Proof related to derivative and Big O notation

[songoku]

Homework Statement

Please see below

Relevant Equations

Limit

Derivative

My attempt:
Since $\frac{f(a+h)-f(a-h)}{2h}-f'(a)=O(h^2)$ as $h \to 0$, then:

$$\lim_{h \to 0} \frac{\frac{f(a+h)-f(a-h)}{2h}-f'(a)}{h^2} < \infty$$

So

$$\lim_{h \to 0} \frac{\frac{f(a+h)-f(a-h)}{2h}-f'(a)}{h^2} = \lim_{h \to 0} \frac{f'(a)-f'(a)}{h^2}=0 < \infty$$

Because the value of the limit is zero, it is true that $\frac{f(a+h)-f(a-h)}{2h}-f'(a)=O(h^2)$ as $h \to 0$

Is this even correct?

Thanks

 

[Hill]

This step:
$$\lim_{h \to 0} \frac{\frac{f(a+h)-f(a-h)}{2h}-f'(a)}{h^2} = \lim_{h \to 0} \frac{f'(a)-f'(a)}{h^2}$$
is incorrect.

 

[songoku]

This step:
$$\lim_{h \to 0} \frac{\frac{f(a+h)-f(a-h)}{2h}-f'(a)}{h^2} = \lim_{h \to 0} \frac{f'(a)-f'(a)}{h^2}$$
is incorrect.

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

I can't do that? Thanks

 

[FactChecker]

Homework Statement: Please see below
Relevant Equations: Limit

Derivative

View attachment 336113

My attempt:
Since $\frac{f(a+h)-f(a-h)}{2h}-f'(a)=O(h^2)$ as $h \to 0$, then:

I'm confused. Isn't this what you are supposed to prove?

$$\lim_{h \to 0} \frac{\frac{f(a+h)-f(a-h)}{2h}-f'(a)}{h^2} < \infty$$

So

$$\lim_{h \to 0} \frac{\frac{f(a+h)-f(a-h)}{2h}-f'(a)}{h^2} = \lim_{h \to 0} \frac{f'(a)-f'(a)}{h^2}=0 < \infty$$

Because the value of the limit is zero, it is true that $\frac{f(a+h)-f(a-h)}{2h}-f'(a)=O(h^2)$ as $h \to 0$

I don't see where you have used the fact that $f$ has continuous derivatives up to order 3 in an open interval containing $a$

Is this even correct?

I think not.

 

[songoku]

I'm confused. Isn't this what you are supposed to prove?

I don't see where you have used the fact that $f$ has continuous derivatives up to order 3 in an open interval containing $a$

Yeah, actually that would be my next question after discussing my really bad approach. How to use the third and/or second derivative to prove the question? I don't see any higher-order derivative than 1 in the question, that's why I started from the limit.

Thanks

 

[renormalize]

How to use the third and/or second derivative to prove the question? I don't see any higher-order derivative than 1 in the question, that's why I started from the limit.

Have you learned about Taylor series? Are you allowed to use them for this problem?

 

[Hill]

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

I can't do that? Thanks

No, on your last step above, instead of $$=\frac{0}{\lim_{h \to 0} h^2}=0$$ you'd get $$=\frac{0}{\lim_{h \to 0} h^2}=\frac 0 0$$

 

[PeroK]

Have you learned about Taylor series? Are you allowed to use them for this problem?

The problem may include functions for which the Taylor series does not exist. This is pure mathematics!

 

[PeroK]

Yeah, actually that would be my next question after discussing my really bad approach. How to use the third and/or second derivative to prove the question? I don't see any higher-order derivative than 1 in the question, that's why I started from the limit.

Thanks

This looks tricky. It's not clear to me why you need derivatives up to order three. You'd think that $f'$ being continuous would be sufficient. If not, then the solution must get quite involved. Those are my initial thoughts.

Possibly start with the continuity of $f'''$? And see what you get starting from there?

 

[Hill]

The existence of three derivatives allows to apply L'Hôpital's rule three times. This is all we need.

