2} $$

and so the change in $C_{r(h)}$ from $r$ to $r+1$ is given by:- $$\text{d2_}~C_{r(h)} = Q_{r+2} -Q_{r} $$ $$+ \sum_{n=1}^{r+2-1} Q_n * (-1)^{r-n+2} \binom{2n-2 + r-n +2}{2n-2} $$ $$- \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{2n-2} $$

Now we use the fact that $(-1)^{r-n+2}=(-1)^{r-n}$ to replace the former by the latter...

$$\text{d2_}~C_{r(h)} = Q_{r+2} -Q_{r} $$ $$+ \sum_{n=1}^{r+2-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n +2}{2n-2} $$ $$- \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{2n-2} $$

Now we peel off the last sum-term of the first sum so that the two sums both have final index $n=r-1$...

I need to think about the difference in the top index value between odd and even streams.

$$\text{d2_}~C_{r(h)} = Q_{r+2} -Q_{r} $$

$$+ Q_{r+2-1} * (-1)^{r-({r+2-1})} \binom{2({r+2-1})-2 + r-({r+2-1}) +2}{2({r+2-1})-2} $$

$$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n +2}{2n-2} $$ $$- \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{2n-2} $$

Rationalize the first binomial term...

$$\text{d2_}~C_{r(h)} = Q_{r+2} -Q_{r} $$ $$+ Q_{r+ 1} * (-1)^{-1} \binom{ 2r+1}{2r} $$

$$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n +2}{2n-2} $$ $$- \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{2n-2}$$

Now merge two sums into one...

$$\text{d2_}~C_{r(h)} = Q_{r+2} -Q_{r} $$

$$+ Q_{r+ 1} * (-1) \binom{ 2r+1}{2r} $$ $$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \left[ \binom{2n-2 + r-n +2}{2n-2} - \binom{2n-2 + r-n }{2n-2} \right]$$

It is demonstrable that $\binom{N+2}{K}-\binom{N}{K}=((N+2)(N+1)-1)\binom{N}{K}$ so...

$$\left[ \binom{2n-2 + r-n +2}{2n-2} - \binom{2n-2 + r-n }{2n-2} \right]$$

$$=\left[ \left((2n-2 + r-n +2)(2n-2 + r-n +1)-1 \right) * \binom{2n-2 + r-n }{2n-2} \right]$$

then

$$=\left[ \left((n + r )(n + r- 1)-1 \right) * \binom{2n-2 + r-n }{2n-2} \right]$$

then again

$$=\left[ \left(n^2 + r^2 +2nr -n - r-1 \right) * \binom{2n-2 + r-n }{2n-2} \right]$$

So... $$\text{d2_}~C_{r(h)} = Q_{r+2} -Q_{r} + Q_{r+ 1} * (-1) \binom{ 2r+1}{2r} $$

$$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} * \left(n^2 + r^2 +2nr -n - r-1 \right) * \binom{2n-2 + r-n }{2n-2} $$

Unfortunately this does not look very tractable either :-(.

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Using the Double Factorial formulae for $K()$ and $E()$

Given the failure so far of the previous methods it might be worthwhile exploring the use of the alternative formula for $K()$ and $E()$, i.e. those which use Double Factorial terms (see wikipedia page for Elliptic Integrals).

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( need to check ALL the above !!! )

see Pascal's Extended Triangle by Ken Ward

to be continued...

Answer 3

The identity holds for $0<x<1$ and can be easily established using the imaginary transformation formula for elliptic integral $E(m) $. We have $$E(-m) =\int_{0}^{\pi/2}\sqrt{1+m\sin^2t}\,dt=\int_{0}^{\pi/2}\sqrt{1+m-m\sin^2t}\,dt=\sqrt{1+m}E\left(\frac{m} {1+m}\right)$$ where $m\in(0,1)$.

Putting $m=4x/(1-x)^2$ we can see that $m/(1+m)=4x/(1+x)^2$ and thus we get $$E\left(\frac{-4x}{(1-x)^2}\right)=\frac{1+x}{1-x}E\left(\frac{4x}{(1+x)^2}\right)$$ This gives us the first and last expressions of the desired equality. To get the middle expression you need the Gauss/Landen transformation.

