Kitonum

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17 years, 227 days

MaplePrimes Activity


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@Carl Love  The widest range for the parameter  m  is easy to find. It is determined by the slopes  m  of two tangents to the circle c2 passing through  J .

interface(version);

`Standard Worksheet Interface, Maple 2018.2, Windows 10, October 23 2018 Build ID 1356656`

(1)

Physics:-Version();

"C:\Program Files\Maple 2018\lib\update.mla", `2018, October 24, 4:22 hours, version in the MapleCloud: unable to determine, version installed in this computer: not installed`

(2)

libname;

"C:\Program Files\Maple 2018\lib"

(3)

restart;

e:=inttrans:-laplace(x(t),t,s)

laplace(x(t), t, s)

(4)

latex(e,'output'='string')

"{\it laplace} \left( x \left( t \right) ,t,s \right) "

(5)
 

 

Download latex_of_laplace_nov_21_2024_Maple_2018.2.mw

@EugeneKalentev  Yes, I accidentally missed  x  when I was typing the code in Maple. The correct answer was given by vv . 

@Blanc But you omitted thу  assuming  option in your solution. Just copy the code from my answer as text and paste it into your worksheet.

@Blanc  It's strange that it doesn't work in Maple 2023. Try adding the  real  option. Upload your worksheet here.

restart;
f := (x, n) -> 2^n/(1 + n*2^n*x^2) - 2*x^2*(2^n)^2*n/(1 + n*2^n*x^2)^2;
solve(f(x,n)>0, x, parametric=full) assuming real, n::integer;
Eq:=simplify(f(x,n));
solve(Eq>0, x) assuming  real, n>0;
solve(Eq>0, x) assuming  real, n=0;
solve(Eq>0, x) assuming  real, n<0;


Also try using the  useassumptions  option:

restart;
f := (x, n) -> 2^n/(1 + n*2^n*x^2) - 2*x^2*(2^n)^2*n/(1 + n*2^n*x^2)^2;
solve(f(x,n)>0, x, parametric=full, useassumptions) n::integer;
Eq:=simplify(f(x,n));
solve(Eq>0, x, useassumptions) assuming  n>0;
solve(Eq>0, x, useassumptions) assuming  n=0;
solve(Eq>0, x, useassumptions) assuming  n<0;

 

Here is another solution for my example. It is based on a parametric representation of the curve  11*x^3-9*x^2*y+27*x*y^2+7*y^3-18*x^2-24*x*y+18*y^2=0. Well-known formulas are used. The results coincide with the results obtained by @Rouben Rostamian  :

restart;
P:=(x,y)->11*x^3-9*x^2*y+27*x*y^2+7*y^3-18*x^2-24*x*y+18*y^2:
solve(P(x,y)=0, {x(t),y(t)});
assign(%);
DP:=simplify(diff([x,y], t));
D2P:=simplify(diff(DP, t));
T:=solve({x=0,y=0}, explicit);
t1:=eval(t,T[1]); t2:=eval(t,T[2]);
simplify(rationalize(eval(DP[2]/DP[1], t=t1)));
simplify(rationalize(eval(DP[2]/DP[1], t=t2)));
expand(rationalize(eval((DP[1]*D2P[2]-DP[2]*D2P[1])/DP[1]^3, t=t1)));
expand(rationalize(eval((DP[1]*D2P[2]-DP[2]*D2P[1])/DP[1]^3, t=t2)));

         

 

@Rouben Rostamian  Thank you very much for your solution! 

@Rouben Rostamian  

I tried to apply your method to the following example of an implicitly defined real-valued function. The graph shows that in the neighborhood of the singular point (0,0) there is no single-valued function y(x) , but there are 2 intersecting smooth curves. What do the results (2/3 and infinity) mean geometrically?

