## 724 Reputation

8 years, 107 days

## applyrule and hypergeom...

Maple

1. This seems wrong:

```applyrule(x::symbol+y::symbol = 0, a+b);
a + b
```

2. Should these two match f(a)?

```applyrule((h::anything)(a) = 0, f(a));
f(a)
applyrule(x::f(x1::anything) = 0, f(a));
f(a)
```

3. hypergeom will give an error if the arguments are not lists, so how to write a pattern that will match hypergeom(anything, anything, a)? This works but is... skittish:

```applyrule('''hypergeom'''(x1::anything, x2::anything, a) = 0, hypergeom([], [], a));
0
```

## 2d input glitch...

 >
 (1)

Copy the string, paste into a new cell, delete and retype the opening quote.

 >

Copy the string from the first cell, add f() around it.

 >

## 2d input glitch...

Spawned from here. The code itself is perfectly fine, but if the definition of TD is in the same cell with the rest of the code, TD gives an error. The same definition in a separate cell works.

 >
 >
 >
 >
 (1)

## diff(u(x,y),x,y)=x+y...

```eqs := [D[1, 2](u)(x, y) = x+y, u(x, 0) = x, u(0, y) = y]:

pdsolve(eqs[1], u(x, y));
u(x, y) = _F2(x)+_F1(y)+(1/2)*x^2*y+(1/2)*x*y^2

pdsolve(eqs[1 .. 2], u(x, y)); # indeterminate
u(x, y) = invfourier((y^2*undefined+signum(y)*infinity+y*(limit(-(2*I)*Pi*y/s1,
s1 = 0))+undefined)*Dirac(s1)+2*Pi*(limit((I*s1+y)/s1, s1 = 0))*Dirac(1, s1), s1, x)

pdsolve(eqs, u(x, y)); # doesn't satisfy the DE
u(x, y) = (Int(1, tau1 = 0 .. y)+1)*(-_F1(0)+x-_F2(0)+y)

(value@eval)(eqs, u = unapply(rhs(%), [x, y]));
[1 = x+y, -_F1(0)+x-_F2(0) = x, (y+1)*(-_F1(0)-_F2(0)+y) = y]
```

The first pdsolve is fine, but not the other two.

## diff(int(f(x,t), t=0..x),x)...

```diff(int(1/(1+exp(1/(x-t))), t = 0 .. x), x);
diff(int(1/(1+exp(1/(x-t))), t = 0 .. x), t)

simplify(%);
0
```

diff returns a derivative wrt the bound variable t.

This is probably tricky because diff needs to evaluate the limit of f(x, t) at t=x, and it must distinguish between the case of the left limit (x>0) and of the right limit (x<0).

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