Where am I fault? Find g(x) when $\lim_{h\rightarrow 0}\frac{g(x+h)-2g(x)+g(x-h)}{h^2}=20x^3+6x.$

Question

I got some problems here and I can't understand where am I fault. The task says:

if $g$ is 2 times differentiable $,g'(0)=g(0)=1,$$$g''(x)=\lim_{h\rightarrow 0}\frac{g'(x)-g'(x-h)}{h},$$ and also $$20x^3+6x=\lim_{h\rightarrow 0}\frac{g(x+h)-2g(x)+g(x-h)}{h^2},$$ Prove that $$g(x)=x^5+x^3+x+1.$$

My Attempt: $$\lim_{h\rightarrow 0}\frac{g(x+h)-2g(x)+g(x-h)}{h^2}$$ $$=\lim_{h\rightarrow 0}\frac{g(x+h)-g(x)+g(x-h)-g(x)}{h^2}$$ $$=\lim_{h\rightarrow 0}\frac{g(x+h)-g(x)}{h^2}+\frac{g(x-h)-g(x)}{h^2}$$ $$=\lim_{h\rightarrow 0}\frac{1}{2}\frac{g'(x+h)-g'(x)}{h}+\frac{1}{2}\frac{g'(x-h)-g'(x)}{-(-h)}$$ $$=\lim_{h\rightarrow 0}\frac{1}{2}\frac{g'(x+h)-g'(x)}{h}-\frac{1}{2}\frac{-(g'(x)-g'(x-h))}{-h}$$ $$=\lim_{h\rightarrow 0}\frac{1}{2}\frac{g'(x+h)-g'(x)}{h}-\frac{1}{2}\frac{g'(x)-g'(x-h)}{h}$$ $$=\frac{g''(x)}{2}-\frac{g''(x)}{2}=0.$$ Where I can't find $g(x).$

Answer 1

It is better to use L'Hospital's Rule here. We are given that $g$ is twice differentiable. So we have \begin{align} 20x^{3} + 6x &= \lim_{h \to 0}\frac{g(x + h) - 2g(x) + g(x - h)}{h^{2}}\notag\\ &= \lim_{h \to 0}\frac{g'(x + h) - g'(x - h)}{2h}\text{ (via L'Hospital's Rule)}\notag\\ &= \frac{1}{2}\left(\lim_{h \to 0}\frac{g'(x + h) - g'(x)}{h} + \frac{g'(x - h) - g'(x)}{-h}\right)\notag\\ &= \frac{1}{2}\{g''(x) + g''(x)\}\notag\\ &= g''(x)\notag \end{align} Hence upon integrating the above equation we get $$g'(x) = 5x^{4} + 3x^{2} + A$$ for any constant $A$. Since $g'(0) = 1$ it means that $A = 1$. Further integration gives $$g(x) = x^{5} + x^{3} + x + B$$ for any constant $B$. Since $g(0) = 1$ it follows that $$g(x) = x^{5} + x^{3} + x + 1$$ The mistake in your attempt is in applying L'Hospital's Rule. Here the limit is taken as $h \to 0$ and hence the differentiation for L'Hospital Rule needs to be done with respect to $h$ (and not with respect to $x$ as you have done). Moreover as I have shown in my solution there is no need to split the $2g(x)$ term separately as $g(x) + g(x)$.

Answer 2

Taylor expand both $g(x+h)$ and $g(x-h)$: $$g(x+h)=g(x)+\frac{g'(x)}{1!}h+\frac{g''(x)}{2!}h^2+O(h^3)$$ $$g(x-h)=g(x)-\frac{g'(x)}{1!}h+\frac{g''(x)}{2!}h^2+O(h^3)$$

Add both equations:

$$g(x+h)-2g(x)+g(x-h)=g''(x)h^2+O(h^3)$$

$$g''(x)+O(h)=\frac{g(x+h)-2g(x)+g(x-h)}{h^2}$$

Neglect $O(h)$, by taking the limit $h\to 0$, and you have the result.

Considering this result, we can replace the limit with $g''(x)$ and integrate twice.

$$g''(x)=20x^3+6x$$ $$g'(x)=5x^4+3x^2+c_1$$ $$g(x)=x^5+x^3+c_1x+c_2$$

If you use $g'(x=0)$ and $g(x=0)$ you will see that $c_1=g'(x=0)=1$ and $c_2=g(x=0)=1$.