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Ðàçëîæåíèå îñíîâíûõ ýëåìåíòàðíûõ ô-öèé ïî ôîðìóëå Ìàêëîðåíà. 1)f(x)=ex, f(0)=1, f(k)(x)=ex, f(k)(0)=1, ex=1+x+x?/2!+…+xn/n!+o(xn), x®0; 2)f(x)=sinx, f(0)=0, f’(x)=cosx, f’’(x)=-sinx, f’’’(x)=-cosx, f(IV)(x)=sinx,…; f(k)(x)={(-1)msinx, k=2m {(-1)m-1cosx, k=2m-1 m=1,2,…; f(2m-1)(0)=(-1)m-1 ïîëàãàÿ n=2m ïîëó÷èì sinx=x-x3/3!+x5/5!-…+(-1)n-1x2m-1/(2m-1)!+o(x)2m,x®0; cosx=1-x?/2!+x4/2!-x6/6!+….+(-1)mx2m/(2m)!+o(x2m+1),x®0; 4)f(x)=ln(1+x)…f(0)=ln1=0, f’(x)=1/(1+x), f’’(x)=-1/(1+x)?, f’’’(x)=2/(1+x)3…,f(k)(x)=(-1)k-1(k-1)/(1+x)k ;f(k)(0)=(-1)k-1?(k-1)! Ïîäñòàâèì â ôîðìóëó Òåéëîðà ? l(1+x)=x-x?/2+x3/3+..+(-1)n-1xn/n+o(xn),x®0 ; 5)f(x)=(1+x)b f(0)=1, f’(x)=b(1+x)b-1, f’’(x)=b(b-1)(1+x)b-2; f(k)(x)=b(b-1)…(b-k+1)(1+x)b-k ;f(k)(0)=b(b-1)…(b-k+1); (1+x)b=1+b?x+b(b-1)x?/2!+…+b(b-1)…(b-n+1)xn/n!+o(xn), x®0