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C_R

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These are questions asked by C_R

Who invented the inverse model approach...

C_R 3712 MapleSim
causal inverse-kinematics acausal
June 26 2022
0 3

Inverse kinematics can be done in several ways (this webinar gives a very good overview https://www.youtube.com/watch?v=X0OZ9EM6dns). A effective and simple method is to run a model in reverse direction. This can’t be done with causal modeling tools, where information flow is fixed by design (https://de.maplesoft.com/support/help/MapleSim/view.aspx?path=MapleSimUserGuide/Chapter01).

Inverse kinematics, which is possible with acausal modeling tools, is only an example for running a model in the reverse (i.e., inverted) direction.

Without success, I tried to find a reference who first came up with that elegant approach.

Anyone knows more?

is(EllipticK(1) = infinity)...

C_R 3712 Maple 2022
infinity elliptic-integral
June 06 2022
2 4

 I came across this question while trying to verify the equality of expressions containing elliptic integrals.

 

That's what Maple returns for EllipticK(1)

EllipticK(1)

Error, (in EllipticK) numeric exception: division by zero

  

However

evalf(EllipticK(1.))

Float(infinity)

 (1)

limit(EllipticK(k), k = 1, left)

infinity

 (2)

indicate infinity.

Float(infinity) = infinity

Float(infinity) = infinity

 (3)

is(Float(infinity) = infinity)

true

 (4)

NULL

Download EllpticK(1).mw

I have no experience with elliptic integrals. Can I assume in this case that infinity is correct?

Why does Maple not simplify to infinity?...

C_R 3712 Maple
integration simplification
May 29 2022
2 2

Evaluating this integral

Int(1/(sqrt(1-x)*sqrt(-x^2+1)), x = 0 .. 1); ((proc (x) options operator, arrow; x end proc) = value)(%)

Int(1/((1-x)^(1/2)*(-x^2+1)^(1/2)), x = 0 .. 1) = infinity*(Pi+1)

 (1)

simplify(Int(1/((1-x)^(1/2)*(-x^2+1)^(1/2)), x = 0 .. 1) = infinity*(Pi+1))

Int(1/((1-x)^(1/2)*(-x^2+1)^(1/2)), x = 0 .. 1) = infinity

 (2)

NULL

produces an infinite product as output. Why does Maple not automatically simplify to infinity. Can the extra information (1+pi) be of any use?


Download Infinity_times_something.mw

 

Why is simplify ineffective with these c...

C_R 3712 Maple
simplify assumptions arctan
May 15 2022
1 2

The expression

f := arctan(y, x)+arctan(-y, x)

arctan(y, x)+arctan(-y, x)

 (1)

simplifies to zero in the real range if y=0 is excluded.

x < 0, y::real, y <> 0; `assuming`([simplify(f), is(f = 0)], [%])

0, true

 (2)

x >= 0, y::real, y <> 0; `assuming`([simplify(f), is(f = 0)], [%])

0, true

 (3)

Combining the above assumptions as attempted bellow does not simplify to zero

x::real, y::real, y <> 0; `assuming`([simplify(f), is(f = 0)], [%])

arctan(y, x)+arctan(-y, x), FAIL

 (4)

`or`(x < 0, x >= 0), y::real, y <> 0; `assuming`([simplify(f), is(f = 0)], [%])

arctan(y, x)+arctan(-y, x), FAIL

 (5)

Or(x < 0, x >= 0), y::real, y <> 0; `assuming`([simplify(f), is(f = 0)], [%])

arctan(y, x)+arctan(-y, x), FAIL

 (6)

`and`(-infinity <= x, x <= infinity), y::real, y <> 0; `assuming`([simplify(f), is(f = 0)], [%])

arctan(y, x)+arctan(-y, x), FAIL

 (7)

interface(version)

`Standard Worksheet Interface, Maple 2022.0, Windows 10, March 8 2022 Build ID 1599809`

 (8)

 

  

Download assuming_and_differently_combined_inequalities.mw

is(-arctan(-x, y) = arctan(x, y))?...

C_R 3712 Maple
arctan
May 07 2022
1 3

In the positive range Maple confirms that this is true.
In the real range Maple fails to provide an answer (see attachments).

Is this identity correct?

restart

kernelopts(version)

`Maple 2022.0, X86 64 WINDOWS, Mar 8 2022, Build ID 1599809`

 (1)

NULL

is(arctan(-x) = -arctan(x))

true

 (2)

`assuming`([is(arctan(x, y) = -arctan(-x, y))], [x::real, y::real])

FAIL

 (3)

`assuming`([is(arctan(x, y) = -arctan(-x, y))], [x::positive, y::positive])

true

 (4)

`assuming`([simplify(arctan(-x, y)+arctan(x, y), trig)], [x::real, y::real])

arctan(-x, y)+arctan(x, y)

 (5)

But

plot3d([arctan(-x, y)+arctan(x, y)], x = -1000000 .. 1000000, y = -1000000 .. 1000000, title = arctan(-x, y)+arctan(x, y))

  

On a unit circle

x = cos(alpha), y = sin(alpha)

x = cos(alpha), y = sin(alpha)

 (6)

subs(x = cos(alpha), y = sin(alpha), arctan(-x, y)+arctan(x, y))

arctan(-cos(alpha), sin(alpha))+arctan(cos(alpha), sin(alpha))

 (7)

`assuming`([simplify(%)], [alpha::real])

arctan(-cos(alpha), sin(alpha))+arctan(cos(alpha), sin(alpha))

 (8)

`assuming`([simplify(%)], [alpha::positive])

arctan(-cos(alpha), sin(alpha))+arctan(cos(alpha), sin(alpha))

 (9)

`assuming`([simplify(%)], [-Pi < alpha and alpha < Pi])

arctan(-cos(alpha), sin(alpha))+arctan(cos(alpha), sin(alpha))

 (10)

plot(arctan(-cos(alpha), sin(alpha))+arctan(cos(alpha), sin(alpha)), alpha = -2*Pi .. 2*Pi, axes = boxed, color = red)

  

NULL


Download arctan_xy_simplify.mw

and another maybe related case where simplification does not work

arctan_xy_simplify_2.mw

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