# Equation Evaluation Problem in Mathematica

by S_David
Tags: equation, evaluation, mathematica
 P: 601 Hello, I have the following line in Mathematica: Print[Pout = (2^-Q*E^(A/2))/SNR \!$$\*UnderoverscriptBox[\(\[Sum]$$, $$q = 0$$, $$Q$$]Binomial[Q, q] $$\*UnderoverscriptBox[\(\[Sum]$$, $$n = 0$$, $$Ne + q$$] FractionBox[ SuperscriptBox[$$(\(-1$$)\), $$n$$], $$a[n]$$] Re[ \*FractionBox[$$Meq[\(- \*FractionBox[\(A + \((2*Pi*I*n)$$\), $$2*SNR$$]\)]\), FractionBox[$$A + \((2*Pi*I*n)$$\), $$2*SNR$$]]]\)\)] But the problem is that for different values of SNR, the result will be the same all the time. Why is that happening? A, Q, and Ne are all constants. Thanks in advance
 PF Gold P: 472 What are some examples of what A Q and Ne are so i can try it? and what is meq?
P: 601
 Quote by Hepth What are some examples of what A Q and Ne are so i can try it? and what is meq?
Try these expressions:
A=23
Q=15
Ne=21
Meq[s_]:=1/(1-s)
Regards

 PF Gold P: 472 Equation Evaluation Problem in Mathematica and also, the a[n] function or array? but before that make sure youre clearing any variables youre reusing. restarting the kernel does that.
P: 601
 Quote by Hepth and also, the a[n] function or array? but before that make sure youre clearing any variables youre reusing. restarting the kernel does that.
Assume
a[n]=1
How clear all variables? I have many of them.

Regards
 P: 601 I have the same problem again in the following code: gA = 10; M = 1; Ne = 1; If[M >= 1, m = M, m = 0]; For[SNRdB = 0, SNRdB <= 10, SNRdB++, SNR = 10^(SNRdB/10); Print[F1 = \!$$\*UnderoverscriptBox[\(\[Sum]$$, $$r = m$$, $$M$$]$$\*UnderoverscriptBox[\(\[Sum]$$, $$i = 0$$, $$M - r$$]$$\*UnderoverscriptBox[\(\[Sum]$$, $$j = 0$$, $$r + i$$]$$\*UnderoverscriptBox[\(\[Sum]$$, $$k = 0$$, $$j*\((Ne - 1)$$\)] \*SuperscriptBox[$$(\(-1$$)\), $$i + j$$]*Binomial[M, r]* Binomial[M - r, i]*Binomial[r + i, j]* \*SuperscriptBox[$$E$$, FractionBox[$$\(-j$$*SNR\), $$gA$$]]* \*SuperscriptBox[$$( \*FractionBox[\(SNR$$, $$g$$])\), $$k$$]\)\)\)\)]] For[SNRdB = 0, SNRdB <= 10, SNRdB++, SNR = 10^(SNRdB/10); Print[F2 = 1 - \!$$\*UnderoverscriptBox[\(\[Sum]$$, $$r1 = m$$, $$M - 1$$]$$\*UnderoverscriptBox[\(\[Sum]$$, $$i1 = 1$$, $$M - r1$$]$$\*UnderoverscriptBox[\(\[Sum]$$, $$j1 = 1$$, $$r1 + i1$$]$$\*UnderoverscriptBox[\(\[Sum]$$, $$k1 = 0$$, $$j1*\((Ne - 1)$$\)] \*SuperscriptBox[$$(\(-1$$)\), $$i1 + j1$$]*Binomial[M, r1]* Binomial[M - r1, i1]*Binomial[r1 + i1, j1]* \*SuperscriptBox[$$E$$, FractionBox[$$\(-j1$$*SNR\), $$gA$$]]* \*SuperscriptBox[$$( \*FractionBox[\(SNR$$, $$g$$])\), $$k1$$]\)\)\)\)]] 1-1/\[ExponentialE]^(1/10) 1-\[ExponentialE]^-1/10^(9/10) 1-\[ExponentialE]^-1/10^(4/5) 1-\[ExponentialE]^-1/10^(7/10) 1-\[ExponentialE]^-1/10^(3/5) 1-\[ExponentialE]^-1/Sqrt[10] 1-\[ExponentialE]^-1/10^(2/5) 1-\[ExponentialE]^-1/10^(3/10) 1-\[ExponentialE]^-1/10^(1/5) 1-\[ExponentialE]^-1/10^(1/10) 1-1/\[ExponentialE] 1 1 1 1 1 1 1 1 1 1 1 Why is the second part constant, even though it is a dependent on SNR? Regards
 PF Gold P: 472 Because of your sums and their indices. To fix this, add: If[M > 1, m = M, m = 0]; before the second For loop. It changes the greater than equal to to just a greater than. That way the second sum in the second for loop doesn't go from 1 to 0 (1 to M-r1 == M-m==0)
 PF Gold P: 472 As for your first question, I DO get something different each time I change SNR.
P: 601
 Quote by Hepth As for your first question, I DO get something different each time I change SNR.
Really? How is that? Try to put the SNR in a For[] loop, and tell me what will happen. Because I am using a For[] loop actually.

