vv

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MaplePrimes Activity


These are replies submitted by vv

The problem was caused when I tested "Maple for Screen Readers" (for using the old GUI). At the end, I turned off the "Screen Readers" mode, but for some reason, in the Gui configuration file the entry  "AccessibleMode=true" was left unchanged. After setting manually to "false", the display is normal again.

@dharr Thank you. So, I am not alone.

@dharr Windows 10. So, have you tried to export as .mpl or .tex ?

@acer As this site shows, there are many Maple users interested in typesetting. 
In this case I think It would be a good idea for Maple to include support for LaTeX (see e.g. Geogebra).

@janhardo Why don't you post your code?

@Carl Love The limit is for sqrt(n/3) * f(n,x)  as requested, not for f(n,x) (which is obviouosly 0).

limit( sqrt(n/3)*asympt('rsolve'({f(0) = a, f(n) = sin(f(n - 1))}, f(n)), n), n=infinity );

        1

@Alfred_F  Maple is only able to guess the limit:

term1 := sqrt(1 - t);

(1-t)^(1/2)

(1)

term3 := sum(t^(k^2), k = 0 .. infinity);

sum(t^(k^2), k = 0 .. infinity)

(2)

limit(term1*term3, t=1, left)

limit((1-t)^(1/2)*(sum(t^(k^2), k = 0 .. infinity)), t = 1, left)

(3)

# evalf(limit(term1*term3, t=1, left));

term3a:=simplify(term3) assuming t<1,t>0;

(1/2)*JacobiTheta3(0, t)+1/2

(4)

limit(term1*term3a, t=1, left);

limit((1-t)^(1/2)*((1/2)*JacobiTheta3(0, t)+1/2), t = 1, left)

(5)

evalf(%)

.8862269255

(6)

identify(%);

(1/2)*Pi^(1/2)

(7)
 

 

Download Jacobi-vv.mw

@Carl Love 

arctan(2/n^2) = arctan(n+1) - arctan(n-1)

Telescoping sum ==>
sum( 2/n^2), n=1..N )  = -Pi/4 + arctan(N) + arctan(N+1) --> -Pi/4 + 2*Pi/2 = 3*Pi/4.
 

@Alfred_F The point (1,1) does not matter, the monotone convergence condition is a.e. (almost everywhere).

@Alfred_F Here are higher precision computations; you may play with Digits and epsilon.

restart;
N:=10^12:
Digits:=50:
exact:=evalf(ln(2026/2025)):
J1:=evalf(Int(N*x^N/(x^N+x+2024), x = 0 .. 1, method=_Dexp, epsilon=1e-20)); 
J2:=evalf(Int(t^(1/N)/(t + t^(1/N) + 2024), t = 0 .. 1, method=_Dexp, epsilon=1e-20));
err1=J1-exact;
err2=J2-exact;

      J1 := 0.00049370526798868160514093629790805923985216893049776
      J2 := 0.00049370526798868160514093619869170079706955297924341
      err1 = -4.9352240248370621056973318812232002987*10^(-16) 
      err2 = -4.9352240248380542692817597073827128422*10^(-16)   

If you are using the good old Derive, be careful with numeric computations because it has a very special implementation with fractions instead of the standard floating-point numbers.

@mmcdara The problem with such transforms is that expr and S(expr) could be totally different; just plot expr - S(expr) for  expr:=sin(6-2*sin(x)) + cos(6*sin(-3+x));

You already have a function z(t) in the system. Do you mean another function, say Z(t) = exp(int(y(u),u=0..t)) ?  Maybe u=t..0 ?

@mmcdara The correct mathematical formulation  AREA  ∈  {4, 8}  was given. The and/or expression corresponds to a  common language and it is permitted. I don't see any reason to be too pedantic.

@mmcdara The standard formula for the area of a triangle in coordinate geometry uses abs:
Area of a triangle - Wikipedia

@mmcdara  abs is necessary because we want an unoriented area (Area(ABC) = Area(ACB)).
See the examples:

A,B,C:=[0, 3*t-3], [1, t+1], [2, t+1]; # oriented area = 8 (at t=10)

A,B,C:=[0, 3*t*(1/5)-3], [1, t+1], [2, t+1]; # oriented area = -4 (at t=10)

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