DarkMath

125 Reputation

4 Badges

4 years, 31 days

MaplePrimes Activity


These are replies submitted by DarkMath

@tomleslie I re-read it again and I still think that my question is not answered there.

The dot product of a free vector (cartesian) with a rooted vector (spherical): no error is given when theta=pi/4, but throws an error when theta=pi/6. If this behaviour is explained, could you please kindly point me to where it is written exactly then?

@tomleslie This question is actually inspired from an expample on that page. I thought during the dot product, it converts the free vector into rooted vector using the rooted vectors root, but apparently, it doesn't and I am confused. 

@vv @Carl Love Thanks a lot. I really like this way of solving. Much better than writing a for loop 

@Carl Love Makes sense. I hope Maplesoft integrates this in the next version

@Carl Love In @Axel Vogt 's example above. It shouldn't be related to (explicit,allsolutions), though. Since there is only one root, but still the integral is left unevaluated. Don't you think so?

@Carl Love Are those ineffective epecially against differential equations? or ineffective in general?

@Axel Vogt Could this be a bug? or something else going on?

@acer I was just about to ask how to get all the solutions for a given interval. 

Thanks for the example. It is weird that I didn't see those keywords under solve's documentation. Did I miss something?

I tried this 

solve({sin(x)},x,real,explicit,allsolutions) assuming -3/2*Pi<=x<=3/2*Pi

But this gives 

{x=Pi_Z14~}

I take Z14 is just an arbitrary constant here, but this means that it didn't take the assumption into acount. 

So, using assume like this is not allowed in solve?

@nm This is exactly what I initially did to verify the result 3. I was expecting Maple to handle this easily, though. 

@vv Quite interesting example. But in the sin(x) case in between -3/2Pi and 3/2Pi, there are three real-valued roots r1=-Pi, r2=0 and r3=+Pi. With scale and multiple value rules of Dirac delta, the integrand can be written as

1/abs(cos(+Pi))*Dirac(x-Pi) + 1/abs(cos(0))*Dirac(x) + 1/abs(cos(-Pi))*Dirac(x+Pi)  

Integrating this from -3/2Pi to 3/2Pi gives 3. 

This is no different than 

int(Dirac(x^2-3^2),x=-10..10)

which is the integration of 

1/(2*abs(-3))*Dirac(x+3) + 1/(2*abs(3))*Dirac(x-3)

from -10 to 10, and Maple correctly calculates this to 1/3.

So if Maple can handle the Dirac(x^2-3^2) case successfully, I would expect it to handle sin(x) as well. Don't you think?

 

@vv I don't think anywhere I called it "function". Nevertheless, in the definition of the dirac delta, as long as the argument is a smooth real-valued function, and the end points are not singular, this use of dirac delta is perfectly valid. 

1 2 Page 2 of 2