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Question Number 62679 by ajfour last updated on 24/Jun/19

Let f be defined in the neighborhood of x and that f ′′(x) exists. Prove that lim_(h→0) ((f(x+h)+f(x−h)−2f(x))/h^2 )=f ′′(x) .

$${Let}\:{f}\:{be}\:{defined}\:{in}\:{the}\:{neighborhood} \\ $$$${of}\:{x}\:{and}\:{that}\:{f}\:''\left({x}\right)\:{exists}. \\ $$$${Prove}\:{that}\:\: \\ $$$$\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{f}\left({x}+{h}\right)+{f}\left({x}−{h}\right)−\mathrm{2}{f}\left({x}\right)}{{h}^{\mathrm{2}} }={f}\:''\left({x}\right)\:. \\ $$

Commented by kaivan.ahmadi last updated on 24/Jun/19

hi sir, check the quoestion please

$${hi}\:{sir},\:{check}\:{the}\:{quoestion}\:{please} \\ $$

Commented by ajfour last updated on 24/Jun/19

Its correct, Sir.

$${Its}\:{correct},\:{Sir}. \\ $$

Commented by MJS last updated on 24/Jun/19

I think it should be h→0?

$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:{h}\rightarrow\mathrm{0}? \\ $$

Commented by kaivan.ahmadi last updated on 24/Jun/19

if h→0 (0/0) =^(Hop) lim_(h→0 ) ((f′(x+h)−f′(x−h))/(2h)) =^(Hop) lim_(h→0) ((f′′(x+h)+f′′(x−h))/2)=((2f′′(x))/2)=f′′(x) this way is true i had a mistake in previous solution ((d/dh)(2f(x))=0)

$${if}\:\:{h}\rightarrow\mathrm{0} \\ $$$$\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\overset{{Hop}} {=}\:{lim}_{{h}\rightarrow\mathrm{0}\:} \:\frac{{f}'\left({x}+{h}\right)−{f}'\left({x}−{h}\right)}{\mathrm{2}{h}}\:\overset{{Hop}} {=} \\ $$$${lim}_{{h}\rightarrow\mathrm{0}} \:\:\frac{{f}''\left({x}+{h}\right)+{f}''\left({x}−{h}\right)}{\mathrm{2}}=\frac{\mathrm{2}{f}''\left({x}\right)}{\mathrm{2}}={f}''\left({x}\right) \\ $$$${this}\:{way}\:{is}\:{true} \\ $$$${i}\:{had}\:{a}\:{mistake}\:{in}\:{previous}\:{solution} \\ $$$$ \\ $$$$\left(\frac{{d}}{{dh}}\left(\mathrm{2}{f}\left({x}\right)\right)=\mathrm{0}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Mr X pcx last updated on 24/Jun/19

hospital theorem let u(h)=f(x+h)+f(x+h)−2f(x) v(h) =h^2 lim_(h→0) u(h)=lim_(h→0) v(h)=0 u^′ (h) =f^′ (x+h)+f^′ (x+h)∫→2f^′ (x) u^(′′) (h) =f^(′′) (x+h)+f^(′′) (x+h)→2f^(′′) (x) v^′ (h)=2h →0 v^(′′) (h) =2 ⇒ lim_(h→0) ((f(x+h)+f(x+h)−2f(x))/h^2 ) =((2f^(′′) (x))/2) =f^(′′) (x) .

$${hospital}\:{theorem}\:{let} \\ $$$${u}\left({h}\right)={f}\left({x}+{h}\right)+{f}\left({x}+{h}\right)−\mathrm{2}{f}\left({x}\right) \\ $$$${v}\left({h}\right)\:={h}^{\mathrm{2}} \\ $$$${lim}_{{h}\rightarrow\mathrm{0}} {u}\left({h}\right)={lim}_{{h}\rightarrow\mathrm{0}} {v}\left({h}\right)=\mathrm{0} \\ $$$${u}^{'} \left({h}\right)\:={f}^{'} \left({x}+{h}\right)+{f}^{'} \left({x}+{h}\right)\int\rightarrow\mathrm{2}{f}^{'} \left({x}\right) \\ $$$${u}^{''} \left({h}\right)\:={f}^{''} \left({x}+{h}\right)+{f}^{''} \left({x}+{h}\right)\rightarrow\mathrm{2}{f}^{''} \left({x}\right) \\ $$$${v}^{'} \left({h}\right)=\mathrm{2}{h}\:\rightarrow\mathrm{0} \\ $$$${v}^{''} \left({h}\right)\:=\mathrm{2}\:\Rightarrow \\ $$$${lim}_{{h}\rightarrow\mathrm{0}} \:\:\frac{{f}\left({x}+{h}\right)+{f}\left({x}+{h}\right)−\mathrm{2}{f}\left({x}\right)}{{h}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{f}^{''} \left({x}\right)}{\mathrm{2}}\:={f}^{''} \left({x}\right)\:. \\ $$

Answered by mr W last updated on 24/Jun/19

f′′(x)=lim_(h→0) ((f ′(x+(h/2))−f ′(x−(h/2)))/h) f′′(x)=lim_(h→0) ((((f(x+h)−f(x))/h)−((f(x)−f(x−h))/h))/h) f′′(x)=lim_(h→0) ((f(x+h)+f(x−h)−2f(x))/h^2 )

$${f}''\left({x}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{f}\:'\left({x}+\frac{{h}}{\mathrm{2}}\right)−{f}\:'\left({x}−\frac{{h}}{\mathrm{2}}\right)}{{h}} \\ $$$${f}''\left({x}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}}−\frac{{f}\left({x}\right)−{f}\left({x}−{h}\right)}{{h}}}{{h}} \\ $$$${f}''\left({x}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{f}\left({x}+{h}\right)+{f}\left({x}−{h}\right)−\mathrm{2}{f}\left({x}\right)}{{h}^{\mathrm{2}} } \\ $$

Commented by ajfour last updated on 24/Jun/19

Thanks Sir, nice way!

$${Thanks}\:{Sir},\:{nice}\:{way}! \\ $$

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