C# pattern - is it same as in image?

Question

2.75/5 (4 votes)

See more:

Im not really sure about that is it same pattern i wrote in c# as in this image?
code:

C#

-Math.Cos(x1) * Math.Cos(x2) * Math.Pow(Math.E, Math.Pow(-(x1 - Math.PI), 2) - Math.Pow((x2 - Math.PI),2));

Right pattern:
http://s18.postimage.org/btxjslcaf/wzor.png[^]

Thank u you for help

Posted 13-May-12 4:31am

trucha13657

Updated 13-May-12 18:50pm

Shahin Khorshidnia

v2

Add a Solution

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Clifford Nelson · Answer 1 · 2012-05-13T05:26:00

Solution 1

Acutally made error in putting the "-" inside the Math.Pow(-(x1 - Math.PI), 2), and the approach is not the best. I would also not use the Pow for square because of perfomance, and would use the Exp function:

v

ar VALUE = - Math.Cos(x1) * Math.Cos(x2) * Math.Exp(-(x1 -nu) *(x1 - Math.PI) -(x2 -nu) * (x2 -Math.PI));

Posted 13-May-12 5:26am

Clifford Nelson

Updated 13-May-12 5:29am

v2

Comments

Maciej Los 13-May-12 11:59am

Does the pattern (from image) is: cos(x1)*cos(x2)-(((-x1-Pi)^2)-(-x2-Pi)^2))

VJ Reddy 13-May-12 12:15pm

Good answer. 5!

Clifford Nelson 13-May-12 12:16pm

You missed a piece of the equation:
-cos(x1)*cos(x2)(e ^(-(x1-Pi)^2)-(x2-Pi)^2))

Maciej Los 13-May-12 12:34pm

I'm a blind man ;)
Of course, 5!

Shahin Khorshidnia 14-May-12 0:49am

"Acutally made error in putting the - inside the Math.Pow(-(x1 - Math.PI), 2), and the approach is not the best."... What kind of error?!

Clifford Nelson 14-May-12 2:21am

"Math.Pow(-(x1 - Math.PI), 2)" should have been "- Math.Pow((x1 - Math.PI), 2)"

Shahin Khorshidnia 14-May-12 3:02am

You mean a logical error. Ok. Thanks

trucha13657 15-May-12 9:07am

So the final pattern is:
- Math.Cos(x1) * Math.Cos(x2) * Math.Exp(-(x1 -nu) *(x1 - Math.PI) -(x2 -nu) * (x2 -Math.PI));

? ;)