Taylor series for getting different formulas

In summary, the conversation discusses the use of Taylor series and the elementary definition of derivative to establish the equation \frac{\partial u}{\partial t}(t+k/2, x)= (u(t+k,x)-u(t,x))/k + O(k^2). The speaker has attempted various combinations of adding and subtracting Taylor series, but has not been able to get an exact solution with an O(k^2) term. Another speaker suggests using the elementary definition of derivative as a limit and breaking down the equation into smaller parts.
  • #1
ericm1234
73
2
I am trying to establish why, I'm assuming one uses taylor series,
[itex] \frac{\partial u}{\partial t}[/itex](t+k/2, x)= (u(t+k,x)-u(t,x))/k + O(k^2)

I have tried every possible combination of adding/subtracting taylor series, but either I can not get it exactly or my O(k^2) term doesn't work out (it's O(k^1) or O(k^3) )
 
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  • #2
hi ericm1234! :smile:

no you don't need taylor, just use the elementary definition of derivative (as a limit) …

perhaps it's more obvious if you write (u(t+k,x)-u(t,x)) as (u(t+k,x)-u(t+k/2,x)) + (u(t+k/2,x)-u(t,x)) ? :wink:
 

What is a Taylor series and why is it useful?

A Taylor series is a mathematical representation of a function as an infinite sum of terms. It is useful because it allows us to approximate complicated functions with simpler ones and can be used to find derivatives and integrals of functions.

How do you find a Taylor series for a specific function?

To find a Taylor series for a specific function, you need to first determine the function's derivatives at a specific point. Then, you can use the Taylor series formula, which is: f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ... , where a is the point at which you are trying to find the Taylor series.

What is the difference between a Taylor series and a Maclaurin series?

A Taylor series is a representation of a function at any point, while a Maclaurin series is a special case of a Taylor series where the point a is 0. In other words, a Maclaurin series is a Taylor series centered at 0.

How accurate is a Taylor series approximation?

The accuracy of a Taylor series approximation depends on the number of terms included in the series. The more terms you include, the more accurate the approximation will be. However, for functions that are not infinitely differentiable, the Taylor series may not accurately represent the function.

Can Taylor series be used to find exact values of functions?

No, Taylor series are infinite series and can only provide approximations of functions. They can get closer and closer to the exact value, but they will never be exactly equal to the function.

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