Regarding your previous post, I have doubts that I have something wrong in the mathematical equations. So, I will double check them and see what happen then.

Thank you
 PF Gold P: 472 FOR: A = 23; Q = 15; Ne = 21; Meq[s_] := 1/(1 - s) a[n_] = 1; For[SNRdB = 0, SNRdB <= 10, SNRdB++, SNR = 10^(SNRdB/10); Print[ Pout = Refine[(1.0) (2^-Q*E^(A/2))/SNR $$\left.\left.\left.\sum _{q=0}^Q \text{Binomial}[Q,q]\sum _{n=0}^{\text{Ne}+q} \frac{(-1)^n}{a[n]}\text{Re}\left[\frac{\text{Meq}\left[-\frac{A+(2*\text{Pi}*I*n)}{2*\text{SNR}}\right]}{\frac{A+(2*\text{Pi}*I*n)}{2*\text{SNR}}}\right]\right]\right]\right]$$ The changes I made were adding the Refine to simplify the complex stuff, and multiplying by 1.0 to give me a real value. I also added the SNR changing in the for loop. Looks like it changes when SNR does. Or did I do something wrong? EDIT: oops, heres my output: 343.991 424.207 520.658 635.431 770.336 926.63 1104.7 1303.75 1521.56 1754.43 1997.27
PF Gold
P: 472
 Quote by S_David Regarding your previous post, I have doubts that I have something wrong in the mathematical equations. So, I will double check them and see what happen then. Thank you
What I mean is that for your given values, you have a sum over
$$\sum _{\text{r1}=m}^{M-1} \sum _{\text{i1}=1}^{M-\text{r1}} \sum _{\text{j1}=1}^{\text{r1}+\text{i1}} \sum _{\text{k1}=0}^{\text{j1}*(\text{Ne}-1)}$$
but you have defined :
If[M >= 1, m = M, m = 0];
and M IS 1, so m=M=1;
$$\sum _{\text{r1}=m}^{M-1} \sum _{\text{i1}=1}^{M-\text{r1}}$$
$$M=1;m=1;\sum _{\text{r1}=m}^{M-1} \sum _{\text{i1}=1}^{M-\text{r1}} 1==0$$
 P: 601 Yes, now the fake code is working. I said fake because I gave you fake parameters, so the values you got are not the expected one, because there are not in the range between 0 and 1 as it must be by definition. When I did a slight change toward the real parameters I got the expected results as following: A = 23; Q = 15; Ne = 21; Meq[s_] := 1/(1 - 0.5 s)^4 ; a[n_] = If[n == 0, 2, 1]; For[SNRdB = 0, SNRdB <= 10, SNRdB++, SNR = 10^(SNRdB/10); Print[Pout = Refine[(1.0) (2^-Q*E^(A/2))/SNR \!$$\*UnderoverscriptBox[\(\[Sum]$$, $$q = 0$$, $$Q$$]Binomial[Q, q] $$\*UnderoverscriptBox[\(\[Sum]$$, $$n = 0$$, $$Ne + q$$] FractionBox[ SuperscriptBox[$$(\(-1$$)\), $$n$$], $$a[n]$$] Re[ \*FractionBox[$$Meq[\(- \*FractionBox[\(A + \((2*Pi*I*n)$$\), $$2*SNR$$]\)]\), FractionBox[$$A + \((2*Pi*I*n)$$\), $$2*SNR$$]]]\)\)]]] 0.142877 0.246246 0.390748 0.564677 0.738295 0.875494 0.956523 0.989842 0.998584 0.999898 0.999997 But when I turn my attention to my real, relatively long code, I faced with the same problem again, although I used the same procedure as you described. I don't know why. Best Regards