symsum equation not matching

I am solving an equation to get the result in terms of symbol 'SI'. This is the given equation

but answer in terms of SI i am getting is not correct. I have used following code:

A=symsum(   2*m/( SI*SI*(NI-SI+1 )),m,1,SI/2   ) +  symsum( 1/ ( SI*SI*(NI-SI+1 )),m,SI/2+1,SI )

which becomes

A = 1/(2*SI*(NI - SI + 1)) + (SI + 2)/(4*SI*(NI - SI + 1))

solving for SI with A=0 gives result as

ans =
 root(z^3 - (z^2*(2*NI + 2))/2 - z/2 - 1, z, 1)
 root(z^3 - (z^2*(2*NI + 2))/2 - z/2 - 1, z, 2)
 root(z^3 - (z^2*(2*NI + 2))/2 - z/2 - 1, z, 3)

when i use vpa(ans), i get 3 equations with long terms (without z) but still answers does not match the given result shown below in diagram (NI is substituted by SL/u in the document so in square root term is actually (2/3)*NI-11/36

Any help will be highly appreciated.

Answer 1

[Walter Roberson]

No, the SI you give is not a solution to that summation being 0.

2/(SI^2*(NI-SI+1)) is independent of m, so the first term of A is

 2/(SI^2*(NI-SI+1)) * Sum(m, m=1..SI/2)

Sum(m, m=1..SI/2) for positive SI is ((SI/2)+1)*(SI/2)/2 . So you can calculate the first term as 2/(SI^2*(NI-SI+1)) * ((SI/2)+1)*(SI/2)/2 which works out as (SI+2)/(4*SI*(NI-SI+1))

1/(SI*(NI-SI+1)) is independent of m, so the second term of A is

1/(SI*(NI-SI+1)) * Sum(1, m=SI/2+1..SI)

which is 1/(SI*(NI-SI+1)) * (SI - (SI/2+1) + 1) = 1/(SI*(NI-SI+1)) * SI/2 which works out as 1/(2*(NI-SI+1))

so A = (SI+2)/(4*SI*(NI-SI+1)) + 1/(2*(NI-SI+1)) which works out as

 A = (3*SI+2)/(4*SI*(NI-SI+1))

Note: this assumes that SI/2 is a positive integer! If it is not then the summation bounds need to be more carefully defined!

Now, for (3*SI+2)/(4*SI*(NI-SI+1)) to be 0, the numerator must be 0, which requires that 3*SI+2 is 0, which would require that SI = -2/3.

SI = -2/3 violates the assumption that SI/2 is a positive integer.

Therefore there is no solution for A = 0 -- not unless you very carefully define what it means to sum with non-integer boundaries or what it means to sum from 1 to a negative number in the bounds.

Exception: if NI is infinite but SI is finite then A can be 0.

[Masood Abbasi]

Thanks Walter for the reply, but still not getting output as

I found in the document that SI should be solved for minimum value of A (no value of A was mentioned) so i used 0 for A but no luck.

[Masood Abbasi]

Solving for SI with SL=12 and u=4 (NI=3=SL/u) gives different answers (like two roots) quite different from what i get from documented equation below

[Walter Roberson]

To minimize, take

A = (3*SI+2)/(4*SI*(NI-SI+1))

and differentiate with respect to A, equate to 0 and solve for SI.

You get the two possibilities,

SI = -2/3+(1/3)*sqrt(10+6*NI)
SI = -2/3-(1/3)*sqrt(10+6*NI)

the second of those cannot be a positive integer as required, so concentrate on the first one.

With some work you can show that SI is an integer under the case that NI = 6*P^2+4*P-1 for some positive integer P -- in which case SI = 2*P, which also satisfies the condition that SI/2 is a positive integer as required for the summations to be correctly defined.

The value at the minima is A = 1/(8*P^2)

For the situation where NI is not 6*P^2+4*P-1 for some positive integer P then I think SI would have to be in the range ceil(-2/3+(1/3)*sqrt(10+6*NI) - 2) to floor(-2/3+(1/3)*sqrt(10+6*NI) + 2) .

This value of SI is not the one from the formula you show.

If we examine the formula you are given,

SI = sqrt(2/3*NI-11/36)-1/6

and as before we say that SI must be an even integer so that SI/2 is integer so that the bounds of summation SI/2 and SI/2+1 are clearly defined; then let SI = 2 * p for some integer p, and then

solve(sqrt(2/3*NI-11/36)-1/6 = 2*p, NI)

This gives NI = 6*p^2+p+1/2

which is never an integer for any integer p. Though to be fair, nothing in the problem statement requires that NI is an integer.

If we substitute these values into A we get

A = (3*p+1)/(24*p^3-4*p^2+6*p)

for large p this approaches 1/(8*p^2), the value I found earlier. None the less, my analysis suggests that the formula you gave is not correct.

So a question to ask here is whether NI is expected to be integer, or if it is expected to be 1/2 greater than an integer.

[Masood Abbasi]

Thanks Walter, as you said, the given solution in document does not seems to be the correct solution.