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By Charles Jordan

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Example text

Log(l--t) = f, s G $ = -log(l-f) if x>O. In this way we obtain again formula (3). A second integration will give t+(l----t)log(1-f) 1 5 t”z=2 x ( x - 1 ) t + ( l - 4 ) Zog(1-q G x ( x’- 1 ) = i f x>l. Third method. Sometimes it is possible to obtain the generating function of f(x) by performing directly the summation ii f(x) t” = u(f). -. l-at Example. Let f(x) = cos 8x. cos IYX is the real part of eiSX: therefore a(t) will be the real part of 1 l--e’:+ f This is easily determined and we find 17.

I 2’; Ay Hence 11 Expansion of a function by syntbolical methods. We have z+1 x Ex~(l+A)~=s I y ) Av v=o since the operation p performed on f(z) gives f(x) for z=O if h=l Tberefore this is the symbolical expression of Newton’s formula (1) f(x) = f(O) +(;I Af(o) + [;) A’fW +-a.. Hitherto we have only defined operations combined by polynomial relations (except in the case of E-n): therefore the above demonstration assumes that x is a positive integer. But the significance of the operation Ex is obvious for any value of x; indeed if h=l we always have I ” I Pf(0) = f ( x ) .

Remark: f(x) may be obtained by solving directly the above linear difference equation of the second order with constant coefficients (Q 165). 30 This method is known as that of indeterminate coefficients. Third method. Expansion by the binomial theorem. Examples. (1$-f)” (3) . (4) (*At)” = y;) f” = f, (---iI” (A”) f” = i. +x) f” Example 1. f(l+f)(l-f)-3 = f(l+f) SO (y2) fl and finally the coefficient of fx will he f(x) = ((:z:, + ( x:2)i = 3. Example 2. (l+r+fz+ , , . + f”‘)” = (1-fm+l)” (l-f)-” = (5) = z.

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