 

restart;
P:=(x,y)->11*x^3-9*x^2*y+27*x*y^2+7*y^3-18*x^2-24*x*y+18*y^2:
plots:-implicitplot(P(x,y), x=-1..3, y=-1..3, gridrefine=3, scaling=constrained);

x0:=0: y0:=0:
D[1](P)(0,0), D[2](P)[2](0,0);
 
implicitdiff(P(x,y), y, x);
subs(x=x0, %);
limit(%, y=y0);

implicitdiff(P(x,y), y, x, x);
subs(x=x0, %);
limit(%, y=y0, left), limit(%, y=y0, right);
 

 

0, 0

 

-(11*x^2-6*x*y+9*y^2-12*x-8*y)/(-3*x^2+18*x*y+7*y^2-8*x+12*y)

 

-(9*y^2-8*y)/(7*y^2+12*y)

 

2/3

 

4*(-495*x^5-167*x^4*y-186*x^3*y^2-2178*x^2*y^3+1013*x*y^4+357*y^5+876*x^4+1896*x^3*y-264*x^2*y^2-2280*x*y^3+876*y^4-144*x^3-48*x^2*y+336*x*y^2-144*y^3-624*x^2-832*x*y+624*y^2)/(-27*x^6+486*x^5*y-2727*x^4*y^2+3564*x^3*y^3+6363*x^2*y^4+2646*x*y^5+343*y^6-216*x^5+2916*x^4*y-10656*x^3*y^2+4104*x^2*y^3+7896*x*y^4+1764*y^5-576*x^4+5184*x^3*y-10320*x^2*y^2+3744*x*y^3+3024*y^4-512*x^3+2304*x^2*y-3456*x*y^2+1728*y^3)

 

4*(357*y^5+876*y^4-144*y^3+624*y^2)/(343*y^6+1764*y^5+3024*y^4+1728*y^3)

 

-infinity, infinity

(1)
 

 

Download implicit.mw

@AHSAN  Your problems with integration are related to the fact that for some values ​​of the parameter  k  the denominator of the expression  Pre  is 0 and the function is not defined (its graph goes to infinity). Therefore, we can first find such values ​​of the parameter  k  and use the corresponding assumption on  k  when integrating:

restart; 
Pre := -(6*(x-1))*(k-1)*x/(((x-1)*k-x)^2*(k+1)); 
sol := solve(denom(Pre) = 0, x); 
solve({sol[1] >= 0, sol[1] <= 1}, k); 
int(Pre, x = 0 .. 1)  assuming k > 0;

                

@AHSAN  I don't understand why you integrate the expression  Pre ? What does it have to do with your problem  "Need to plot maximum relation by considering Pre_max alog vertical axis and k along x axis or horizontal axis" ?

@mmcdara 

`(Probability(Total)=50)` = (Probability(Total <=50)+Probability(Total >=50)) - 1;

                 

@nm  The answer is not necessarily the greatest common divisor of the numbers 45 and 63. The answer can be any common divisor of this numbers.

@Aixleft math 

restart;
solutions := solve({v1^2 = 2*g*(h-h1), (1/2)*g*t^2 = h2, v1*t+(1/2)*g*t^2 = h1}, {h, t, v1}, explicit): 
selected_solutions := select(p->op(1,eval(t,p))<>-1, [solutions])[]; 

     

@Raafat  Maybe you want these points to be connected by line segments:

S:=seq(P(1,2,n), n=0..10):
A:=plots:-pointplot([S][1..4], color=red, symbol=solidcircle, symbolsize=15):
B:=plot([S][1..4], linestyle=3, color=blue):
plots:-display(A,B, view=[0..110,0..16]);

                  

 

@GFY  Maple 2018.2 solves this system without any problems. Try to specify the list of unknown functions in the code:


 

restart

couple := diff(eta(t), `$`(t, 2))+`&omega;__2`*eta(t)+B*(diff(varphi(t), `$`(t, 2))) = 0, diff(varphi(t), `$`(t, 2))+`&omega;__2`^2*varphi(t)+beta*(diff(eta(t), `$`(t, 2))) = 0

diff(diff(eta(t), t), t)+omega__2*eta(t)+B*(diff(diff(varphi(t), t), t)) = 0, diff(diff(varphi(t), t), t)+omega__2^2*varphi(t)+beta*(diff(diff(eta(t), t), t)) = 0

(1)

dsolve([couple], [eta(t), varphi(t)])

{eta(t) = (1/2)*(exp((1/2)*(-2*omega__2*((1+omega__2*(4*B*beta+omega__2-2))^(1/2)-omega__2-1)*(B*beta-1))^(1/2)*t/(B*beta-1))*(1+omega__2*(4*B*beta+omega__2-2))^(1/2)*_C2+_C2*exp((1/2)*(-2*omega__2*((1+omega__2*(4*B*beta+omega__2-2))^(1/2)-omega__2-1)*(B*beta-1))^(1/2)*t/(B*beta-1))*omega__2+exp(-(1/2)*(-2*omega__2*((1+omega__2*(4*B*beta+omega__2-2))^(1/2)-omega__2-1)*(B*beta-1))^(1/2)*t/(B*beta-1))*(1+omega__2*(4*B*beta+omega__2-2))^(1/2)*_C1+_C1*exp(-(1/2)*(-2*omega__2*((1+omega__2*(4*B*beta+omega__2-2))^(1/2)-omega__2-1)*(B*beta-1))^(1/2)*t/(B*beta-1))*omega__2-exp(-(1/2)*2^(1/2)*((B*beta-1)*(omega__2+1+(4*B*beta*omega__2+omega__2^2-2*omega__2+1)^(1/2))*omega__2)^(1/2)*t/(B*beta-1))*(1+omega__2*(4*B*beta+omega__2-2))^(1/2)*_C3-exp((1/2)*2^(1/2)*((B*beta-1)*(omega__2+1+(4*B*beta*omega__2+omega__2^2-2*omega__2+1)^(1/2))*omega__2)^(1/2)*t/(B*beta-1))*(1+omega__2*(4*B*beta+omega__2-2))^(1/2)*_C4+_C3*omega__2*exp(-(1/2)*2^(1/2)*((B*beta-1)*(omega__2+1+(4*B*beta*omega__2+omega__2^2-2*omega__2+1)^(1/2))*omega__2)^(1/2)*t/(B*beta-1))+_C4*omega__2*exp((1/2)*2^(1/2)*((B*beta-1)*(omega__2+1+(4*B*beta*omega__2+omega__2^2-2*omega__2+1)^(1/2))*omega__2)^(1/2)*t/(B*beta-1))-_C2*exp((1/2)*(-2*omega__2*((1+omega__2*(4*B*beta+omega__2-2))^(1/2)-omega__2-1)*(B*beta-1))^(1/2)*t/(B*beta-1))-_C1*exp(-(1/2)*(-2*omega__2*((1+omega__2*(4*B*beta+omega__2-2))^(1/2)-omega__2-1)*(B*beta-1))^(1/2)*t/(B*beta-1))-_C3*exp(-(1/2)*2^(1/2)*((B*beta-1)*(omega__2+1+(4*B*beta*omega__2+omega__2^2-2*omega__2+1)^(1/2))*omega__2)^(1/2)*t/(B*beta-1))-_C4*exp((1/2)*2^(1/2)*((B*beta-1)*(omega__2+1+(4*B*beta*omega__2+omega__2^2-2*omega__2+1)^(1/2))*omega__2)^(1/2)*t/(B*beta-1)))/beta, varphi(t) = _C1*exp(-(1/2)*(-2*omega__2*((1+omega__2*(4*B*beta+omega__2-2))^(1/2)-omega__2-1)*(B*beta-1))^(1/2)*t/(B*beta-1))+_C2*exp((1/2)*(-2*omega__2*((1+omega__2*(4*B*beta+omega__2-2))^(1/2)-omega__2-1)*(B*beta-1))^(1/2)*t/(B*beta-1))+_C3*exp(-(1/2)*2^(1/2)*((B*beta-1)*(omega__2+1+(4*B*beta*omega__2+omega__2^2-2*omega__2+1)^(1/2))*omega__2)^(1/2)*t/(B*beta-1))+_C4*exp((1/2)*2^(1/2)*((B*beta-1)*(omega__2+1+(4*B*beta*omega__2+omega__2^2-2*omega__2+1)^(1/2))*omega__2)^(1/2)*t/(B*beta-1))}

(2)

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