C_R

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6 years, 73 days

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These are questions asked by C_R

So far I have only noticed them for special functions.

For the case of Airy in 2D: Is possible the get for the input Ai(x) the output Ai(x) and not AiryAi(x)?

Sometimes this information is not needed. Example

(Int = int)(1/x, x = a .. 2, 'AllSolutions')

I had a look at help(interface).

I need tutoring on some calculus basics.

y(x)

y(x)

(1)

diff(y(x), x)

diff(y(x), x)

(2)

(Int = int)(diff(y(x), x), x)

Int(diff(y(x), x), x) = y(x)

(3)

(Int = int)(diff(y(x), x), x = a .. b)

Int(diff(y(x), x), x = a .. b) = int(diff(y(x), x), x = a .. b)

(4)

The right-hand side below is the desired evaluation.

(`@`(value, rhs))(Int(diff(y(x), x), x = a .. b) = int(diff(y(x), x), x = a .. b)) = eval(int(diff(y(x), x), x), x = b)-(eval(int(diff(y(x), x), x), x = a))

int(diff(y(x), x), x = a .. b) = y(b)-y(a)

(5)

Download Eval_definite_integral.mw

Maple does integrate the indefinite integral but not a definite version of it.

I assume that this is not possible without additional assumptions on y(x).

I tried to assume the properties continuous and differentiable.

Anything else that can be done/assumed to force evaluation the way I want?

Here is an example of how such integrals can come about

I try to understand why Maple throws 2 times an error but on a third attempt (with the same input) output is returned. Is that a new mechanism of suppression of error or warning messages and returning output up to a point where evaluation cannot further be performed.

With -sin instead of cos Maple returns output immediately. Does this mean that there is no information to the user available (yet) for this particular case?

three_times_entering_the_same.mw

Eval(`ϕ`(t), t = 0) = `ϕ__0`, Eval(diff(`ϕ`(t), t), t = 0) = 0

Eval(varphi(t), t = 0) = varphi__0, Eval(diff(varphi(t), t), t = 0) = 0

(1)

`~`[value](convert({Eval(diff(varphi(t), t), t = 0) = 0, Eval(varphi(t), t = 0) = varphi__0}, D))

{eval((D(varphi))(t), t = 0) = 0, varphi(0) = varphi__0}

(2)

map(value, convert({Eval(diff(varphi(t), t), t = 0) = 0, Eval(varphi(t), t = 0) = varphi__0}, D))

{eval((D(varphi))(t), t = 0) = 0, varphi(0) = varphi__0}

(3)

convert({Eval(diff(varphi(t), t), t = 0) = 0, Eval(varphi(t), t = 0) = varphi__0}, D); `~`[value](%)

{varphi(0) = varphi__0, (D(varphi))(0) = 0}

(4)

NULL

I have expected elementwise and map to be effective. Why aren't they for this example?

(Evaluation at a point (see help(D)): In output (3) it is the first time that I see a composition of the D operator with the Eval function (vertical bar). Is that an undocumented feature or part of an answer that I could not figure out myself?)

Download Convert_inert_IC_to_D_notation.mw

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