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hi everyone , i need your point of view in my question,any help would be appreciated in advance .

we have a discrete function named g(t) and a continous function f(t) in in convolution integral just like this :
int(f(t-x)*g(x),x=0..t) ; 
we have just g(x) in some special points int the interval (0..t) , thus i need to convert this integral to a series.
how should i do this ? can anyone help or any idea ? i need at first a mathematical solution or idea about how to do this and then, how to do this in a software ?
tnx again.

I'd like to pay attention to the recent article "The Misfortunes of a Trio of Mathematicians Using Computer Algebra Systems. Can We Trust in Them?"

In particular, the authors consider the integral

int(abs(exp(2*Pi*Ix)+exp(2*Pi*I*y)),[x=0..1,y=0..1]),

stating "Both Mathematica and Maple return zero as the answer to this calculation. Yet this cannot be correct, because the integrand is clearly positive and nonzero in the indicated region". Unfortunately, they give only the Mathematica command to this end.

Of course, the integral under consideration is complicated so the the simple-minded trials

int(evalc(abs(exp((2*Pi*I)*x)+exp((2*Pi*I)*y))), [x = 0 .. 1, y = 0 .. 1]);

and

VectorCalculus:-int(evalc(abs(exp((2*Pi*I)*x)+exp((2*Pi*I)*y))), [x,y]=Rectangle( 0 .. 1, 0 .. 1));

fail. However,this can be found with Maple (I think with Mathematica too.) in such a way.

 

A := evalc(abs(exp((2*Pi*I)*x)+exp((2*Pi*I)*y)))

((cos(2*Pi*x)+cos(2*Pi*y))^2+(sin(2*Pi*x)+sin(2*Pi*y))^2)^(1/2)

(1)

NULL

B := simplify(A, trig)

(2*cos(2*Pi*x)*cos(2*Pi*y)+2+2*sin(2*Pi*x)*sin(2*Pi*y))^(1/2)

(2)

op(B)[1]

2*cos(2*Pi*x)*cos(2*Pi*y)+2+2*sin(2*Pi*x)*sin(2*Pi*y)

(3)

combine(op(B)[1], x)

2*cos(2*Pi*x-2*Pi*y)+2

(4)

C := eval(B, op(B)[1] = combine(op(B)[1], x))

(2*cos(2*Pi*x-2*Pi*y)+2)^(1/2)

(5)

int(C, [x = 0 .. 1, y = 0 .. 1])

4/Pi

(6)

``

 

Download int.mw

 

 

Hello,

I wanted to get the following function integrated with Maple 17:

I used the following command to get a numerical solution for my choise of limits:

Unfortunately, I just get the Integral itself back. Also, taking pi/2 as the upper limit for x does improve anything. I also tried the AllSolutions option but it does not help as well as first trying to integrate only over y and leaving x as it is in order to get at least one integration done.

I hope someone can help. I do not necessarily need to solve it for myself, so if someone gets the result and posts it here, I'm okay with that. Anyway, being able to calculate it myself would be even better.

1) Consider the followin Integral:

int(cosh(a*x)*cos(b*x),x)

How can I tell maple to give the results in terms of hyperbolic and harmonic funtions?
The maple returns the answer in term of exponentials with imaginary powers.
Also, I could not use the "simplify" command to turn the answer returned by maple into an answer in terms of what I want! Please see the code below.


2) Is there a way that I can update the maple integration database manaully?

 

Example.mw

Below is the function that I have.

 

f := (t-1)^(1/3)

p:=2;

b[n] := 2/p*(Int(f*sin(2*Pi*n*t/p), t = 0 .. p))

 

I also included a picture below to show what it is doing. Some help would be greatly appreciated. All I need to know is why maple doesn't want to evaluate bn?

 

Maple Code

 

Hi everyone!

I wander whether maple can solve the integral of trigonometric series with parametal N, the number of sereis, and how. The formation is showed as below. N is a  variable and 'm' belongs to 'k', 'n' belongs to 'l'.

the intergral of series and the orthogonality conditions

A := int(int(sum(sum(cos(2*k*Pi*x/a)*(1-cos(2*l*Pi*y/b))*(1-cos(2*m*Pi*x/a))*(1-cos(2*n*Pi*y/b)), k = 1 ..N), l = 1 .. N), x = 0 .. a), y = 0 .. b)

orthogonality codition 1:

OrthCondition1 := int(sum(cos(2*k*Pi*x/a)*cos(2*m*Pi*x/a), k = 1 .. N), x = 0 .. a) = (1/2)*a

orthogonality codition 2:

OrthCondition2 := int(sum(cos(2*l*Pi*x/b)*cos(2*m*Pi*x/b), l = 1 .. N), x = 0 .. b) = (1/2)*b;

 

See AAAA.mw

The alpha and beta are 'randomly' chosen, both >0, to produce some values, hence a value for the target function to optimize.

I am having trouble to evaluate the expression in the middle of the calculation.

