Trying to code a tricky double summation

Question

Michael Vaughan 2020-9-24

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/599143-trying-to-code-a-tricky-double-summation

评论： Ameer Hamza 2020-9-24

采纳的回答： Ameer Hamza

So I'm trying to write code that calculates the following:

I have two functions of the same two variables, call them D(x,y) and T(x,y)

I'm trying to write a function that accepts inputs x,y,z and outputs:

So, there are two summations, the outer one takes the sum from i=0 up to min(x,y), and the inner sum takes the summation of terms from k=0 up to x-i

The inner sum isn't as complicated as it looks, it has a (-1)^k at the start and then just multiples the functions D(a,b) and T(a,b)^(z/2) where a and b depend on both where you are in the indexing of your outer summation and inner summation

This is a much harder proram than I have written before, so tips, insights, general advice, and even completed code are welcome!! Thanks for your time

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Answer 1

Ameer Hamza 2020-9-24

1
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/599143-trying-to-code-a-tricky-double-summation#answer_499741

在 MATLAB Online 中打开

Something like this should work

s = 0;
for i=0:min(x,y)
    for k=0:x-1
        s = s + (-1)^k * D(2*x-2*k-2*i, 2*y+2-2*i) * T(2*x-2*k-2*i, 2*y+2-2*i)^(z/2)
    end
end

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Michael Vaughan 2020-9-24

How come when I run the code:

function sum = Sum(x,y,z)

s = 0;

for i=0:min(x,y)

for k=0:x-i

s = s + (-1)^k * QuantumDimension(2*x-2*k-2*i, 2*y+2-2*i) * Twist(2*x-2*k-2*i, 2*y+2-2*i)^(z/2);

end

s

end

I get the output:

s =

((1/q^7 - q^7)*(1/q^(7/2) - q^(7/2))^2)/((q - 1/q)*(1/q^(1/2) - q^(1/2))^2) - ((1/q^4 - q^4)*(1/q^(7/2) - q^(7/2)))/((q - 1/q)*(1/q^(1/2) - q^(1/2))) + ((1/q^5 - q^5)*(1/q^(3/2) - q^(3/2))*(1/q^(7/2) - q^(7/2)))/((q - 1/q)*(1/q^(1/2) - q^(1/2))^2) - ((1/q^6 - q^6)*(1/q^(5/2) - q^(5/2))*(1/q^(7/2) - q^(7/2)))/((q - 1/q)*(1/q^(1/2) - q^(1/2))^2) - ((1/q^8 - q^8)*(1/q^(7/2) - q^(7/2))*(1/q^(9/2) - q^(9/2)))/((q - 1/q)*(1/q^(1/2) - q^(1/2))^2)

s =

((1/q^3 - q^3)*(1/q^(5/2) - q^(5/2)))/((q - 1/q)*(1/q^(1/2) - q^(1/2))) - ((1/q^4 - q^4)*(1/q^(7/2) - q^(7/2)))/((q - 1/q)*(1/q^(1/2) - q^(1/2))) + ((1/q^5 - q^5)*(1/q^(5/2) - q^(5/2))^2)/((q - 1/q)*(1/q^(1/2) - q^(1/2))^2) + ((1/q^7 - q^7)*(1/q^(7/2) - q^(7/2))^2)/((q - 1/q)*(1/q^(1/2) - q^(1/2))^2) - ((1/q^4 - q^4)*(1/q^(3/2) - q^(3/2))*(1/q^(5/2) - q^(5/2)))/((q - 1/q)*(1/q^(1/2) - q^(1/2))^2) + ((1/q^5 - q^5)*(1/q^(3/2) - q^(3/2))*(1/q^(7/2) - q^(7/2)))/((q - 1/q)*(1/q^(1/2) - q^(1/2))^2) - (2*(1/q^6 - q^6)*(1/q^(5/2) - q^(5/2))*(1/q^(7/2) - q^(7/2)))/((q - 1/q)*(1/q^(1/2) - q^(1/2))^2) - ((1/q^8 - q^8)*(1/q^(7/2) - q^(7/2))*(1/q^(9/2) - q^(9/2)))/((q - 1/q)*(1/q^(1/2) - q^(1/2))^2)

s =

((1/q^3 - q^3)*(1/q^(5/2) - q^(5/2)))/((q - 1/q)*(1/q^(1/2) - q^(1/2))) - ((1/q^2 - q^2)*(1/q^(3/2) - q^(3/2)))/((q - 1/q)*(1/q^(1/2) - q^(1/2))) - ((1/q^4 - q^4)*(1/q^(7/2) - q^(7/2)))/((q - 1/q)*(1/q^(1/2) - q^(1/2))) + ((1/q^3 - q^3)*(1/q^(3/2) - q^(3/2))^2)/((q - 1/q)*(1/q^(1/2) - q^(1/2))^2) + ((1/q^5 - q^5)*(1/q^(5/2) - q^(5/2))^2)/((q - 1/q)*(1/q^(1/2) - q^(1/2))^2) + ((1/q^7 - q^7)*(1/q^(7/2) - q^(7/2))^2)/((q - 1/q)*(1/q^(1/2) - q^(1/2))^2) - (2*(1/q^4 - q^4)*(1/q^(3/2) - q^(3/2))*(1/q^(5/2) - q^(5/2)))/((q - 1/q)*(1/q^(1/2) - q^(1/2))^2) + ((1/q^5 - q^5)*(1/q^(3/2) - q^(3/2))*(1/q^(7/2) - q^(7/2)))/((q - 1/q)*(1/q^(1/2) - q^(1/2))^2) - (2*(1/q^6 - q^6)*(1/q^(5/2) - q^(5/2))*(1/q^(7/2) - q^(7/2)))/((q - 1/q)*(1/q^(1/2) - q^(1/2))^2) - ((1/q^8 - q^8)*(1/q^(7/2) - q^(7/2))*(1/q^(9/2) - q^(9/2)))/((q - 1/q)*(1/q^(1/2) - q^(1/2))^2)

I would much rather just get one output..... why is it giving me s 3 times??

Michael Vaughan 2020-9-24

Thank you very much. I'm now trying to give this function a call handle,

I type S1 = @Sum1.m

Where Sum1.m is the code:

function sum = Sum1(x,y,z)

s = 0;

for i=0:min(x,y)

for k=0:x-i

s = s + (-1)^k * QuantumDimension(2*x-2*k-2*i, 2*y+2-2*i) * Twist(2*x-2*k-2*i, 2*y+2-2*i)^(z/2);

end

s

When I type in S1(3,2,1) I get:

Undefined function 'Sum1.m' for input arguments of type 'double'.

What is going wrong here?? THanks for your patience

Ameer Hamza 2020-9-24

在 MATLAB Online 中打开

To make a function handle type this

S1 = @Sum1
S1(3,2,1)

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