Taylor series for getting different formulas

ericm1234

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) )

tiny-tim

hi ericm1234!

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)) ?

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ericm1234	Taylor series for getting different formulas Tweet 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) )

tiny-tim Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor	hi ericm1234! 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)) ?