高専の数学解説

第5章 5.20

与式においてh = x − aとおくと

左辺=lim⁡h→01h{f(a+h+12h)+f(a+h+32h)−2f(a+h)}\text{左辺} = \lim_{h\to0}\frac{1}{h}\left\{f\left(a+h+\frac{1}{2}h\right) + f\left(a+h+\frac{3}{2}h\right) - 2f(a+h)\right\}

=lim⁡h→01h{f(a+32h)+f(a+52h)−2f(a+h)}= \lim_{h\to0}\frac{1}{h}\left\{f\left(a+\frac{3}{2}h\right) + f\left(a+\frac{5}{2}h\right) - 2f(a+h)\right\}

=lim⁡h→01hf(a+32h)+lim⁡h→01hf(a+52h)−2lim⁡h→01hf(a+h)= \lim_{h\to0}\frac{1}{h}f\left(a+\frac{3}{2}h\right) + \lim_{h\to0}\frac{1}{h}f\left(a+\frac{5}{2}h\right) - 2\lim_{h\to0}\frac{1}{h}f(a+h)

h′=32h、 h′′=52h とおくとh' = \frac{3}{2}h、\ h'' = \frac{5}{2}h \text{ とおくと}

=lim⁡h→0123h′{f(a+h′)−f(a)}+lim⁡h→0125h′′{f(a+h′′)−f(a)}−2lim⁡h→01hf(a+h)+lim⁡h→0123h′f(a)+lim⁡h→0125h′′f(a)= \lim_{h\to0}\frac{1}{\frac{2}{3}h'}\{f(a+h')-f(a)\} + \lim_{h\to0}\frac{1}{\frac{2}{5}h''}\{f(a+h'')-f(a)\} - 2\lim_{h\to0}\frac{1}{h}f(a+h) + \lim_{h\to0}\frac{1}{\frac{2}{3}h'}f(a) + \lim_{h\to0}\frac{1}{\frac{2}{5}h''}f(a)

=32f′(a)+52f′(a)−2lim⁡h→01hf(a+h)+lim⁡h→01hf(a)+lim⁡h→01hf(a)= \frac{3}{2}f'(a) + \frac{5}{2}f'(a) - 2\lim_{h\to0}\frac{1}{h}f(a+h) + \lim_{h\to0}\frac{1}{h}f(a) + \lim_{h\to0}\frac{1}{h}f(a)

=82f′(a)−2lim⁡h→01h{f(a+h)−f(a)}= \frac{8}{2}f'(a) - 2\lim_{h\to0}\frac{1}{h}\{f(a+h)-f(a)\}

=4f′(a)−2f′(a)=2f′(a)= 4f'(a) - 2f'(a) = 2f'(a)

解説: r31bn1z