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How integrate this function 

``

int((1/2)/(s^(1/2)*GAMMA(1/2)*(t^(1/2)-s^(1/2)))*(s^6), s = 0 .. t);

signum(t)^6*infinity

(1)

``

``

Thank you for helpping

Download qu_integral.mwqu_integral.mw


hi.how i can allocate infinite value for a parameter such as N ,which is attached below ( N := infinite) .i encounter with error.please see it and help

thanks..

 

restart; Digits := 55; a := 0; m := 1; n := 1; x[0] := 0; a := 0; h1 := proc (theta) options operator, arrow; cos(n*theta) end proc; h3 := proc (theta) options operator, arrow; cos(n*theta) end proc; h2 := proc (theta) options operator, arrow; sin(n*theta) end proc; N := infinite; `#mover(mi("z"),mo("ˆ"))` := z+z__c; z__c := (1/2)*h; `λ__c` := E__c*`υ__c`/(-`υ__c`^2+1); `μ__c` := E__c/(2*(1+`υ__c`)); `η__c` := E__c/(-`υ__c`^2+1); `λ__m` := E__m*`υ__m`/(-`υ__m`^2+1); `μ__m` := E__m/(2*(1+`υ__m`)); `η__m` := E__m/(-`υ__m`^2+1); E__m := 42.9*10^9; `υ__m` := .325; `ρ__m` := 6020; E__c := 170*10^9; `υ__c` := .25; `ρ__c` := 4640; h := 10^(-9); R := 10*h; L := 20*R; `ℓ` := 0; f := 0; `ε__r` := 0; `ε` := 8.8541878176*10^(-12)*`ε__r`; f__z := 0; `f__θ` := 0; f__x := 0; lambda := proc (`#mover(mi("z"),mo("ˆ"))`) options operator, arrow; (`λ__m`-`λ__c`)*(`#mover(mi("z"),mo("ˆ"))`/h)^N+`λ__c` end proc; mu := proc (`#mover(mi("z"),mo("ˆ"))`) options operator, arrow; (`μ__m`-`μ__c`)*(`#mover(mi("z"),mo("ˆ"))`/h)^N+`μ__c` end proc; rho := proc (`#mover(mi("z"),mo("ˆ"))`) options operator, arrow; (`ρ__m`-`ρ__c`)*(`#mover(mi("z"),mo("ˆ"))`/h)^N+`ρ__c` end proc; eta := proc (`#mover(mi("z"),mo("ˆ"))`) options operator, arrow; (`η__m`-`η__c`)*(`#mover(mi("z"),mo("ˆ"))`/h)^N+`η__c` end proc; `D__1,0` := int(eta(`#mover(mi("z"),mo("ˆ"))`), z = -z__c .. h-z__c); `D__1,1` := int(eta(`#mover(mi("z"),mo("ˆ"))`)*z, z = -z__c .. h-z__c); `D__1,2` := int(eta(`#mover(mi("z"),mo("ˆ"))`)*z^2, z = -z__c .. h-z__c); `D__3,0` := int(lambda(`#mover(mi("z"),mo("ˆ"))`), z = -z__c .. h-z__c); `D__3,1` := int(lambda(`#mover(mi("z"),mo("ˆ"))`)*z, z = -z__c .. h-z__c); `D__3,2` := int(lambda(`#mover(mi("z"),mo("ˆ"))`)*z^2, z = -z__c .. h-z__c); `D__5,0` := int(lambda(`#mover(mi("z"),mo("ˆ"))`), z = -z__c .. h-z__c); `D__5,1` := int(lambda(`#mover(mi("z"),mo("ˆ"))`)*z, z = -z__c .. h-z__c); `D__5,2` := int(lambda(`#mover(mi("z"),mo("ˆ"))`)*z^2, z = -z__c .. h-z__c); `I__1,0` := int(rho(`#mover(mi("z"),mo("ˆ"))`), z = -z__c .. h-z__c); `I__1,1` := int(rho(`#mover(mi("z"),mo("ˆ"))`)*z, z = -z__c .. h-z__c); `I__1,2` := int(rho(`#mover(mi("z"),mo("ˆ"))`)*z^2, z = -z__c .. h-z__c); with(Student[Calculus1])

[AntiderivativePlot, AntiderivativeTutor, ApproximateInt, ApproximateIntTutor, ArcLength, ArcLengthTutor, Asymptotes, Clear, CriticalPoints, CurveAnalysisTutor, DerivativePlot, DerivativeTutor, DiffTutor, ExtremePoints, FunctionAverage, FunctionAverageTutor, FunctionChart, FunctionPlot, GetMessage, GetNumProblems, GetProblem, Hint, InflectionPoints, IntTutor, Integrand, InversePlot, InverseTutor, LimitTutor, MeanValueTheorem, MeanValueTheoremTutor, NewtonQuotient, NewtonsMethod, NewtonsMethodTutor, PointInterpolation, RiemannSum, RollesTheorem, Roots, Rule, Show, ShowIncomplete, ShowSolution, ShowSteps, Summand, SurfaceOfRevolution, SurfaceOfRevolutionTutor, Tangent, TangentSecantTutor, TangentTutor, TaylorApproximation, TaylorApproximationTutor, Understand, Undo, VolumeOfRevolution, VolumeOfRevolutionTutor, WhatProblem]

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infinite.mw

I want to obtain the taylor series of a function say sin(x) at x=0 up to infinity. I mean that I don't want a trauncated series. I tried using "series" and "taylor" but they just give the truncated series.

