Rozwiń funkcję w szereg Mariusz:

	1
f(t)=
	√1−2xt+t2

1
	= (1−2xt+t2)−1/2
√1−2xt+t2

	−(1/2) n
(1+t(t−2x))−1/2 = ∑n=0∞		tn(t−2x)n

(1+t(t−2x))^−1/2 =

	−(1/2) n		n k
∑n=0∞		tn(∑k=0n		tk(−1)n−k(2x)n−k)

	−(1/2) n	n k
(1+t(t−2x))−1/2 = ∑n=0∞∑k=0n			(−1)n−k(2x)n−ktn+k

Rozwińmy teraz ten symbol Newtona aby pozbyć się tego ułamkowego argumentu

−(1/2) n		1   1   1   1   (− )(− −1)(− −2)*...*(− −(n−1))   2   2   2   2
	=
		n!

−(1/2) n		(−1)n		1*3*5*...*(2n−1)
	=		*
		2n		n!

−(1/2) n		(−1)n		1*3*5*...*(2n−1)*2*4*6*...*2n
	=		*
		2n		n!*2*4*6*...*2n

−(1/2) n		(−1)n		(2n)!
	=		*
		2n		n!*2nn!

−(1/2) n		(−1)n		(2n)!
	=		*
		22n		n!*n!

−(1/2) n		(−1)n		2n n
	=		*
		22n

(1+t(t−2x))^−1/2 =

	2n n	n k		(−1)n
∑n=0∞∑k=0n			*		*(−1)n−k(2x)n−ktn+k
				22n

(1+t(t−2x))^−1/2 =

	2n n	n k		(−1)k		2n
∑n=0∞∑k=0n			*		*		xn−ktn+k
				22n		2k

(1+t(t−2x))^−1/2 =

	2n n	n k		(−1)k
∑n=0∞∑k=0n			*		xn−ktn+k
				2n+k

(1+t(t−2x))^−1/2 =

	2m−2k m−k	m−k k		(−1)k
∑m=0∞∑k=0floor(m/2)			*		xm−2ktm
				2m

(1+t(t−2x))^−1/2 =

	(2m−2k)!	(m−k)!
∑m=0∞∑k=0floor(m/2)
	(m−k)!(m−k)!	k!(m−2k)!

	(−1)k
*		xm−2ktm
	2m

(1+t(t−2x))^−1/2 =

	(2m−2k)!	1		(−1)k
∑m=0∞∑k=0floor(m/2)			*		xm−2ktm
	(m−k)!	k!(m−2k)!		2m

(1+t(t−2x))^−1/2 =

	(2m−2k)!	1		(−1)k
∑m=0∞∑k=0floor(m/2)			*		xm−2ktm
	(m−2k)!	k!(m−k)!		2m

(1+t(t−2x))^−1/2 =

	(2m−2k)!	m!
∑m=0∞∑k=0floor(m/2)
	m!(m−2k)!	k!(m−k)!

	(−1)k
*		xm−2ktm
	2m

(1+t(t−2x))^−1/2 =

	2m−2k m	m k		(−1)k
∑m=0∞∑k=0floor(m/2)			*		xm−2ktm
				2m

Tylko jak uzasadnić przejście z (1+t(t−2x))^−1/2 =

	2n n	n k		(−1)k
∑n=0∞∑k=0n			*		xn−ktn+k
				2n+k

do (1+t(t−2x))^−1/2 =

	2m−2k m−k	m−k k		(−1)k
∑m=0∞∑k=0floor(m/2)			*		xm−2ktm
				2m

X: