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Axel Vogt

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20 years, 221 days
Munich, Bavaria, Germany
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These are replies submitted by Axel Vogt

what is the actual question...

Axel Vogt 5821
January 27 2024
0 0

@WA573 

So what is your actual question - how to compute the limit for y in your posted image? Using Maple or manually?

hm ......

Axel Vogt 5821
January 16 2024
0 0

Roughly you want f(x,y) = 4 and f(x,y) = 5. Could that be a typo?

sorry...

Axel Vogt 5821
January 13 2024
0 0

@Preben Alsholm It was thought for Hahn Hahn, not as critics on your post, I correct that. Sorry for being unclear.

it always has negative values as well...

Axel Vogt 5821
January 11 2024
0 0

@Preben Alsholm  @Hahn Hahn

Adapting Preben's view that task is to consider a line A + B*s, 0 < B, 0 < s.
Maple finds that A is negativ. Hence the line is negativ until it is zero, always:

> 

# https://www.mapleprimes.com/questions/237643-Is-This-Expression-Greater-Or-Smaller-Than-Zero-Evalb--Or-Is
> 

restart; # kernelopts(version);

> 

Q:=(a - c)^2*y*(3*b*y - 1)/(2*(2*b*y - 1)^2) +
  (-a*(a + c)/(2*b) + (a - c)^2/(8*b) + s*(a - c)^2/(8*b^2));
theAssumptions:=(2*c < a, a/c < 2*b*y, 0 < a, 0 < b, 0 < c, 0 < s, 0 < y);

(1/2)*(a-c)^2*y*(3*b*y-1)/(2*b*y-1)^2-(1/2)*a*(a+c)/b+(1/8)*(a-c)^2/b+(1/8)*s*(a-c)^2/b^2

  

2*c < a, a/c < 2*b*y, 0 < a, 0 < b, 0 < c, 0 < s, 0 < y

 (1)
> 

A:=eval(Q, s=0); B:=(Q-A)/s; ``;
'Q=A+B*s', '0<B'; map(is, [%]) assuming theAssumptions: op(%);

(1/2)*(a-c)^2*y*(3*b*y-1)/(2*b*y-1)^2-(1/2)*a*(a+c)/b+(1/8)*(a-c)^2/b

  

(1/8)*(a-c)^2/b^2

  

``

  

Q = B*s+A, 0 < B

  

true, true

 (2)

 

Hence Q = A + B*s is an increasing line and thus 0 < Q for 0 < s iff 0 <= A

However Maple says that is not the case (signum = -1 means 'negative'):

 

> 

'signum(A)'; ``=% assuming theAssumptions;

signum(A)

  

`` = -1

 (3)
> 

 

Download MP_237643_always_has_negatives.mw

example for false...

Axel Vogt 5821
January 10 2024
0 0

example for false

> 

# https://www.mapleprimes.com/questions/237643-Is-This-Expression-Greater-Or-Smaller-Than-Zero-Evalb--Or-Is
> 

restart; kernelopts(version);

`Maple 2023.2, X86 64 WINDOWS, Nov 24 2023, Build ID 1762575`

 (1)
> 

Q:=(a - c)^2*y*(3*b*y - 1)/(2*(2*b*y - 1)^2) +
  (-a*(a + c)/(2*b) + (a - c)^2/(8*b) + s*(a - c)^2/(8*b^2));
theAssumptions:=(2*c < a, a/c < 2*b*y, 0 < a, 0 < b, 0 < c, 0 < s, 0 < y); #nops([%]);

(1/2)*(a-c)^2*y*(3*b*y-1)/(2*b*y-1)^2-(1/2)*a*(a+c)/b+(1/8)*(a-c)^2/b+(1/8)*s*(a-c)^2/b^2

  

2*c < a, a/c < 2*b*y, 0 < a, 0 < b, 0 < c, 0 < s, 0 < y

 (2)
> 

Bsp:=[a = 103/49 + sqrt(25776)/98, b=1/4,c=1, y=8];
'eval([theAssumptions], %)': map(is, %) = '%';

[a = 103/49+(6/49)*179^(1/2), b = 1/4, c = 1, y = 8]

  

[true, true, true, true, true, false, true] = eval([theAssumptions], [a = 103/49+(6/49)*179^(1/2), b = 1/4, c = 1, y = 8])

 (3)
> 

# hence the assumptions are satisfied, except for 0 < s
> 

'eval(Q, Bsp)': '%'= simplify(%);
plot(rhs(%), s=0 .. 4);

eval(Q, Bsp) = (144/2401)*(s-1)*(9*179^(1/2)+130)

  

 
> 

# thus (for that example values) the assertion is false for 0 < s <=1 and correct for 1 < s
> 

 

Download MP_237643_Example.mw

the 2nd task...

