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## Main Question or Discussion Point

Sorry, I mistakenly reported my own post last time. But later I realized that these limits do work. So, I'm posting this again.

I'm using these limits to check second-order differentiability:

$$\lim_{h\rightarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$

And,

$$\lim_{h\rightarrow 0}\frac{f(x-2h)-2f(x-h)+f(x)}{h^2}$$

I think that when these limits are equal, then the function is twice differentiable.

Now, I checked it for this function:

$$f(x)=\frac{x^2}{2},x\geq0$$

$$=\frac{-x^2}{2},x<0$$.

Now, this function is differentiable once at ##x=0##. But after the first differentiation, its derivative comes out to be ##|x|## which isn't differentiable at ##x=0##. So, this function is not twice differentiable at ##x=0##. So, those limits should give different values.

Now,$$\lim_{h\rightarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$ At ##x=0##,

$$=\lim_{h\rightarrow 0}\frac{f(2h)-2f(h)+f(0)}{h^2}$$

$$=\lim_{h\rightarrow 0}\frac{\frac{(2h)^2}{2}-2\frac{h^2}{2}+0}{h^2}$$

$$=2-1=1$$

And, at ##x=0##,

$$\lim_{h\rightarrow 0}\frac{f(x-2h)-2f(x-h)+f(x)}{h^2}$$

$$=\lim_{h\rightarrow 0}\frac{f(-2h)-2f(-h)+f(0)}{h^2}$$

$$=\lim_{h\rightarrow 0}\frac{\frac{-(-2h)^2}{2}+2\frac{-h^2}{2}+0}{h^2}$$

$$=-2+1=-1$$

So, these limits are not equal, which means ##f(x)## is not twice differentiable at ##x=0##.

So, I think this limit works. Is there some way to prove that a function is twice differentiable only if these two limits are equal?

I'm using these limits to check second-order differentiability:

$$\lim_{h\rightarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$

And,

$$\lim_{h\rightarrow 0}\frac{f(x-2h)-2f(x-h)+f(x)}{h^2}$$

I think that when these limits are equal, then the function is twice differentiable.

Now, I checked it for this function:

$$f(x)=\frac{x^2}{2},x\geq0$$

$$=\frac{-x^2}{2},x<0$$.

Now, this function is differentiable once at ##x=0##. But after the first differentiation, its derivative comes out to be ##|x|## which isn't differentiable at ##x=0##. So, this function is not twice differentiable at ##x=0##. So, those limits should give different values.

Now,$$\lim_{h\rightarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$ At ##x=0##,

$$=\lim_{h\rightarrow 0}\frac{f(2h)-2f(h)+f(0)}{h^2}$$

$$=\lim_{h\rightarrow 0}\frac{\frac{(2h)^2}{2}-2\frac{h^2}{2}+0}{h^2}$$

$$=2-1=1$$

And, at ##x=0##,

$$\lim_{h\rightarrow 0}\frac{f(x-2h)-2f(x-h)+f(x)}{h^2}$$

$$=\lim_{h\rightarrow 0}\frac{f(-2h)-2f(-h)+f(0)}{h^2}$$

$$=\lim_{h\rightarrow 0}\frac{\frac{-(-2h)^2}{2}+2\frac{-h^2}{2}+0}{h^2}$$

$$=-2+1=-1$$

So, these limits are not equal, which means ##f(x)## is not twice differentiable at ##x=0##.

So, I think this limit works. Is there some way to prove that a function is twice differentiable only if these two limits are equal?