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(71B) Euler-Taylor method for the HP-71B
12-14-2019, 08:44 AM
Post: #10
RE: (71B) Euler-Taylor method for the HP-71B
(12-13-2019 02:12 PM)Albert Chan Wrote:  Δy' ≈ y'(x+h, y + h y'(x,y)) - y'(x,y)
→ y ≈ y + h y' + ½ h² y'' ≈ y + h y' + ½ h Δy'

50 D1=FND(X,Y)
60 X=X+H @ Y=Y+H*D1
70 Y=Y+H/2*(FND(X,Y)-D1)

Smile Thanks. You're lucky man, you have more spare time than me. It is an end-year-rush in power industry and today (Saturday) is a work day in Hungary (in this December two Saturdays are workdays - these are "relocated" to increase the Christmas holiday period).

So, you're lucky and thanks for the substitutions and the result - is looks like a Predictor-Corrector method generally, so I have a little comment - idea - on it (and yes, I have no time until 21 Dec to check it):

When you calculated the slope in the line 70 and you use the Euler estimation of Y, you get an estimated value of the slope in X+dX. But when Y new value is calculated, you have a possibility to improve this slope if you use the second order estimated Y value in X+dX. Maybe repeating the line 70 with the most precise Y value gives a better estimation?!

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RE: (71B) Euler-Taylor method for the HP-71B - Csaba Tizedes - 12-14-2019 08:44 AM

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