Christian Wolinski

MaplePrimes Activity


These are replies submitted by Christian Wolinski

How effective is series approximation?

@Carl Love Visibly, one more definition is required to solve numerically...

@Carl Love Unexpected. I get unevaluated with maple 2017.

@Glowing Obviously, integers have infinite precision. Your "error" in the code is to use "1014" instead of "1014.".

@Axel Vogt 

restart;
 P1 := 1007;
 P2 := 1014;
 P3 := 1014.1;
 P2 - P1;
 P3 - P1;
 evalf(P2 - P1, 2);
 evalf(P3 - P1, 2);
 
 restart;
 P1 := 1007.;
 P2 := 1014.;
 P3 := 1014.1;
 P2 - P1;
 P3 - P1;
 evalf(P2 - P1, 2);
 evalf(P3 - P1, 2);

@taro You can replace identical(w^sigma) with identical(w)^Non(integer) as convert/parfrac will not accept non integer powers of the variable. Also, You are correct. If You negate the first element in the list then you will see the type of which  occurences will be substituted. In this case it is Non(Non(identical(w^sigma))) = identical(w^sigma). The items in the second element in the list will present without substitution, preserved. Note, the substitutions are applied only to the face of the expression passed and the result returned. If you want to see how it works out You may want to use print:

frontend(proc(E,a,v) print('args'); convert('args'); end ,[e_n_1b2,parfrac, w^sigma],[{Non}(identical(w^sigma)), {}]);
 

 

@HS I expect it is because in your equation mod is applied only to the rhs: "Q2 = P2 mod p". If you want to campare mod p then apply modulus to all elements compared. "(Q2 = P2) mod p" will apply to all elements.

Does evalb((Q2 = P2) mod p); work?

Are we to assume that k stands for K and i stands for I?
Also notice you are using e^(-k*h), e^(-k*z). That should be exp(-k*h) and exp(-k*z).

Have you made any progress with this equation?

I am assuming you are trying to compare dates to determine which comes first. Depending on ordering DayCount is positive or negative. Maybe you can use this.

@Stretto Maybe it will be of use. If you want to color code the entire vertical of a bar to your coloring scheme then I suppose you must subdivide that bar (yourself). Matrixplot will not do that I believe.

@acer Did not see that.

One question:

solve(And(y < -1, y < x, 0 < (2*x)/(x^2 - 1), x < 1), y);
solve(And(y < -1, y < x, 0 < (2*x)/(x^2 - 1), x < 1), x);
solve(And(y < -1, y < x, 0 < (2*x)/(x^2 - 1), x < 1), x) assuming y<-1;

Is there some other way to induce the answer in the above?

How do you know that:
 

 

pdsolve({
x*diff(f(x,y),x)+y*diff(f(x,y),y) = x*y*ln(x*y),
x^2*diff(f(x,y),x,x)+y^2*diff(f(x,y),y,y) = x*y
});


gives the complete solution?

@arashghgood 

 map(`@`(abs, lhs - rhs), EQ)

this will compute the absolute values of the errors of the equations in EQ.

map(`@`(evalf, subs), sols, ...

this will substitute each solution in sols into the above, and use evalf on it.

fnormal will annihilate numeric error smaller than 10^(-(Digits-4)).

Solutions in sols generate error no greater than 10^(-56).

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