# Using a Taylor expansion to prove equality

• LogarithmLuke
What happens when you add the two expansions?In summary, the problem is showing that the difference between the central difference operator for the second order derivative and the actual second order derivative is of order ##h^2##. This can be done by expanding ##u(x+h)## and ##u(x-h)## using the Taylor expansion, and then adding them together.
LogarithmLuke
Homework Statement
Use Taylor expansion to show that for ##u \in C^4([0,1]) ## $$max |\partial^+\partial^-u(x) - u''(x)| = \mathcal{O}(h^2)$$ For ##x \in [0,1]## and where the second order derivative ##u''## can be approximated by the central difference operator defined by $$\partial^+\partial^-u(x) = \frac{u(x+h) - 2u(x) + u(x-h)}{h^2} \approx u''(x)$$
Relevant Equations
$$u(x+h) = u(x) + hu'(x) \frac{h^2}{2!}u''(x) + ... + \frac{h^{k-1}}{(k-1)!}u^{(k-1)}(x) + \mathcal{O}(h^k)$$
Homework Statement: Use Taylor expansion to show that for ##u \in C^4([0,1]) ## $$max |\partial^+\partial^-u(x) - u''(x)| = \mathcal{O}(h^2)$$ For ##x \in [0,1]## and where the second order derivative ##u''## can be approximated by the central difference operator defined by $$\partial^+\partial^-u(x) = \frac{u(x+h) - 2u(x) + u(x-h)}{h^2} \approx u''(x)$$
Homework Equations: $$u(x+h) = u(x) + hu'(x) \frac{h^2}{2!}u''(x) + ... + \frac{h^{k-1}}{(k-1)!}u^{(k-1)}(x) + \mathcal{O}(h^k)$$

Mentor note:
Thread moved from Precalc section, as it is well beyond precalculus types of questions. @LogarithmLuke, please post your questions in the appropriate forum section.

I know what the Taylor expansion of ##u## around ##x+h## looks like, but I don't know how to evaluate ##u′′(x)## other than setting it equal to the approximation ##\partial^+\partial^-u(x)## which makes the left side equal ##0##

Last edited by a moderator:
LogarithmLuke said:
Homework Statement: Use Taylor expansion to show that for ##u \in C^4([0,1]) ## $$max |\partial^+\partial^-u(x) - u''(x)| = \mathcal{O}(h^2)$$ For ##x \in [0,1]## and where the second order derivative $u''$ can be approximated by the **central difference operator** defined by $$\partial^+\partial^-u(x) = \frac{u(x+h) - 2u(x) + u(x-h)}{h^2} \approx u''(x)$$
Homework Equations: $$u(x+h) = u(x) + hu'(x) \frac{h^2}{2!}u''(x) + ... + \frac{h^{k-1}}{(k-1)!}u^{(k-1)}(x) + \mathcal{O}(h^k)$$

I know what the Taylor expansion of ##u## around ##x+h## looks like, but I don't know how to evaluate ##u''(x)## other than setting it equal to the approximation ##\partial^+\partial^-u(x)## which makes the left side equal ##0##.

Hint: if you know how to expand ##u(x + h)## you must know how to expand ##u(x -h)## as well!

## 1. What is a Taylor expansion?

A Taylor expansion is a mathematical series that represents a function as an infinite sum of terms, each of which is a derivative of the function evaluated at a specific point.

## 2. How is a Taylor expansion used to prove equality?

A Taylor expansion can be used to approximate the value of a function at a specific point. By setting the approximated value equal to the actual value of the function, we can prove equality.

## 3. What are the steps involved in using a Taylor expansion to prove equality?

The steps involve in using a Taylor expansion to prove equality are:
1. Choose a function to be approximated
2. Choose a point around which to expand the function
3. Write out the Taylor series expansion of the function
4. Set the approximate value equal to the actual value of the function
5. Simplify and rearrange to prove equality.

## 4. What are the limitations of using a Taylor expansion to prove equality?

A Taylor expansion can only be used to prove equality for infinitely differentiable functions. Additionally, the approximation may not be accurate if the function has a large curvature at the chosen point of expansion.

## 5. How is a Taylor expansion different from other methods of proving equality?

A Taylor expansion is a specific method of approximation that uses derivatives of a function to represent it as a series. Other methods of proving equality may involve direct substitution, algebraic manipulation, or other mathematical techniques.

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