 

[fresh_42]

My understanding is that you are asked to show the equivalence of your definition of a derivative (I can only guess that this is the usual definition as a limit) and Weierstraß's formula
$$
f(a+h)=f(a)+ J_a(h) + r(h)
$$
where $J_a(h)=f'(a)\cdot h$ is the Jacobi matrix of $f$ that is the derivarive at $x=a.$ $r(h)$ is the remainder that tends faster to zero than $h,$ i.e. $\lim_{h \to 0}\dfrac{r(h)}{h}=0.$

You do not need a Taylor series for that. It is just the linear approximation of $f$ by its derivative $f'$ at $x=a$. All higher-order terms of the Taylor series are in $O(h^2)$ or $r(h).$ You only need the separation of the linear term and all the rest. Taylor would be an overkill since there would be nothing left to prove. Use the definitions of a derivative and a limit, that's all.

 

[PeroK]

Yeah, actually that would be my next question after discussing my really bad approach. How to use the third and/or second derivative to prove the question? I don't see any higher-order derivative than 1 in the question, that's why I started from the limit.

Thanks

What is the context of this question? I was wondering whether this is in connection with the formal development of Taylor series? In any case, I've found Taylor's Theorem:

https://en.wikipedia.org/wiki/Taylor's_theorem#Taylor's_theorem_in_one_real_variable

 

[songoku]

Have you learned about Taylor series? Are you allowed to use them for this problem?

Yes I have and I am allowed to use them for this problem. It is just I don't know how to use them

No, on your last step above, instead of $$=\frac{0}{\lim_{h \to 0} h^2}=0$$ you'd get $$=\frac{0}{\lim_{h \to 0} h^2}=\frac 0 0$$

Oh ok, I see it now.

What is the context of this question? I was wondering whether this is in connection with the formal development of Taylor series? In any case, I've found Taylor's Theorem:

https://en.wikipedia.org/wiki/Taylor's_theorem#Taylor's_theorem_in_one_real_variable

This is an exercise for practice for test as part of Calculus course, covering topics starting from function (domain, range, etc) up until Taylor (including limit, derivative, integration, ode).

My understanding is that you are asked to show the equivalence of your definition of a derivative (I can only guess that this is the usual definition as a limit) and Weierstraß's formula
$$
f(a+h)=f(a)+ J_a(h) + r(h)
$$
where $J_a(h)=f'(a)\cdot h$ is the Jacobi matrix of $f$ that is the derivarive at $x=a.$ $r(h)$ is the remainder that tends faster to zero than $h,$ i.e. $\lim_{h \to 0}\dfrac{r(h)}{h}=0.$

You do not need a Taylor series for that. It is just the linear approximation of $f$ by its derivative $f'$ at $x=a$. All higher-order terms of the Taylor series are in $O(h^2)$ or $r(h).$ You only need the separation of the linear term and all the rest. Taylor would be an overkill since there would be nothing left to prove. Use the definitions of a derivative and a limit, that's all.

$$1) \lim_{h\to 0} \frac{f(a+h)-f(a)}{h}=f'(a)$$
$$2) \lim_{h\to 0} \frac{f'(a+h)-f'(a)}{h}=f"(a)$$
$$3) \lim_{h\to 0} \frac{f"(a+h)-f"(a)}{h}=f^{3} (a)$$
$$4) \lim_{h\to 0} \frac{f(a+h)-f(a-h)}{2h}=f'(a)$$
$$5) \lim_{h\to 0} \frac{f(a+h)-f(a-h)}{h^2}=f"(a)$$

Do you mean that is the starting point? Thanks

 

[fresh_42]

$$1) \lim_{h\to 0} \frac{f(a+h)-f(a)}{h}=f'(a)$$
$$2) \lim_{h\to 0} \frac{f'(a+h)-f'(a)}{h}=f"(a)$$
$$3) \lim_{h\to 0} \frac{f"(a+h)-f"(a)}{h}=f^{3} (a)$$
$$4) \lim_{h\to 0} \frac{f(a+h)-f(a-h)}{2h}=f'(a)$$
$$5) \lim_{h\to 0} \frac{f(a+h)-f(a-h)}{h^2}=f"(a)$$

Do you mean that is the starting point? Thanks

Yes.