It is best here to use related integrals $$I(a, b) = \int_{0}^{\pi/2}\frac{dx}{\sqrt{a^2\cos^2x+b^2\sin^2x}}=\int_{b}^{a}\frac{dt}{\sqrt{(a^2-t^2)(t^2-b^2)}},\\L(a, b) =(a^2-b^2)\int_{0}^{\pi/2}\frac{\sin^2x}{\sqrt{a^2\cos^2x+b^2\sin^2x}}\,dx=\int_{b}^{a}\sqrt{\frac{a^2-t^2}{t^2-b^2}}\,dt\tag{1}$$ The equivalence between both the forms (trigonometric and algebraic) is established using the substitution $$t^2=a^2\cos^2x+b^2\sin^2x$$ The first of the integrals above is related to $K(m) $ via $$K(m) =I(1,\sqrt{1-m})\tag{2}$$ and the second one is related to $E(m) $ by $$E(m)=K(m) - L(1,\sqrt{1-m})\tag{3}$$ The integral $I(a, b) $ remains invariant under the AGM transformation $$a\to\frac{a+b}{2}=a_1,b\to\sqrt{ab}=b_1$$ and this can be proved using the substitution $$\sin x =\frac{2a\sin t} {a+b+(a-b) \sin^2t} $$ given by Gauss. The second integral $L(a, b) $ satisfies the following AGM transformation $$L(a, b) =\frac{a^2-b^2}{2}I(a_1,b_1)+2L(a_1,b_1)\tag{4}$$ The above result can be established by using substitution $$u=\frac{1}{2}\left(t+\frac{ab}{t}\right)$$ in the algebraic form for $L(a, b) $. Both the integral substitutions described above require a good deal of labor and care and the details are not presented here. You can get more details in my blog posts starting with this one.

Now we can put $m=4x/(1+x)^2$ in $(3)$ to get $$E\left (\frac{4x}{(1+x)^2}\right)=K\left(\frac{4x}{(1+x)^2}\right)-L\left(1,\frac{1-x}{1+x}\right)$$ Note further that $L(ca, cb) = cL(a, b) $ and hence the above can be written as $$(1+x)E\left(\frac{4x}{(1+x)^2}\right) =(1+x)K\left(\frac{4x}{(1+x)^2}\right)-L(1+x,1-x)$$ Putting $a=1+x,b=1-x$ in $(4)$ we can see that $$L(1+x,1-x)=2xI(1,\sqrt{1-x^2})+2L(1,\sqrt{1-x^2})$$ From $(3)$ we note that $$I(1,\sqrt{1-x^2})=K(x^2),L(1,\sqrt{1-x^2})=K(x^2)-E(x^2)$$ and hence we arrive at $$(1+x)E\left(\frac{4x}{(1+x)^2}\right)=(1+x)K\left(\frac{4x}{(1+x)^2}\right)-2xK(x^2)-2K(x^2)+2E(x^2)$$ And $$K\left(\frac{4x}{(1+x)^2}\right) = I\left(1,\sqrt{1-\frac{4x}{(1+x)^2}}\right)=I\left(1,\frac{1-x}{1+x}\right)=(1+x) I(1+x,1-x)$$ which can be further written as $$(1+x)I(1,\sqrt{1-x^2})=(1+x)K(x^2)$$ Combining everything we finally get $$(1+x)E\left(\frac{4x}{(1+x)^2}\right)=(x^2-1)K(x^2)+2E(x^2)$$ as desired.

Answer 4

Only the proof of second equality is suffice to justify the relation of the formula. For this purpose, simply use the transformation as shown below.(Please note that this is just one of the ways among various transformation techniques)

\begin{equation} \begin{split} &\sin\psi=\frac{(1+k) \sin\theta}{1+k\sin^2\theta}\,\,\,(1) \\ \end{split} \end{equation} \begin{equation} \begin{split} &\cos\psi=\frac{\cos\theta\sqrt{1-k^2\sin^2\theta}}{1+k\sin^2\theta}\,\,\,(2) \\ \end{split} \end{equation} in Jacobi's elliptic integrals.

