MaplePrimes - answers and comments on Question, Getting a weird result with integration of an absolute value

absolute value

Pseudomodo — Tue, 17 Jan 2012 00:44:14 Z

Why are you taking the absolute value in your assignment to ErTime? Alternatively, if you take the absolute value of the error, then why do you expect it to integrate to zero?

As shown by your first plot, at first one curve lies above, and then this switches. And ErTime may be defined as one minus the other. What's your objection to defining ErTime as the simple difference -- allowing for and retaining the sign -- without taking the absolute value?

The presence of the sign of ErTime at every point in time is what allows for error "one way" (red above green) to be compensated for later on by error "the other way" (green above red).

If you remove the absolute value from the definition of ErTime, then all the plots should make sense. The very last plot, of the integral of ErTime from zero to K, will tend to zero as K the upper limit of initegration tends to infinity. And the slope of that last plot will change sign at the same point in time that the red and green curves cross over in the first plot -- namely at the point at which ErTime itself the becomes zero, and incidentally in agreement with the middle plot.

Simplification of TotEr?

Markiyan Hirnyk — Tue, 17 Jan 2012 01:04:01 Z

I have doubt about the simplification of TotEr. Here are the arguments. Let us consider

> TotEr := int(ErTime, t = 0 .. infinity);

You apply the ?expand command to this output. Such usage is not described in Maple Help. It can produce a wrong result or NULL. For example,

> expand(limit(exp(-x)/(exp(-x)+sin(x)*exp(-2*x)), x = infinity));

On the other hand,

> eval(ErTime, [k[0] = 3, k[2] = .1, x[2](0) = 30, k[1] = .2]);

|1.000000000 + 0.06666666667 exp(-0.2 t) x[1](0) - 1.000000000 exp(-0.2 t) +

3.333333333 (-0.3 exp(0.1 t) + 0.02 x[1](0) exp(-0.1 t) - 0.3 exp(-0.1 t)

+ 0.30 - 0.02 x[1](0)) exp(-0.1 t)|
and

> TotEr := int(eval(ErTime, [k[0] = 3, k[2] = .1, x[2](0) = 30, k[1] = .2]), t = 0 .. infinity);

Float(infinity)

exact vs floating-point

pagan — Tue, 17 Jan 2012 19:17:22 Z

Using 64bit Maple 15.01 on Windows 7, I get

int(simplify(eval(ErTime,
                       [k[0]=3,k[1]=2/10,k[2]=1/10,x[1](0)=0,x[2](0)=30])),
    t = 0 .. infinity);

                                          5

It looks like the floating-point values k[1]=0.2 and k[2]=0.1 are causing the problem and producing the infinite result that Markiyan showed.

I have not pinpointed the particular place that thing are going astray with the float values for the parameters.

Thank you for your help. I don't expect

TommySnowman — Tue, 17 Jan 2012 16:14:12 Z

Thank you for your help.

I don't expect the integral to be zero. Any confusion is probably due to my lack of clarity though, apologies about that.

Rather, the spurious result that maple gives me (almost certainly due to my own spurious use of the expand function) suggests that the integral, even with the absolute values, gives zero error in the specific case where k_0 = k_2 x_{2}(0). The plots I showed demonstrate that this can't be true.

As for why I am interested in having the absolute value of the error - I only want zero error in the case where the two curves are identical. Not where the area before and after crossing cancel out. I want this to be the case because otherwise my approximation would appear to be a good predictor of the exact result for all times when it is not.

thanks

Pseudomodo — Tue, 17 Jan 2012 23:33:02 Z

@TommySnowman Thanks for clearing that up. I misunderstood which you expected.

Hi, thanks for your help. I agree

TommySnowman — Tue, 17 Jan 2012 16:21:20 Z

Hi, thanks for your help.

I agree that the issue is probably my use of the expand function, although I disagree

that the TotEr should be infinite. I think Maple is making an error here - I mean look

at the following:

> int(subs([k[0] = 3, k[1] = .2, k[2] = .1, x[1](0) = 0, x[2](0) = 30], ErTime),

 t = 0 .. 100);
                          4.999546030

> int(subs([k[0] = 3, k[1] = .2, k[2] = .1, x[1](0) = 0, x[2](0) = 30], ErTime),

t = 0 .. 1000);
                          5.000000098

> int(subs([k[0] = 3, k[1] = .2, k[2] = .1, x[1](0) = 0, x[2](0) = 30], ErTime),

t = 0 .. 10000);
                          5.000000998

> int(subs([k[0] = 3, k[1] = .2, k[2] = .1, x[1](0) = 0, x[2](0) = 30], ErTime),

t = 0 .. 100000);
                          5.000009998

> int(subs([k[0] = 3, k[1] = .2, k[2] = .1, x[1](0) = 0, x[2](0) = 30], ErTime),

t = 0 .. 10000000);
                          5.000999998

> int(subs([k[0] = 3, k[1] = .2, k[2] = .1, x[1](0) = 0, x[2](0) = 30], ErTime),

t = 0 .. 1000000000);
                          5.099999998

> int(subs([k[0] = 3, k[1] = .2, k[2] = .1, x[1](0) = 0, x[2](0) = 30], ErTime),

t = 0 .. 100000000000);
                          15.00000000

This seems very odd to me, and I can see no way in which the function would jump

back into 'life' like this.