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This question is related to the recent post
http://www.mapleprimes.com/questions/211460-Series-Of-Bessel-Functions

1. Consider the following fast convergent series:

f:=n->(-1)^(n+1)*1/(n+exp(n));
S1:=Sum(f(n),n=1..infinity);
evalf(S1);
S2:=Sum(f(2*n-1)+f(2*n),n=1..infinity);
evalf(S2);

As expected, the sum of the series is obtained very fast (with any precision), same results for S1 and S2.


2. Now change the series to a very slowly convergent one:

f:=n->(-1)^(n+1)/sqrt(n+sqrt(n));

evalf(S1) is computed also extremely fast, because the acceleration algorithm works here perfectly.
But evalf(S2) demonstrates a bug:

Error, (in evalf/Sum1) invalid input: `evalf/Sum/infinite` expects its 2nd argument, ix, to be of type name, but received ...


3. Let us take another series:

f:=n->(-1)^(n+1)/sqrt(n+sqrt(n)*sin(n));

Now evalf(S1) does not evaluate numerically and evalf(S2) ==> same error.
Note that I do not know whether this series is convergent or not, but the same thing happens for the obviously convergent series

f:=n->(-1)^(n+1)/sqrt(n^(11/5)+n^2*sin(n));

(because it converges slowly (but absolutely) and the acceleration fails).
I would be interested to know a method to approximate (in Maple) the sum of such series.

Edit. Now I know that the mentioned series 

converges (but note that Leibniz' test cannot be used).

I'm having trouble evaluating an expression with an infinite sum of bessel functions. The expression is:

or, in mathematical notation:

The program doesn't seem to be able to solve the expression and return a value. I only get the answer

When I try to evaluate a simple infinite sum like 

the answer is a value. Breaking up the expression in components and evaluating each one works in some cases, e.g. the expression 

which returns a value. However, the expression

fails, and returns

Is there a trick to evaluating infinite sums with indices appearing inside a function in the summation expression? Or am I doing something wrong?

hi,i am studying the maple most recent.But when calculating function integral,I ran into trouble.I hope to get your help.Here is the code I wrote, but it runs a very long time. How to effectively reduce the integration time?

restart;
with(student);
assume(n::integer);
Fourierf := proc (sigma, a, b, N) local A, A0, B, T, S, Ff; T := b-a; A0 := int(sigma, t = a .. b); A := int(sigma*sin(n*Pi*t/T), t = a .. b); B := int(sigma*cos(n*Pi*t/T), t = a .. b); S := sum(A*sin(n*Pi*t/T)+B*cos(n*Pi*t/T), n = 1 .. N)+(1/2)*A0; Ff := unapply(S, t) end proc;

f := proc (t) options operator, arrow; piecewise(t < .13*2.6 and 0 <= t, 100*t/(.13*2.6), .13*2.6 <= t and t < 2.6, 100, 2.6 <= t and t < 2.6*1.1, 0) end proc;

sigma := f(t);
a := 0;
b := 1.1*2.6;
s1 := unapply((Fourierf(sigma, a, b, 500))(t)/uw0, t);

s2 := unapply((Fourierf(sigma, a, b, 500))(t)/ua0, t);
A1 := (2*n+1)^2*Pi^2*(C3+1+sqrt(4*C1*C2*C3+C3^2-2*C3+1))/(8*C1*C2-8);
A2 := (2*n+1)^2*Pi^2*(C3+1-sqrt(4*C1*C2*C3+C3^2-2*C3+1))/(8*C1*C2-8);
g := -C2*Cww*(diff(s1(x), `$`(x, 2)))+Caa*(diff(s2(x), `$`(x, 2))+(n+1/2)^2*Pi^2*(diff(s2(x), x)));
f1 := -(1/2)*(n+1/2)^2*Pi^2*sqrt(4*C1*C2*C3+C3^2-2*C3+1)+C2*Cww*((D@@1)(s1))(0)-Caa*((D@@1)(s2))(0)+(n+1/2)^2*Pi^2*(C2-(1/2)*C3+1/2);

CN := ((2*(int(exp(-A1*x)*g, x = 0 .. t)-f1))*exp(A1*t)-(2*(int(exp(-A2*x)*g, x = 0 .. t)-f1))*exp(A2*t))/((n+1/2)^3*Pi^3*sqrt(4*C1*C2*C3+C3^2-2*C3+1));
ua := sum(CN*sin((n+1/2)*Pi*z), n = 0 .. 100);

 

 

Before version 2016 Maple was incredibly good at evaluating an infinite power series and returning a simple function, e.g. 1/(3x+2).  Now version 2016 just returns the input sum expression with no change.  Is there some new command to get the old results?

Hi!

