# An numeric Inconsistency in mathematica

Here are my mathematica command and output:

FourierTransform[((a*Cos[2*c*\[Pi]*x]*Sin[b*\[Pi]*x])/(\[Pi]*x))^3, x,2*Pi*f, FourierParameters -> {-1, 1}]

(1/(256 \[Pi]))a^3 (-9 (2 b + c - 2 f)^2 Sign[2 b + c - 2 f] -
3 (6 b + c - 2 f)^2 Sign[6 b + c - 2 f] -
3 (2 b + 3 c - 2 f)^2 Sign[-2 b - 3 c + 2 f] -
3 (2 b - 3 c + 2 f)^2 Sign[
2 b - 3 c + 2 f] - (6 b - 3 c + 2 f)^2 Sign[6 b - 3 c + 2 f] +
9 (-2 b + c - 2 f)^2 Sign[2 b - c + 2 f] +
3 (-6 b + c - 2 f)^2 Sign[6 b - c + 2 f] -
3 (-6 b + c + 2 f)^2 Sign[-6 b + c + 2 f] -
9 (-2 b + c + 2 f)^2 Sign[-2 b + c + 2 f] -
9 (2 b + c + 2 f)^2 Sign[2 b + c + 2 f] -
3 (6 b + c + 2 f)^2 Sign[
6 b + c + 2 f] + (-6 b + 3 c + 2 f)^2 Sign[-6 b + 3 c + 2 f] +
3 (-2 b + 3 c + 2 f)^2 Sign[-2 b + 3 c + 2 f] +
3 (2 b + 3 c + 2 f)^2 Sign[
2 b + 3 c + 2 f] + (6 b + 3 c + 2 f)^2 Sign[
6 b + 3 c + 2 f] - (6 b + 3 c - 2 f)^2 Sign[-6 b \[Pi] -
3 c \[Pi] + 2 f \[Pi]])


for example for first statement (Sign[b + 2 c - 2 f]) for a=1, b=3 , c=20 , we should have a sign function like this Sign[43 - 2 f] but as I put this number in first command I don't get any Sign[43 - 2 f] in my output

(or for Sign[2 b + 3 c + 2 f] for a=1, b=3 , c=20 => Sign[66+ 2 f])

FourierTransform[((1*Cos[2*20*\[Pi]*x]*Sin[3*\[Pi]*x])/(\[Pi]*
x))^3, x, 2*Pi*f, FourierParameters -> {-1, 1}]

(1/(256 \[Pi]))(-(129 - 2 f)^2 Sign[-129 + 2 f] +
3 (123 - 2 f)^2 Sign[-123 + 2 f] -
3 (117 - 2 f)^2 Sign[-117 + 2 f] + (111 - 2 f)^2 Sign[-111 + 2 f] -
3 (49 - 2 f)^2 Sign[-49 + 2 f] + 9 (43 - 2 f)^2 Sign[-43 + 2 f] -
9 (37 - 2 f)^2 Sign[-37 + 2 f] + 3 (31 - 2 f)^2 Sign[-31 + 2 f] -
3 (31 + 2 f)^2 Sign[31 + 2 f] + 9 (37 + 2 f)^2 Sign[37 + 2 f] -
9 (43 + 2 f)^2 Sign[43 + 2 f] +
3 (49 + 2 f)^2 Sign[49 + 2 f] - (111 + 2 f)^2 Sign[111 + 2 f] +
3 (117 + 2 f)^2 Sign[117 + 2 f] -
3 (123 + 2 f)^2 Sign[123 + 2 f] + (129 + 2 f)^2 Sign[129 + 2 f])


what is my mistake?

• Try restarting he kernel and doing it again (you probably have old definitions around)... I do get the Sign[43 - 2 f] for {a -> 1, b -> 3, c -> 20} – bill s Oct 18 '16 at 22:53
• Unfortunately it doesn't work again , I restart my kernel with Evaluation->Quit Kernal and here is my result -> obrazki.elektroda.pl/6442679300_1476858409.png – Ehsan Zakeri Oct 19 '16 at 6:28
• What symbol are you using to get that little x -- multiplication can be done with space or with * – bill s Oct 19 '16 at 13:16

## 1 Answer

Copy this line:

fft = FourierTransform[((a*Cos[2*c*\[Pi]*x]*Sin[b*\[Pi]*x])/(\[Pi]*x))^3, x, 2*Pi*f, FourierParameters -> {-1, 1}]


Now evaluate at the points:

fft /. {a -> 1, b -> 3, c -> 20}


You will see that the very first Sin term is Sin[43 - 2 f].