We have the differential equation
We simulate it with 64-bits floating point. For a large value of n the calculated solution will be:
My attempt was to first solve it by hand:
The two roots: r_1=1/2 r_2=1/3
y_0=C+D=1 y_1=C 1/2+D 1/3=1/2 (1-D)1/2+D 1/3=1/2
1/2-1/2 D+D 1/3=1/2
-3/6 D+D 2/6=0 D=0 C=1
I don't see where the floating point inequarcy would lead to a wrong anwer for large n. Can you show where it does that?
A second order inhomoegeneous differential equation has the solution
If one are given two start values one would obtain that it is:
If the equation is simulated on a pc with 64-bits floating point one would get for large n that the computed result would be
From the assignment we see that C=0 and D=1 and A as a guessed solution is 1. How can you show that this would create overflow?© BrainMass Inc. brainmass.com October 10, 2019, 8:10 am ad1c9bdddf
Why does (1/2)^n appear as 0 for large values of n?
What is causing the overflow in this other expression for large values of n?