 

I wonder how is it possible to numerically evaluate two-dimensional sum, something like this:

sum

Hi everyone. This problem is driving me nuts. I'm pretty sure it's a glitch but I'm not sure how to solve it. I'm trying to do some data analysis with Maple:

(as a side note, even if I remove the for loop but don't execute the restart command the error remains, however if I get rid of the for loop and execute the restart command it is fine.)

Any help would be greatly appreciated. As it stands this is really driving me insane.

I get the maple result as a product and sum off terms

 

x= (a+b*c)/d

i want to convert it to

x= a/d+b*c/d

i try to use convert(x,?)  

 

Hi,

 

I am trying to use a simple sum ( f:=t,n->sum(MyFunction(k,t),k=1..n) ), but the sum function doesn't "work" :

Error, (in MTM:-sum) invalid input: ArrayTools:-AddAlongDimension expects its 2nd argument, DimToAdd, to be of type posint, but received k = 1 .. n

 

As far as I can understand, I have loaded the package MTM, and now, the sum has been redefined to do something else, and now I can't use a "normal" sum anymore. Or can I ?

Does someone know how to use the "normal" sum without unloading the MTM package ?

 

Thanks

 

How do I multiply the 4x into the summation to get  (Sum(4*n*a[n]*x^(n), n = 0 .. infinity))  and same idea for the 3rd third?

Also, how do I go from   Sum(a[n-2]*x^(n-2), n = 2 .. infinity)  to  Sum(a[n]*x^(n), n = 0 .. infinity)  by manipulating the indices?

 

I have the following expression.

 

What I want to do is take the derivative of this expression with respect to,

 

Where L is somewhere between 1 and n. 

Using the diff operator, I would like to get something like,

 

But I don't.  Instead, I get zero.  I want to leave n undefined so that the summation isn't automatically expanded out to alot of terms, thereby making it much more difficult to read.  This is especially true as my formulas become more complicated.

 

Thanks!!

 

Dear all;

 

Hello everybody, I need your help to dispaly some values obtained using my function f. When I run the code there is no results obtained. Many thanks.

restart:

# The vectors e(i) satify the folowing conditions
e(0)*e(1)=e(n-1) assuming  1<n;
e(0)*e(0)=e(2):
e(1)*e(1)=e(n-1) assuming  1<n: :
e(2)*e(1)=e(n) assuming  1<n:
#
for i from 1  to n-1 do
e(i)*e(0)=e(i+1);
end do:

# We define the function f
f:=e(0)->e(0)+(n-3)*e(1);
f:=e(1)->(n-2)*e(1);
for i from 2  to 3 do
f:=e(i)->(n+i-3)*e(i)+(i-1)*(n-3)*e(n-3+i);
end do:

for i from 4 to n do
f:=e(i)->(n+i-3)*e(i)
end do:

# We define the two vectors
x:=sum(alpha(k)*e(k),k=0..n);
y:=sum(beta(k)*e(k),k=0..n);

#Question : I would like to compute the following  but there is no display of the solution. 
(x*y);
f(x*y);
f(x);
f(y);
x*f(y);
f(x)*y;
f(x*y)- f(x)*y-x*f(y);

Hello,

I have a vector with 4 components. Each component is a polynom.

N := Vector[row](4, {(1) = -(1/2)*t^3+t^2-(1/2)*t, (2) = (3/2)*t^3-(5/2)*t^2+1, (3) = -(3/2)*t^3+2*t^2+(1/2)*t, (4) = (1/2)*t^3-(1/2)*t^2});
N[1]+N[2]+N[3]+N[4];
sum(N[i], i = 1..4);

I don't why the function sum doesn't work in this case. If i do the sum like this N[1]+N[2]+N[3]+N[4];, it works but it doesn't work with the use of the sum function.

Thank you for your help.

Good evening,

I am trying to solve the following sum equation:

 

sum((factorial(n)/factorial(2*n))^n, n = 1 .. infinity)

 

but I require the step-by-step solution as to how it is done, but can't seem to find that option. 

 

If anone could help me out, that would be great.

 

Kind regards,

A

 

Cn ={ 1, n = 0 ,                                                  }

      {Xn−1[sum of] k=0   C(k)C(n−1−k) , otherwise.  }

 

looking to complete the definition of catalan so that catalan(n) returns Cn whenever n is a non-negative integer. usin g the definition above...any help appreciated

 

catalan:=
proc(n::TYPE)
description "Print the n'th Catalan number.";
option remember;
---MORE STUFF HERE---
end proc; # catalan

I'm begginer and I don't know, whether this equation can be solved in Maple:

equation

The only variable is N.

i=1,2,3,...,n

n=the nearest smaller natural number from the second expression.

The equation sides probably won't be equal precisely, also I'm looking for some N when equation side's values are close enough.

Can you help me with programming this into Maple?

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