Differantiation proof question

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In summary, if f(x) is differentiable twice at x_0, then f''(x_0) can be found by taking the limit as h approaches 0 of the quotient of the second derivative of f at x_0+h and x_0-h, divided by 2h. This can also be calculated using L'Hôpital's Rule by taking the limit as h approaches 0 of the average of the second derivatives of f at x_0+h and x_0-h.
  • #1
transgalactic
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f(x) is differentiable twice at x_0
prove that:

[tex]
f''(x_0 ) = \mathop {\lim }\limits_{x \to \infty } {{f(x_0 + h) - 2f(x_0 ) + f(x_0 - h)} \over {h^2 }}
[/tex]
 
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  • #2
If g(x) is differentiable at x_0, what is the definition of g'(x_0) ?
Now apply this to g := f'.
 
  • #3


transgalactic said:
f(x) is differentiable twice at x_0
prove that:

[tex]
f''(x_0 ) = \mathop {\lim }\limits_{x \to \infty } {{f(x_0 + h) - 2f(x_0 ) + f(x_0 - h)} \over {h^2 }}
[/tex]

The limit should be as h approaches 0. You have it as x approaches [itex]\infty[/itex], which makes no sense, and besides, x doesn't even appear in the limit expression.
 
  • #4
this is the correct function:
[tex]

f''(x_0 ) = \mathop {\lim }\limits_{h \to \infty } {{f(x_0 + h) - 2f(x_0 ) + f(x_0 - h)} \over {h^2 }}

[/tex]

i don't know what means g'(x_0)
if g(x) is differentiable at x_0
??
 
  • #5
transgalactic said:
this is the correct function:
[tex]

f''(x_0 ) = \mathop {\lim }\limits_{h \to \infty } {{f(x_0 + h) - 2f(x_0 ) + f(x_0 - h)} \over {h^2 }}

[/tex]
No it's still not the correct function... the limit should be [itex]h \to 0[/itex] and not [itex]h \to \infty[/itex].

i don't know what means g'(x_0)
if g(x) is differentiable at x_0
??

Come on, you should know the definition of the derivative! Look it up in your textbook if you need to.

Have you never seen the justification for the definition (with the first order approximation of the function and stuff)?
 
  • #6
i know tailor series approximations but its not telling
what means g'(x_0)
if g(x) is differentiable at x_0
??
 
  • #7
Surely you have seen

[tex]g'(x_0) = \lim_{h \to 0} \frac{g(x_0 + h) - g(x_0)}{h}[/tex]
or equivalently
[tex]g'(x_0) = \lim_{x \to x_0} \frac{g(x) - g(x_0)}{|x - x_0|}[/tex]
?
 
  • #8
its a proof for a differentiability of a functions
so if its differential twice at point x_0 then

[tex]
f''(x_0) = \lim_{x \to x_0} \frac{f'(x) - f'(x_0)}{|x - x_0|}
[/tex]
and i input f'(x_0) expression
 
  • #9
You better use the other one (with h) :P
 
  • #10
i tried to solve it but i can't get to the asked expression
[tex]
g'(x_0) = \lim_{h \to 0} \frac{g(x_0 + h) - g(x_0)}{h}\\
[/tex]
[tex]
g''(x_0) = \lim_{h \to 0} \frac{g'(x_0 + h) - g'(x_0)}{h}\\
[/tex]
[tex]
g''(x_0) = \lim_{h \to 0} \frac{\frac{g(x_0 + 2h) - g(x_0+h)}{h} - \frac{g(x_0 + h) - g(x_0)}{h}}{h}\\
[/tex]
[tex]
g''(x_0) = \lim_{h \to 0} \frac{{g(x_0 + 2h) - g(x_0+h)} - {g(x_0 + h) - g(x_0)}}{h^2}\\
[/tex]
[tex]
g''(x_0) = \lim_{h \to 0} \frac{{g(x_0 + 2h) - 2g(x_0+h)} { - g(x_0)}}{h^2}\\
[/tex]
??
 
  • #11
It is also true that the derivative can be calculated as
[tex]g'(x_0)= \lim_{h\rightarrow 0}\frac{g(x_0+ h/2)- g(x_0- h/2)}{h}[/tex]
(this is the "centered" formula)_
Try using that.
 
  • #12
It seems to me a more direct (and easy) method for tackling this problem involves applying L'Hôpital's Rule: the top approaches 0 as h approaches 0, as does the bottom. We know f is differentiable, so differentiate the top and bottom with respect to h. We can now consider the following limit (if it exists):

[tex]\lim_{h \rightarrow 0} \frac{f'(x_{0}+h) - f'(x_{0}-h)}{2h}[/tex]

As f is twice differentiable, and we have both numerator and denominator approaching 0, we can apply L'Hôpital's Rule again to end with:

[tex]\lim_{h \rightarrow 0} \frac{f''(x_{0}+h) + f''(x_{0}-h)}{2}[/tex]

which exists and equals f''(x0).
 

Related to Differantiation proof question

What is differentiation?

Differentiation is a mathematical process used to find the rate of change of a function at a specific point. It involves calculating the slope of a curve at a given point and can be used to find the instantaneous rate of change at that point.

What is a differentiation proof question?

A differentiation proof question is a mathematical problem that requires the use of differentiation to prove a certain statement or find a specific value. These types of questions typically involve finding derivatives, using the rules of differentiation, and solving for unknown variables.

What are the steps for solving a differentiation proof question?

The steps for solving a differentiation proof question are:
1. Identify the function and the variable you are differentiating with respect to.
2. Use the rules of differentiation (power rule, product rule, quotient rule, etc.) to find the derivative.
3. Simplify the derivative and substitute in any given values.
4. Solve for the unknown variable or prove the given statement.

What are the common mistakes made when solving a differentiation proof question?

Some common mistakes made when solving a differentiation proof question include:
- Forgetting to apply the chain rule when differentiating composite functions.
- Misapplying the product or quotient rule.
- Making arithmetic errors when simplifying the derivative.
- Not substituting in the given values correctly.
To avoid these mistakes, it is important to double check your work and practice using the rules of differentiation.

How can I improve my skills in solving differentiation proof questions?

The best way to improve your skills in solving differentiation proof questions is to practice regularly. Start with simple problems and gradually work your way up to more challenging ones. Additionally, make sure to understand the rules of differentiation and how they are applied in different scenarios. Seeking help from a tutor or joining a study group can also be beneficial in improving your skills.

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