Sometimes, the 'chosen' alpha and beta works fine with 'method = _d01amc' for numerical integration. and such way is really "fast".

 

But sometimes, the the 'chosen' alpha and beta will fail. Even when both of them are perfectly defined. and the integral can be easily evaluated using Int() and then evalf().

 

So what's be best way to proceed?

Thanks,

casperyc

Hello.

I have a line integral and i want to calculate and graph this line integral.

The line integral comes from a physical problem, the integrand is good in the interval [0, 2*Pi] and the value of the integrand at zero = the value of the integrand at 2*Pi.

The problem is when i use the trapizidal rule to calculate the line integral.

The results is so bad, i.e.  the value of the line integral at zero not equal to the value of the line integral at 2* Pi.

where is the problem.

Any one could help me.

Attached is my sheet.

On my system if just execute the whole worksheet maple return the last statement's integral un evaluated, however at this point if re-execute the statements from the definition of L2, it looks that maple has evaluated the last statement's integral.

 

What am I missing here, any help?

Regards!!

Strange_beh.mw

I am working on fopurier analysis and below is what I have. The method I used worked fine when I had a piecewise continuous function, but now the bottom int function below is not calculating the integral.

restart;
with(plots);

f := 2*sin(Pi-Pi*exp(-t))

 

A[0] := (1/3)*(Int(f, t = 0 .. 3))

A[0] := (1/3)*(int(f, t = 0 .. 3))   I think this one is evaluating it correctly. The bold and italicized below is the answer from this calulation

A[0] := (2/3)*Si(Pi)-(2/3)*Si(exp(-3)*Pi)

 

a[n] := (2/3)*(Int(f*cos(2*Pi*n*t*(1/3)), t = 0 .. 3))  this prints the integral like I want it to.

a[n] := (2/3)*(int(f*cos(2*Pi*n*t*(1/3)), t = 0 .. 3))  this just reporints the integral, I want it to evaluate it. Am I doing something wrong? Even with the assume(n, integer); it still just prints the integral.

 

 

Is there any help that y'all can provide? My professor is not ver proficent in the Math softwares out there.

 

 While calculating an integral including complex numbers, I have encountered with the output "undefined if a+ib>0". What does this mean?A complex number bigger than zero???

Hi,

I read this thread, but it didnt really help.

test.mw (i noticed an error in the expression pj, now fixed!)

See this file above.

The function needs to be maximized is "loglik", with constraints,

f0>0, sigma>0 and mu is FREE. Only three variables.

 

Using some calculus and by changing variables, I was able to use Gauss–Hermite quadrature, and do some approximation, to get the values:

f0=105.9535

mu=-2.1587

sigma=3.5156

 

I am wondering if Maple can do this straight away. By that, I mean, without doing much further,  how could I get the values? (most efficiently and quickly)

 

Thanks,

 

casper

 

http://www.mapleprimes.com/questions/122813-Is-It-Possible-To-Optimize-This-Kind-Of-Integral

Hi all

I have a mathematical problem and I asked it in various sites but the answers till yet are not correct.

Assume that we have:

T[m]:=t->t^m:
b[n,m]:=unapply(piecewise(t>=(n-1)*tj/N and t<n*tj/N, T[m](N*t-(n-1)*tj), 0), t):

where n,N,tj are known constants. furthermore assume that we want to comute the following integral:

for following approximations:

I have written the following code but it seems to be incorrect:

V1:=Vector([seq(seq(b[n,m](t),m=0..1),n=1..3)]);
V:=evalf(V1.Transpose(V1));

the original program is :

taaylor.mws

I will be so grateful if any one can help me to solve it by maple

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

Maple evaluates

Int(ln(1+x)/(1+x^2),x=0..1);

as

1/2*I*dilog(1/2+1/2*I)-1/2*I*dilog(1/2-1/2*I)+1/4*Pi*ln(2)-Catalan

It would be nice if Maple could simplify this to 1/8*Pi*ln(2). If you evalf the integral and use "identify" then Maple does return 1/8*Pi*ln(2).

MMa directly returns 1/8*Pi*ln(2), which seems preferable.

can i compute this integral so that the answer does not have any complex part ? i just want the integral returns me real function or number .
 i also use evalf(int ... ) and evalf(Int ... ) and two different answers were given ! what is wrong ?
tnx for help in advance .

restart:

int(log(sec(x)+ tan(x)),x);

-I*ln(exp(I*x))*ln((-I*(exp(I*x))^2+I+2*exp(I*x))/((exp(I*x))^2+1))+(1/2)*Pi*ln(-I*(exp(I*x)-I))+I*dilog(-I*exp(I*x))+I*dilog(-I*(I+exp(I*x)))+I*ln(exp(I*x))*ln(-I*(I+exp(I*x)))

(1)

evalf(Int(log(sec(x)+ tan(x)),x=0..10));

3.493589070+16.61151993*I

(2)

evalf(int(log(sec(x)+ tan(x)),x=0..10));

7.035851984+23.35343547*I

(3)

 

 

Download integral.mw

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