Taylor series for getting different formulas


by ericm1234
Tags: formulas, series, taylor
ericm1234
ericm1234 is offline
#1
Feb19-13, 04:08 AM
P: 67
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) )
Phys.Org News Partner Science news on Phys.org
NASA's space station Robonaut finally getting legs
Free the seed: OSSI nurtures growing plants without patent barriers
Going nuts? Turkey looks to pistachios to heat new eco-city
tiny-tim
tiny-tim is offline
#2
Feb19-13, 04:36 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,167
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)) ?


Register to reply

Related Discussions
Shortcut to taylor series of f, given taylor series of g Calculus 2
Inverse Laplace Transfrom - Taylor Series/Asymptotic Series?! Calculus & Beyond Homework 2
check my solution to a power series/taylor series problem Calculus & Beyond Homework 2
Taylor Series using Geometric Series and Power Series Calculus & Beyond Homework 5
[URGENT] Taylor Series without using the built-in MATLAB "Taylor's Function" Math & Science Software 1