how to implement an infinite series?

3 次查看（过去 30 天）

显示更早的评论

Fatma Abdullah 2017-6-26

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/346259-how-to-implement-an-infinite-series

评论： Fatma Abdullah 2017-7-12

在 MATLAB Online 中打开

im trying to implement the function in the attached image but i get that error "

Attempt to reference field of non-structure array.

at the step where (v) is being calculated. i dont know what it means and i dont know how to fix it

if true
clc;
close all;
N=32;
K=14;
c=0.8;
p=1;
pc=p*gamma(1+2/c);
T=399.01923099362968691492508329876;
s=T/(pc*(1+lambda);
syms i j f
nlambda = 50;
lambdavals2 = linspace(0,30,nlambda);
lambdavals = 10.^(lambdavals2/10);
for L = 1 : nlambda
  lambda = lambdavals(L);
 for i=K:N/2
     w=nchoosek(N/2,i) %first series over i
     z=arrayfun(@(j)nchoosek(K+i-1,j)*(-1)^(j),0 :K+i-1, 'uniform',0)%second series over j
     v=symsum((-1)^(f)*(N+j-K-i+1)^(f)*gamma((f+1)*c/2)/(factorial(f)*(p*(T/(pc*(1+lambda))))^((f+1)*c/2)),i ,0, Inf)%third series over f
      pd1temp=w*z*v
       end
     pd1 = c * K * nchoosek(N/2,K) * pd1temp;   %the whole function
     end
end

27 个评论
显示 25更早的评论隐藏 25更早的评论

Walter Roberson 2017-7-4

The denominator of your summation includes rho and s to the power of (f+1)*c/2 with f growing to 0. You could imagine that c might be small, but since it is positive, no matter how small it is, (f+1)*c/2 is going to grow indefinitely large as f approaches infinity. So we have a denominator that includes (rho*s) to an indefinitely large power as f approaches infinity.

Now, if rho*s is greater than 1, then as the power (f+1)*c/2 grows indefinitely large, so would (rho*s)^((f+1)*c/2) . That would lead you to an indefinitely large positive denominator. If the numerator of the fraction is growing more slowly then the implication is that the fraction as a whole would converge to 0, allowing for the possibility of a finite summation.

But if rho*s is less than 1, then as the power (f+1)*c/2 grows indefinitely large, then (rho*s)^((f+1)*c/2) would grow indefinitely small. If the numerator of the fraction is not growing smaller at a faster rate, then this would lead to an indefinitely small denominator which would explode, leading to divergence of the sum.

Can (rho*s) be less than 1? If you work through your equations, rho*s = 399.01923099362968691492508329876 / (gamma(2/c + 1)*(lambda + 1)) . Your c is permitted to range up to infinity, so 2/c can decrease indefinitely towards, so gamma(2/c+1) can head towards gamma(1) = 1. So what about lambda? We do not have an expressed limit on lambda so far; looking at the way lambdavals is constructed, it would appear that lambda would always be at least 1 and could be at least 1000, possibly more, so that gives you 399/1000 which is less than 1, so Yes, rho*s can indeed be less than 1. gamma(2/c+1) would have to be less than about 1/4 to prevent that, and that combination cannot happen for non-negative c.

Maybe we had better have another look at the numerator. The N+ etc looks non-negative, so that to the power f is going to grow indefinitely large. gamma((f+1)*c/2) is going to grow indefinitely large. With those two growing indefinitely large, if the denominator grows indefinitely small, we have divergence.

So back to the denominator then, where I have neglected the factorial(f) . Perhaps that could be shown to grow more quickly than (rho*s)^((f+1)*c/2) falls.... ah yes, it appears that for constant rho*s that is indeed the case, that factorial(f) grows faster than (rho*s)^((f+1)*c/2) decreases. So maybe we do not need to worry? But then there is the question of whether the gamma in the numerator cancels that out... I can see circumstances in which it can...

At the moment it is looking to me as if the sum can be divergent, but it needs more examination.

Walter Roberson 2017-7-11

David, you wrote,

B = (N+k-j-i+1)/(rho*s)^(c/2). [modified]

for c = 2: divergent for |B| > 1

That test looks at any one B value, but the B values change because there are two other levels of summation happening that involve changing some of those values. I just wanted to check some things along those lines:

In the (N+k-j-i+1) component, N and k are fixed in all levels of the summation, but i and j can increase relative to their starting values; an increase of i or j would give a smaller numerator, which would lead to a smaller B value (provided the terms are non-negative, which should be the case given the constraints.) Therefore we should be able to test the first combination for B, and if it gives a value < 1 then it is not possible for any of the other possibilities to end up with B > 1.

The starting combination is i = k and j = 0, which would lead to (N+k-0-k+1) = N+1 . So looking overall, if c == 2 and (N+1)/(rho* s)^(c/2) > 1 then there would be divergence. Taking into account that this is the c==2 case, that could be rewritten as N > (rho*s) -1 would give divergence.

