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Dear all

 

I have a confusion between these symbol

Sum , add and sum

If we consider u(n) is a sequence and n integer

and what is the difference between 

sum( u(n),n=0..infinity)

Sum(u(n),n=0..infinity)

and sum('u(n)', n=0..infinity)

Many thanks

Hello

Any idea about the summation of Fibonacci sequence

 

Fibonacci.mw

 

Best regards

 

Hello everybody.

I have a function:

f(x,y)=GAMMA(y, -ln(x))/GAMMA(y)

seq(sum(f(x, y), y = 0 .. 1), x = 0 .. 5)

 

and I got a error message:

Error, (in ln) numeric exception: division by zero ??
This is normal behavior in seq function or Bug?

 

but  when I'm first calculate the sum sol := sum(f(x, y), y = 0 .. 1) -> x,

and evalf([seq(sol, x = 0 .. 5)]) ->[0., 1., 2., 3., 4., 5.] works fine.

 

Seq-division_by_zero.mw

Mariusz Iwaniuk

Dear all

 

If its possible in  Maple to change the integral of the sum to  the sum of integrals when I calucle the integral of a function series

 

Thank you

Hello guys,

I was just playing around with the Shanks transformation of a power series, when I noticed that polynomials aren't evaluated as I would expect.
I created this minimal working example; the function s should evaluate for z=0 to a[0], however it return simply 0.
Is there something I messed up?

restart

s := proc (n, z) options operator, arrow; sum(a[k]*z^k, k = 0 .. n) end proc;

proc (n, z) options operator, arrow; sum(a[k]*z^k, k = 0 .. n) end proc

(1)

series(s(n, z), z = 0)

series(a[0]+a[1]*z+a[2]*z^2+a[3]*z^3+a[4]*z^4+a[5]*z^5+O(z^6),z,6)

(2)

The value of s in z=0 should be a[0], however it returns 0:

s(n, 0)

0

(3)

s(1, 0)

0

(4)

Download evaluate_sum.mw

 

Thanks for your help,

Sören

Please check this:

N:=3;

sum1 := lcm(N, 0)+lcm(N, 1)+lcm(N, 2)+lcm(N, 3);

sum2 := sum(lcm(N, k), k = 0 .. N);

 

Why is sum2 wrong?

 

Regards,

César Lozada

 

Hey,

 

I'm trying to make a sum of only odd numbers regarding 2 functions and I've come up with what I assume is a terrible way to do it...

So basically what I want to write is this:

 


And this should give me something like this:


This works, what I have now. The thing is I need to be able to add and subtract terms in order to compare with other stuff and it just seems so inefficient right now.

 

Thanks in advance!

 

 

I have encountered a behavior of Maple that I find hard to explain and I am hoping for help. The command

sum(floor((exp(Pi)-Pi)*n)/3^n,n=0..infinity);

was meant to be an example of "High-precision fraud" as in the 1992 paper of Borwein and Borwein, and indeed it gives 29/2 to within 531 digits. But I am unable to make Maple see this; indeed I get with evalf(%,1000)

14.50000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

I find it hard to guess why Maple gets this wrong, actually. The point of the example is that floor((exp(Pi)-Pi)*n)=20*n-1 for many n, but Maple has no problems finding the first failure at n=1112. It must hence be trying something more advanced than just adding up all summands until the tail sum is small enough to satisfy the precision? I guess an alternative approach would indeed be possible since what is being "floored" is relatively simple, but then that seems to be buggy?

Would be very grateful for any assistance!

Best,

Soren Eilers

 

 

 

 

L(1):=[1,2,3]:

L(2):=[A,B,C]:

sum(nops(L(i)),i=1..2):

 

The real result of the sum is : 6.

Maple's result is : 2.

Yes, Maple gives an error.
But it doesn't matter, Maple is the most extraordinary software of the world !!!

I can not find a cumulative sum procedure in maple2016.

I want to compute the cumulative sum of a list or vector.

A:=[1, 2, 3, 4];

cumsum(A); ## or whatever the maple function is... 

[1,3,6,10]

Tom Dean

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

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