If we have Weierstraß $f(a+h)=f(a) +f'(a)\cdot h +r(h)$ with $\lim_{h \to 0}\dfrac{r(h)}{h}=0$ then $f(a-h)=f(a) -f'(a)\cdot h +r(h)$ and $\dfrac{f(a+h)-f(a-h)}{2}=f'(a)\cdot h + r(h)$ where $r(h)$ are possibly different remainders. However, $\lim_{h \to 0}\dfrac{r(h)}{h}=0$ holds for all of them. So writing the remainder as a power series yields
$$
\dfrac{r(h)}{h}=\dfrac{r_0}{h}=\dfrac{r_0}{h}+r_1+r_2h+O(h^2) \stackrel{h\to 0}{\longrightarrow }0
$$
and therefore $r_0=r_1=0$ and $r(h)=r_2h^2+O(h^3)=O(h^2).$

The Weierstraß decomposition formula is equivalent to differentiability.
(e.g. https://de.wikipedia.org/wiki/Weierstraßsche_Zerlegungsformel#Beweis)
I haven't searched for an English version, but the formulas should do, or you use Chrome and let it translate the page to English (right-click), or you look on the internet for an English proof, or you prove it as an exercise, which I assumed this exercise is all about.

 

[PeroK]

This is an exercise for practice for test as part of Calculus course, covering topics starting from function (domain, range, etc) up until Taylor (including limit, derivative, integration, ode).

Use L'Hopital's rule, as suggested above.

 

[songoku]

$$
\dfrac{r(h)}{h}=\dfrac{r_0}{h}=\dfrac{r_0}{h}+r_1+r_2h+O(h^2) \stackrel{h\to 0}{\longrightarrow }0
$$
and therefore $r_0=r_1=0$ and $r(h)=r_2h^2+O(h^3)=O(h^2).$

Sorry I don't understand this part. Why $\dfrac{r(h)}{h}=\dfrac{r_0}{h}?$ Then shouldn't it be $r(h)=r_0$?

And also why $r_0=r_1=0$?

Use L'Hopital's rule, as suggested above.

$$\lim_{h\to 0} \frac{\frac{f(a+h)-f(a-h)}{2h}-f'(a)}{h^2}$$
$$=\lim_{h\to 0}\frac{f(a+h)-f(a-h)-f'(a).2h}{2h^3}$$
$$=\lim_{h\to 0} \frac{f'(a+h)+f'(a-h)-2f'(a)}{6h^2}$$
$$=\lim_{h\to 0} \frac{f"(a+h)-f"(a-h)}{12h}$$
$$=\lim_{h\to 0} \frac{f^3 (a+h)+f^3 (a+h)}{12}$$
$$=\frac{f^3 (a)}{6}$$

Is this what you mean?

Thanks

 

[PeroK]

That's exactly what @Hill and I meant!

 

[Hill]

Sorry I don't understand this part. Why $\dfrac{r(h)}{h}=\dfrac{r_0}{h}?$ Then shouldn't it be $r(h)=r_0$?

And also why $r_0=r_1=0$?$$\lim_{h\to 0} \frac{\frac{f(a+h)-f(a-h)}{2h}-f'(a)}{h^2}$$
$$=\lim_{h\to 0}\frac{f(a+h)-f(a-h)-f'(a).2h}{2h^3}$$
$$=\lim_{h\to 0} \frac{f'(a+h)+f'(a-h)-2f'(a)}{6h^2}$$
$$=\lim_{h\to 0} \frac{f"(a+h)-f"(a-h)}{12h}$$
$$=\lim_{h\to 0} \frac{f^3 (a+h)+f^3 (a+h)}{12}$$
$$=\frac{f^3 (a)}{6}$$

Is this what you mean?

Thanks

Except a small typo in the line before the last.

P.S. This is where the continuity of the third derivative is used.

 

[fresh_42]

Say $r(h)=\sum_{k=0}^\infty r_kh^k.$ Then
$$
0=\lim_{h \to 0}\dfrac{r(h)}{h}=\lim_{h \to 0}\sum_{k=0}^\infty r_kh^{k-1}\stackrel{h\to 0}{=}\lim_{h \to 0}\sum_{k=0}^1 r_kh^{k-1}=\lim_{h \to 0}\left(\dfrac{r_0}{h}+r_1\right)
$$
which can only be if $r_0=r_1=0.$ Thus $r(h)=\sum_{k=2}^\infty r_kh^k=O(h^2).$

 

[songoku]

Thank you very much for all the help and explanation Hill, FactChecker, renormalize, PeroK, fresh_42