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A problem related to Chebyshev Polynomials and the Hypergeometric Function involves transformation from one function to another. The task is to transform the Chebyshev polynomial into its correct scaled Hypergeometric Function. $$ T_{2n} (x) = ( - 1)^n {}_2F_1 \left( { - n,n;\tfrac{1}{2};x^2 } \right). $$ Is there any way for me to make use of the relationship below? $$ T_n (x) = {}_2F_1 \left( { - n,n;\tfrac{1}{2};\tfrac{{1 - x}}{2}} \right). $$

Gary
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Lawrence Calixto
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  • Would [math.se] be a better home for this question? – Qmechanic Mar 19 '21 at 08:47
  • It seems that relating T_n(x) and T_2n(x) can't be done using elementary algebra and any form of recursion. Any hints in arriving the answer given above? – Lawrence Calixto Mar 19 '21 at 08:47
  • @Qmechanic I just love how physicists arrive at a simple and hand-waving solutions. But okay, I will be posting this on mathematics field for contrast too. – Lawrence Calixto Mar 19 '21 at 08:53
  • Exact copy of https://math.stackexchange.com/questions/4067996/chebyshev-polynomials-and-the-hypergeometric-function You may try showing that the hypergeometric function you are given and $T_{2n}(x)$ both satisfy the same ODE with certain initial values, and conclude that they must be equal. – Gary Mar 19 '21 at 11:48
  • Perhaps another way would be to start with the differential equation for $ {}2F_1 \left( { - n,n;\tfrac{1}{2};z } \right)$, then derive from that the differential equation for $( - 1)^n {}_2F_1 \left( { - n,n;\tfrac{1}{2};x^2 } \right)$, and compare to the differential equation for $T{2n}(x)$. – GEdgar Mar 19 '21 at 12:35
  • @GEdgar Does it mean that the starting equation should be the differential equation itself? – Lawrence Calixto Mar 20 '21 at 10:23

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The Chebyshev polynomials are defined by $$ T_n(\cos\theta)=\cos(n\theta). $$ Hence $$ T_{2n}(\cos(\theta))=\cos(2n\theta)=T_n(\cos(2\theta)). $$ We now use the relation $\cos(2\theta)=2\cos(\theta)^2-1$ and write $\cos\theta=x$ to get $$ T_{2n}(x)=T_n(2x^2-1). $$ This should prove the claimed relation.

Somos
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Giulio R
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Let me carry out my suggestion.
Start HERE with the differential equation for ${}_2F_1$:
The solution of $$ z(1-z)F''(z)+\left(\frac12 - z\right)F'(z)+n^2 F(z) = 0,\quad F(0)=1, F'(0)=-2n^2 \tag1$$ is $$ F(z) = {}_2F_1\left(-n,n;\frac12;z\right) . $$ Change variables $z=x^2$. Then we have: The solution of $$ (1-x^2)F''(x)-xF'(x)+4n^2F(x) = 0,\quad F(0)=1,F'(0)=0 \tag2$$ is $$ F(x) = {}_2F_1\left(-n,n;\frac12;x^2\right) . $$

Next, HERE we find that the Chebyshev polynomial $T_m(x)$ satisfies $$ (1-x^2)T_m''(x)-xT_m'(x)+m^2T_m(x)=0 \tag3$$ Put $m=2n$, then $(3)$ becomes $(2)$. Also $T_{2n}(0)=(-1)^{n}, T_{2n}'(0)=0$. Therefore $$ T_{2n}(x) = (-1)^n\;{}_2F_1\left(-n,n;\frac12;x^2\right) . $$

GEdgar
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