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Taylor series for getting different formulas

by ericm1234
Tags: formulas, series, taylor
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ericm1234
#1
Feb19-13, 04:08 AM
P: 71
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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tiny-tim
#2
Feb19-13, 04:36 AM
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tiny-tim's Avatar
P: 26,148
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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