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

$$Letfbedefinedintheneighborhood$$$$ofxandthatf{}^{″}\left(x\right)exists.$$$$Provethat$$$$\underset{h\to 0}{\lim }\frac{f(x+h)+f(x-h)-2f\left(x\right)}{h^2}=f{}^{″}\left(x\right).$$

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

$$hisir,checkthequoestionplease$$

Commented by ajfour last updated on 24/Jun/19

$$Itscorrect,Sir.$$

Commented by MJS last updated on 24/Jun/19

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

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

$$ifh\to 0$$$$\frac{0}{0}$$$$\stackrel{Hop}{=}{lim}_{h\to 0}\frac{f^{\prime }(x+h)-f^{\prime }(x-h)}{2h}\stackrel{Hop}{=}$$$${lim}_{h\to 0}\frac{f^{″}(x+h)+f^{″}(x-h)}{2}=\frac{2f^{″}\left(x\right)}{2}=f^{″}\left(x\right)$$$$thiswayistrue$$$$ihadamistakeinprevioussolution$$$$$$$$(\frac{d}{dh}\left(2f\left(x\right)\right)=0)$$$$$$$$$$$$$$$$$$

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

$$hospitaltheoremlet$$$$u\left(h\right)=f(x+h)+f(x+h)-2f\left(x\right)$$$$v\left(h\right)=h^2$$$${lim}_{h\to 0}u\left(h\right)={lim}_{h\to 0}v\left(h\right)=0$$$$u^{{}^{\prime }}\left(h\right)=f^{{}^{\prime }}(x+h)+f^{{}^{\prime }}(x+h)\int \to 2f^{{}^{\prime }}\left(x\right)$$$$u^{{}^{″}}\left(h\right)=f^{{}^{″}}(x+h)+f^{{}^{″}}(x+h)\to 2f^{{}^{″}}\left(x\right)$$$$v^{{}^{\prime }}\left(h\right)=2h\to 0$$$$v^{{}^{″}}\left(h\right)=2\Rightarrow$$$${lim}_{h\to 0}\frac{f(x+h)+f(x+h)-2f\left(x\right)}{h^2}$$$$=\frac{2f^{{}^{″}}\left(x\right)}{2}=f^{{}^{″}}\left(x\right).$$

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

$$f^{″}\left(x\right)=\underset{h\to 0}{\lim }\frac{f{}^{\prime }(x+\frac{h}{2})-f{}^{\prime }(x-\frac{h}{2})}{h}$$$$f^{″}\left(x\right)=\underset{h\to 0}{\lim }\frac{\frac{f(x+h)-f\left(x\right)}{h}-\frac{f\left(x\right)-f(x-h)}{h}}{h}$$$$f^{″}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)+f(x-h)-2f\left(x\right)}{h^2}$$

Commented by ajfour last updated on 24/Jun/19

$$ThanksSir,niceway!$$

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