am trying to calculate the global error bound for Euler's method, but I am having trouble. I am given the formula |y(t_i)−u_i|<=1L(hM/2+delta/h)(e^(L(t_i−a))−1)+|delta_0|e^(L(t_i−a)) where ui is the Euler approxmation. I am also given M,L,a,delta,delta_0, h. If I am not mistaken this will give me the error for each step, but how do I find the upper bound for the total error?

Aleah Avery 2022-11-20 Answered
I am trying to calculate the global error bound for Euler's method, but I am having trouble. I am given the formula | y ( t i ) u i | 1 L ( h M 2 + δ h ) ( e L ( t i a ) 1 ) + | δ 0 | e L ( t i a ) where u i is the Euler approxmation. I am also given M , L , a , δ , δ 0 , h. If I am not mistaken this will give me the error for each step, but how do I find the upper bound for the total error?
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Answers (1)

Rebecca Benitez
Answered 2022-11-21 Author has 20 answers
sin 1 ( 2 x 1 + x 2 ) = θ sin ( θ ) = 2 x 1 + x 2 π 2 θ π 2
Note that
cos ( θ ) = | 1 x 2 1 + x 2 |
When | x | 1
tan ( θ / 2 ) = sin ( θ ) 1 + cos ( θ ) = 2 x 1 + x 2 1 + 1 x 2 1 + x 2 = x
Therefore,
θ = 2 tan 1 ( x )
When | x | > 1
tan ( θ / 2 ) = sin ( θ ) 1 + cos ( θ ) = 2 x 1 + x 2 1 1 x 2 1 + x 2 = 1 / x
Therefore,
θ = 2 tan 1 ( 1 / x )
Thus,
sin 1 ( 2 x 1 + x 2 ) = { 2 tan 1 ( x ) if  | x | 1 2 tan 1 ( 1 / x ) if  | x | > 1
and
tan 1 ( 1 / x ) = { π 2 tan 1 ( x ) if  x > 0 π 2 tan 1 ( x ) if  x < 0
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My question is the following: Having L small is good for convergence and stability (right?) in the sense that L small implies little change. If L = 0 then the function f ( t , y ) (on the right hand side of the ODE) is constant on y. If L < 1 the function is contracting. So why is L in the denominator of the bound? My intuition (which is not working good) says that the error bound should diminish as L diminshes.

What is wrong with my intuition here?