Axel Vogt 5821
January 07 2024
0 0

dsolve gives a solution for the 2nd task. At least some can be done 'directly': if u is a root of the denominator numerator, a cubic, then the rhs is 0 by definition and the lhs is 0 since that u depends on a and b only and not on x

remark aside...

Axel Vogt 5821
January 06 2024
0 0

If you use uppercase Sum (or expand it by 'allvalues') then the error does not occur

break...

Axel Vogt 5821
January 04 2024
0 0

@Zeineb I will stop here

sketch...

Axel Vogt 5821
January 04 2024
0 0
> 

# https://www.mapleprimes.com/questions/237620-Compute-Exact-Integral-
> 

restart; kernelopts(version);
> 

Int((b - x)^(mu - 1)*(x - a)^(nu - 1), x = a .. b) = (b - a)^(nu + mu - 1)*Beta(mu, nu);
IntegrationTools[Change](%,x = (-a+b)*xi+a, xi): subs(xi=x, %): simplify(%);

% / (b - a)^nu: combine(%):
simplify(%, {b-a=r});

T:=lhs(%); #diff(%, r): combine(%): simplify(%);
r=b-a, 0<x,x<1;

Int((b-x)^(mu-1)*(x-a)^(nu-1), x = a .. b) = (b-a)^(nu+mu-1)*Beta(mu, nu)

  

(b-a)^nu*(Int(x^(nu-1)*((x-1)*(-b+a))^(mu-1), x = 0 .. 1)) = (b-a)^(nu+mu-1)*Beta(mu, nu)

  

Int(x^(nu-1)*(-r*(x-1))^(mu-1), x = 0 .. 1) = r^(mu-1)*Beta(mu, nu)

  

Int(x^(nu-1)*(-r*(x-1))^(mu-1), x = 0 .. 1)

  

r = b-a, 0 < x, x < 1

 (1)
> 

T2:=Int(x^(nu - 1)*(r*(-x + 1))^(mu - 1), x = 0 .. 1);

Int(x^(nu-1)*((1-x)*r)^(mu-1), x = 0 .. 1)

 (2)
> 

T3:=Int(x^(nu - 1)*((-x + 1))^(mu - 1)*r^(mu - 1), x = 0 .. 1);
simplify(%); value(%);

Int(x^(nu-1)*(1-x)^(mu-1)*r^(mu-1), x = 0 .. 1)

  

r^(mu-1)*(Int(x^(nu-1)*(1-x)^(mu-1), x = 0 .. 1))

  

r^(mu-1)*GAMMA(mu)*GAMMA(nu)/GAMMA(mu+nu)

 (3)
> 

 

Download MP_237620_proof_for_GR.mw

reference ?...

Axel Vogt 5821
January 04 2024
0 0

@Zeineb what is the reference for that formula?

suggestion...

Axel Vogt 5821
January 02 2024
0 0

Use map combine and simplify:

remark...

Axel Vogt 5821
December 28 2023
0 0

You sheet presumably should start with "local gamma" and not with "_local(gamma)"

double precision and decimals...

Axel Vogt 5821
December 28 2023
0 0

Converting between double precision = binary representation = Basis 2 and deicmals = Basis 10 is not exact in the sense of plain display, roughly one needs 18 Decimals to get 15 - 16 correct decimal places. This is due to the methods and not an error

Especially for Pi: in double precision (IEEE 754) it is represented by 884279719003555 / 281474976710656

Which gives the results above - where it does not matter to use evalhf or evalf[18].

 

stepwise...

Axel Vogt 5821
December 14 2023
0 0

I prefer to have it in steps (there is no effect for performance here):

theExpression:= eval(op(1,ig)); # the parametric integrand
theRange:= rhs(op(2,ig));       # the range of integration

h:= Y -> unapply( eval(theExpression, y=Y), x);
H:= y -> Int(h(y), theRange);

lprint ?...

Axel Vogt 5821
December 10 2023
0 0

@sand15 It seems that lprint works - you can not use that?

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