What is the reason behind the multiplication of the function's derivative with the step size (and the subsequent addition) in the numerical Euler method? y_(n+1)=y_n+hf(t_n,y_n) I can't figure out why exactly this would work for generating a new value. How can scaling the output of the function f and subsequently adding it to the prior value y_n approximate a new value? Is there some formal or graphical explanation of this? The formula seems somewhat counterintuitive.

Nico Patterson

Nico Patterson

Answered question

2022-11-18

What is the reason behind the multiplication of the function's derivative with the step size (and the subsequent addition) in the numerical Euler method?
y n + 1 = y n + h f ( t n , y n )
I can't figure out why exactly this would work for generating a new value. How can scaling the output of the function f and subsequently adding it to the prior value y n approximate a new value? Is there some formal or graphical explanation of this? The formula seems somewhat counterintuitive.

Answer & Explanation

Laura Fletcher

Laura Fletcher

Beginner2022-11-19Added 22 answers

I would say Euler's method is actually the most intuitive of the numerical methods!
We have a function f that tells us the derivative of a solution to the ODE passing through our current point ( t n , y n ). What is the derivative? In a small enough interval around y n , it's approximately the "rise over run", so if h = t n + 1 t n is 'small' then we can use this approximation:
y ( t n + 1 ) y n t n + 1 t n f ( t n , y n )
y ( t n + 1 ) y n + h f ( t n , y n )
kemecryncqe9

kemecryncqe9

Beginner2022-11-20Added 6 answers

It helped a lot)

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?