Hi everyone,

 

Consider this maple 18 doc: Euler18.mw

 

The code is regular code for Julia sets of the exponential.

 

To see how the Julia set behaves at infinity, I apply the transform mu(z)=1/z.

 

The plot3d command correctly plots the Julia set at an appropriate neighborhood of infinity, but:

1) Axes are not transformed

2) Saving as .eps produces an empty plot, sans the axes (plot is saved correctly, when not applying mu(z))

 

Is there any trick to force the axes to also show correctly with the transformed ranges?

Seems that this misalignment is bothering the .eps renderer, which probably plots the graph in twilight zone, given the false ranges of the untransformed axes.

 

Any ideas on how to force the saveas .eps to work in this case?

 

Many thanks,

Yiannis

Apparently inconsistent behaviour of the BesselJ() function.

Examples: BesselJ(-3, 0)  ... gives 0 (correct)

but BesselJ(-3.0, 0), BesselJ(-3, 0.0)  and BesselJ(-3, 0.0) all give Float(infinity) (wrong! - should be 0.0)

The problem seems to occur for all negative integer values of the first argument (the order) when the second argument is 0 or 0.0.

The following integral
f := u-> int(-1/(x*sqrt(-1+u^2*(x+1)^2*x^2)), x = (1/2)*(-u-sqrt(u^2-4*u))/u .. (1/2)*(-u+sqrt(u^2-4*u))/u);
arised in an applied research. I was asked about its properties:
plot on RealRange(4,infinity), limit(f(u),u=4,right), limit(f(u),u=infinity).
Unfortunately, I lost a file. As far as I remember it, I have had a problem with
the last-named one only:

limit(f(u), u = infinity);


MultiSeries:-limit(f(u), u = infinity);

asympt(f(u), u, 2);

Error, (in asympt) unable to compute series

Hope my colleagues will make progress with it. The assumed value is Pi/2.

I'm trying to determine that f(x) = (a/2)*e^(-a|x|) is a pdf for which I have tried to calcuate the integral from -infinity to +infinity but I am no getting a result that converges(even the wolfram alpha widget said the integral doesn't converge). How do I correcly implement this?

hello everyone. I have an undergradute project i'm currently working on and I'm stuck where I have to use the Differential Transforms Method to solve a problem with boundary conditions at infinity


restart;

Digits := 5;

F[0] := 0; F[1] := 0; F[2] := (1/2)*A; T[0] := 1; T[1] := B; M := 2; S := 1;

for k from 0 to 10 do F[k+3] := (2*(sum((r+1)*F[r+1]*(k+1-r)*F[k+1-r], r = 0 .. k))-T[k]-3*(sum((k+1-r)*(k+2-r)*F[r]*F[k+2-r], r = 0 .. k))-M*(k+1)*F[k+1])*factorial(k)/factorial(k+3);

T[k+2] := (-3*(sum((k+1-r)*F[r]*T[k+1-r], r = 0 .. k))-S*T[k])*factorial(k)/factorial(k+2)

end do; f := 0; t := 0;

for k from 0 to 10 do

f := f+F[k]*x^k;

t := t+T[k]*x^k end do;

print(f);
print(t);

but the problem is that i cant seem to evaluate

or higer diagonal pade-approximant. any help will be greatly appreciated. thank you.

How to find asymptotic behaviour of a function.

For example at infinity

sinh(x) behaves as 1/2*exp(x)

1/sinh(x)  behaves as 2*exp(-x)

exp(-x)*(exp(-x)+1) behaves as exp(-x)

so that it works with a more complex expression.

I got a problem with solving a second order ODE. 

The ODE is :

-V(xi)+(1/2)*xi*(diff(V(xi), xi))+(1/4)*(diff(V(xi), xi, xi))=-(1/2)*k2*(diff(H(xi), xi))-k1*n*X/E+1+k2

where k1,k2,n,X,E  all are constant.

the condition is :

V(xi) tends to 2*xi^2 as xi tends to infinity.

I used 'dsolve' to solve the equation firstly, and got a solution with two constant C1 and C2, I want to use the condition to elimilate C2, so I used limit(sol,xi=infinity)=2*xi^2. But when I used the command 'limit', I can't get the answer.

Could any one help me? 

Many thanks!!!

Surely this is a bug.

> 0^0;
                 1 
> sum( 0^m, m=0..infinity );
                 0

The limit of

G := N -> sum(1/N*numtheory[sigma](k)/k,k=1..N ) 

as N -> infinity should be Pi^2/6. However

eval( G(N), N=infinity );

returns 0.

Can anyone explain this puzzling behaviour?

Is it possible to numerically calculate  the integral

int((-12*y^2+1)*ln(abs(Zeta(x+I*y)))/(4*y^2+1)^3, [y = 0 .. infinity, x = 1/2 .. infinity])

in Maple?

The code

int((-12*y^2+1)*ln(abs(Zeta(x+I*y)))/(4*y^2+1)^3, [y = 0 .. infinity, x = 1/2 .. infinity],numeric,epsilon=0.1)

has been executed on my comp  without any output since this morning.

 

 

 

Hi,

Maybe someone can give me a nice answer without Maple.

I am given a fourier series:
ln|cosx|=Co - sum( (-1)^k/k * cos2kx,k=1..infinity)
and am asked what this tells me about the chevychev series for ln(u).

 

Thanks

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