Carl Love

Carl Love

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12 years, 358 days
Himself
Wayland, Massachusetts, United States
My name was formerly Carl Devore.

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These are answers submitted by Carl Love

The absolute value below is in case you don't know which function is on top. That's trivial in this case, but I wanted to be a bit more general.

F:= (-x^2+9) - (x+3):
abs(int(F, x= `..`(solve(F, x))));

TangentLine:= proc(f::algebraic, at::name= algebraic)
local x, a;
     (x,a):= op(at);
     eval(diff(f,x), x= a)*(x-a) + eval(f,x= a)
end proc:

NormalLine:= proc(f::algebraic, at::name= algebraic)
local x, a;
     (x,a):= op(at);
     eval(f, x= a) - (x-a)/eval(diff(f,x), x= a)
end proc:

f:= ln(3*x)+3; a:= exp(1);
                        f := ln(3 x) + 3
                          a := exp(1)

TL:= TangentLine(f, x= a);

NL:= NormalLine(f, x= a);

plot([f, TL, NL], x= 1/2..10, y= 1/2..10, legend= [f, tangent, normal]);

 

 

The following computation strongly suggests that the value that you seek is undefined:

 MG:= MeijerG([[-.3+eps], []], [[.8, 1.3, -.8, -.3, -1.3], []], 1.):
eval(MG, eps= 10^k) $ k= -6..-2;

eval(MG, eps= -10^k) $ k= -6..-2;

Like this:

Ys:= [1,2]: #List all y values here.
Zs:= [3,4]: #List all z values here.

G:= (y,z)-> sqrt(D[1](x)(y,z)^2*dy^2 + D[2](x)(y,z)^2*dz^2):
zip(G, Ys, Zs);

See the command ?map. Example:

V:= <1,2>;

map(tan, V);

What you are trying to do only makes sense to me if E is sorted. So, assuming that E is sorted, a binary search will be much faster in the long run than the linear searching already mentioned. What you want is

1+ListTools:-BinaryPlace(E,x);

Example:

E:= [0,2,7,15,26,40]:
1+ListTools:-BinaryPlace(E,5);
               3

I'll show you a way to get the reduced row echelon form and leave it up to you to interpret it.

A:= < 1,2,3,1; 4,5,6,1; 7,8,9,1 >:

(How did I get that matrix from your equations?)

MTM:-rref(A);

The above command is equivalent to LinearAlgebra:-ReducedRowEchelonForm but is a lot less to type.

Include

with(Optimization):

at the top of your worksheet.

 

Enclose the _X in single back quote characters, like this: `_X`. On American keyboards, this character is in the upper left corner, under Escape. Anything inside the quotes is treated as a variable, regardless of any other meaning(s) that the characters have.

Use square bracket indexing for subscripting and exponents (via ^) for superscripting. Here's an example:

plot(x^2, x= -1..1, labels= [x[old], x[old]^2], labelfont= [TIMES,ITALIC,16]);

Maple will do the integral easily if you assume x > 0. I also changed the floating-point constants to exact rationals, although this is not strictly necessary.

Int((x-tau)^(1/2-1)*(tau^2+tau^(3/2)*8/3-tau-2*tau^(1/2)), tau= 0..x);

value(%) assuming x > 0;

What do you mean by (x)(x)? Do you mean x times x? That should be either x*x or x^2. Juxtaposition does not imply multiplication. Extraneous parentheses do imply function calls, i.e., (x)(x) is treated as if x were a function with itself as the argument.

The numeric constant 5.6*10^-4 needs to be 5.6*10^(-4) or, better yet, 5.6e-4.

solve(5.6e-4 = x^2/(0.2-x), x);

Change your code to this:

restart:
with(DEtools):
a := diff(y(x), x) = exp(-0.1e-1*x*y(x)^2);
dfieldplot(a, y(x), x = -8 .. 8, y(x) = -8 .. 8, color = exp(-0.1e-1*x*y(x)^2));

I corrected several errors. I changed e(x)^ to exp. I changed xy to x*y(x). There is no implied multiplication by juxtaposition in Maple; xy is treated as a variable distinct from x or y. The exponential function is exp; e has no pre-assigned meaning on input, although it is used in the usual mathematical sense on output. In differential equations, the dependent variable must always be used as a functiony(x) rather than simply y.

It can only be plotted for integer values of k because it involves a negative number to a variable power, which is in general complex. The following plot command will work:

plot([seq([k,RR(k)], k= 2..10)], style= point, symbolsize= 16, view= [2..10, 4..7]);

ceil(fsolve(Pi^2/6 - sum(1/n^2, n= 1..N) = .001, N));

                                        1000

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