When $k'$ is defined as $k'^2=4k/(1+k)^2$, we have the relation, \begin{equation} \begin{split} &\sqrt{1-k'^2\sin^2\psi}=\frac{1-k \sin^2\theta}{1+k\sin^2\theta} \,\,\,(3) \\ \end{split} \end{equation} Now take the derivation of $\sin\psi$ by $\theta$ in (1) and you will obtain, \begin{equation} \begin{split} & \cos\psi\,\frac{d\psi}{d\theta}=(1+k) \frac{\cos\theta(1-k\sin^2\theta)}{(1+k\sin^2\theta)^2}\,\,\,(4) \\\\ \end{split} \end{equation} This can be rewritten as, \begin{equation} \begin{split} & d\psi=(1+k) \frac{\cos\theta(1-k\sin^2\theta)}{(1+k\sin^2\theta)^2}\frac{1}{\cos\psi}\,d\theta \,\,\,(5) \\\\ \end{split} \end{equation}

Next, the formula shown below yields \begin{equation} \begin{split} & \frac{d}{d\theta}\left(\frac{k\,\sin2\theta}{1+k\sin^2\theta}\sqrt{1-k^2\sin^2\theta}\right) +2\sqrt{1-k^2\sin^2\theta}-\frac{1-k^2}{\sqrt{1-k^2\sin^2\theta}} \\ &=\frac{2k\cos2\theta\sqrt{1-k^2\sin^2\theta}}{1+k\sin^2\theta}- \frac{k^3\sin\theta\cos\theta\sin2\theta}{(1+k\sin^2\theta)\sqrt{1-k^2\sin^2\theta}} \\ &-\frac{2k^2\sin\theta\cos\theta\sin2\theta\sqrt{1-k^2\sin^2\theta}}{(1+k\sin^2\theta)^2} +2\sqrt{1-k^2\sin^2\theta}-\frac{1-k^2}{\sqrt{1-k^2\sin^2\theta}} \\ &=(1+k)^2\frac{(1-k\sin^2\theta)^2}{(1+k\sin^2\theta)^2\sqrt{1-k^2\sin^2\theta}}\,\,\,(6) \\ \end{split} \end{equation}

Using(2), (5) and (6), the following relation is obtained. \begin{equation} \begin{split} &\sqrt{1-k'^2\sin^2\psi}\,\,d\psi=(1+k) \left(\frac{1-k\sin^2\theta}{1+k\sin^2\theta}\right)^2 \frac{1}{\sqrt{1-k^2\sin^2\theta}}\,d\theta\,\,\,(7) \\ \end{split} \end{equation} So, from (7) and (6), you will see that \begin{equation} \begin{split} &\sqrt{1-k'^2\sin^2\psi}\,d\psi=\frac{1}{1+k}\times \\ &\left[ \frac{d}{d\theta}\left(\frac{k\,\sin2\theta}{1+k\sin^2\theta}\sqrt{1-k^2\sin^2\theta}\right) +2\sqrt{1-k^2\sin^2\theta}-\frac{1-k^2}{\sqrt{1-k^2\sin^2\theta}} \right]\,d\theta\,\,\,(8) \\ \end{split} \end{equation} Integrateing both sides of (8) yields, \begin{equation} \begin{split} &\int\sqrt{1-k'^2\sin^2\psi}\,d\psi=\frac{1}{1+k}\times \\ &\left[ \left(\frac{k\,\sin2\theta}{1+k\sin^2\theta}\sqrt{1-k^2\sin^2\theta}\right) +2\int\sqrt{1-k^2\sin^2\theta}\,\,d\theta-(1-k^2)\times\int\frac{d\theta}{\sqrt{1-k^2\sin^2\theta}} \right]\,\,\,(9) \\ \end{split} \end{equation} When considering the complete elliptic integral, (9) is simplied to, \begin{equation} \begin{split} &\int^{\pi/2}_0\sqrt{1-k'^2\sin^2\psi}\,d\psi \\ &=\frac{1}{1+k} \left[ 2\int^{\pi/2}_0\sqrt{1-k^2\sin^2\theta}\,\,d\theta -(1-k^2)\int^{\pi/2}_0\frac{d\theta}{\sqrt{1-k^2\sin^2\theta}} \right]\,\,\,(10) \\ \end{split} \end{equation}