Hey folks...
So I'm supposed to numerically integrate this:
exp(-x*y*z), dxdydz. Bounds are 0 to 8 for all 3 vars.
What I've tried so far:
Integrate dx by hand...that's easy enough:
-(1-exp(-8*y*z))/y*z dydz.
Expand exp(-8*y*z), divide by yz, integrate again. You get an alternating series:
[512^n * (-1)^(n+1)]/(n^2 * n!). It converges to the desired accuracy in ~1420 runs...not exactly fast, and not exactly what my prof wants. But it yields the correct result.
Other things:
We want to get rid of the yz in the denominator, so take
x=8*e^u
y=8*e^v <--2 parameter change of variables
Result: 1-exp(-512*exp(u)*exp(v)) dudv; bounds are -infinity to 0 for both.
Now I'm stuck. I'm thinking I need to somehow reduce this to a single integral without the infinite bounds and I'll be golden...too bad I'm running dry on ideas 🙁 Ideally I'm aiming for a single var integration of a well behaved function so I can use the romberg method.
Edit: I thought of:
u= a - ln(b)
v= ln(b)
But then that yields an expression that I evaluate as infinity... booo and it had so much promise!
Thanks,
-Eric
So I'm supposed to numerically integrate this:
exp(-x*y*z), dxdydz. Bounds are 0 to 8 for all 3 vars.
What I've tried so far:
Integrate dx by hand...that's easy enough:
-(1-exp(-8*y*z))/y*z dydz.
Expand exp(-8*y*z), divide by yz, integrate again. You get an alternating series:
[512^n * (-1)^(n+1)]/(n^2 * n!). It converges to the desired accuracy in ~1420 runs...not exactly fast, and not exactly what my prof wants. But it yields the correct result.
Other things:
We want to get rid of the yz in the denominator, so take
x=8*e^u
y=8*e^v <--2 parameter change of variables
Result: 1-exp(-512*exp(u)*exp(v)) dudv; bounds are -infinity to 0 for both.
Now I'm stuck. I'm thinking I need to somehow reduce this to a single integral without the infinite bounds and I'll be golden...too bad I'm running dry on ideas 🙁 Ideally I'm aiming for a single var integration of a well behaved function so I can use the romberg method.
Edit: I thought of:
u= a - ln(b)
v= ln(b)
But then that yields an expression that I evaluate as infinity... booo and it had so much promise!
Thanks,
-Eric