Let's see... for c = 2, s works out as T/(rho*(1+lambda)) so rho*s works out as T/(1+lambda), so N > T/(1+lambda) - 1 . The right hand side is least for larger lambda . We are not given a hard constraint on lambda but according to the code it can be up to 10^(30/10) = 1000, and T is about 400. For simplicity let lambda = 999 so 1+lambda = 1000 and approximate T as 400, so the constraint becomes N > 1/4 - 1 so Yes, there would be a problem . The largest supported lambda would become T/(N+1)-1 which would be about 12 for N = 32 and the given T.

On the other hand, the code is evaluating the summation for multiple lambda, so for the given N and T ranges, this mostly tells us that we can stop going fairly early under this circumstance of c = 2 exactly.

David Goodmanson 2017-7-12

编辑：David Goodmanson 2017-7-12

在 MATLAB Online 中打开

Hi Walter, thanks for your continued interest in this. Unfortunately I seem to be losing it trying to write the expression for B, which should not be that hard. Per the numerator in the original posting, it's actually

B = (N+j-k-i+1)/(rho s)^(c/2).        [modified again]
not
B = (N+k-j-i+1)/(rho s)^(c/2).        [modified]

so i and j come in with opposite sign and you can't make the same conclusion that you have. I apologize if this costs you some time.

When c=2 the gamma functions go away and you get a straightforward geometric series that converges for |B|<1. A worse problem is when c < 2 and |B| is only moderately large. This code shows log10 of the individual terms:

n = 0:63000;
c = 1.8;
B = 3;
y = .434*(n*log(B) + gammaln((n+1)*(c/2)) - gammaln(n+1));
plot(n,y)

and it tops out at around n = 23,000 where the value of the term itself is roughly 1e992, eventually dropping to 1e-20 at around n = 62,600. Not a formula for success, so |B| had better be small. I can't vouch for its accuracy but the integral method does give a normal looking answer here, .2847.

Fatma Abdullah 2017-7-12

在 MATLAB Online 中打开

I've forgot to mention that rho would actually be cancelled as described in the attached img......deeply sorry for not mentioning that earlier

I've tried to code differently as:

if true
  clc;
close all;
M=8;
N=32;
K=14;
c=1.2;
pc=p*gamma(1+2/c);
s1=1;
T=54.199575;
syms f
nlambda = 30;
lambdavals2 = linspace(0,30,nlambda);
lambdavals = 10.^(lambdavals2/10);
pd = zeros(1, nlambda);
for L = 1 : nlambda
lambda = lambdavals(L);
Gsum=0;
S=T*p/(p*gamma(1+2/c)*(1+lambda))
pc=p*gamma(1+2/c);
for i=K:N/2
   i;
   w=nchoosek(N/2,i); %first series over i
   z=0;
   sumjf1=0;
   for j1=0:K+i-1
     z=(nchoosek(K+i-1,j1)*((-1)^j1));%second series over j1
   v1=subs((((-1)^(f))*((N+j1-K-i+1)^(f))*gamma((f+1)*c/2)/(factorial(f)*((T/(gamma(1+2/c)*(1+lambda)))^((f+1)*c/2)))),f,0,inf);%infinite series over f
   v1sum=sum(v1);
    pd1temp=v1sum*z
    sumjf1=sumjf1+pd1temp;
   end
   sumjf=sumjf1*w;
 Gsum=Gsum+sumjf ;
end
   pd1 =1*c* K * nchoosek(N/2,K) * Gsum    %the whole function
    pdtemp = arrayfun(@(i) nchoosek(M,i)*(pd1)^(i)*(1-pd1)^(M-i), s1:M, 'Uniform' ,0);%pd1 is implemented in this function of pd to draw pd against lambda(db)
  pdsum = sum( [pdtemp{:}] ); 
              pd(L) = pdsum;
end
plot(lambdavals2, pd);hold on
axis([0 30 0 1])
end

Fatma Abdullah 2017-7-12

编辑：Fatma Abdullah 2017-7-12

its T/[pc*(1+lambda)] and its the same in the image just a different way of writing ....i might have missed the parenthesis in the code

Fatma Abdullah 2017-7-12

and the value of T is not always 399 it could be any value >0 to thousands .....its calculated according to other equations mentioned in the paper

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Walter Roberson 2017-6-26

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/346259-how-to-implement-an-infinite-series#answer_271959

在 MATLAB Online 中打开

You have

i=13;

which assigns a specific numeric value to i . Then, a few lines later, you have

syms pfa1 i T j f

which overwrites i with a symbolic value, equivalent to doing

i = sym('i');

So now i is symbolic with no value other than itself, and has completely lost track of the fact it had previously been assigned 13.

Then later in your code you assume that i is still 13.

7 个评论
显示 5更早的评论隐藏 5更早的评论

Walter Roberson 2017-6-29

You asked for an infinite sum; that can potentially take a long time and a lot of memory unless convergence can be established.

That is why I have been asking the questions about the constraints: because knowing the constraints can make it much easier to establish whether something converges or not.

Fatma Abdullah 2017-7-3

N is always integer positive even number,so yes N is an integer that is a multiple of 2, k is required to be an integer positive between value>0 and N/2 and could be even or odd so the limits is ]0,N/2] , c is limited between value>0 and infinity including fractions and is required to be positive so the limits of c is ]0,inf[

请先登录，再进行评论。

请先登录，再回答此问题。