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I want to show that aj in this photo, but I am unsure of how to get there I know it is a minimum so it takes differentiation to get there but I am unsure how, any help would be appreciated, thanks enter image description here

$$f(x)=\frac{a_0}{2}T_0(x)+\sum\limits_{j=1}^{\infty } a_j\ T_j(x)\tag{1}$$

$$a_j=\frac{2}{\pi}\int_{-1}^1 f(x)\ T_j(x)\frac{dx}{\sqrt{1-x^2}}\quad (j=0,1,2,...)\tag{2}$$

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  • Best of luck. I'd be interested in knowing what you've tried to this point. – A rural reader May 10 '21 at 17:51
  • Use the orthogonality of the $T_k$. – A rural reader May 10 '21 at 22:15
  • I'm not sure the formulas above are consistent with the last set of formulas at https://functions.wolfram.com/HypergeometricFunctions/ChebyshevTGeneral/25/01/. – Steven Clark May 10 '21 at 22:16
  • The chebyshev-function tag is not relevant here since it's reserved for the Chebyshev functions $\vartheta(x)=\sum\limits_{p\le x}\log(p)$ where $p$ is a prime and $\psi(x)=\sum\limits_{n=1}^x\Lambda(n)$ where $\Lambda(n)=\log(p)$ when $n$ is a prime-power ($n=p^k$). – Steven Clark May 10 '21 at 22:32

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