evaluate assuming odd integers?

[[e-mail]]

Is there a way in Mathematica to evaluate an expression assuming some
varialbe is Odd integer, for example in below integral how can i tell it
that n is odd integer in the Assumptions section of FullSimplify command or
the integral itself??:
In[35]:=
Integrate[Sin[n*Pi*(z/L)], {z, 0, L}]
FullSimplify[%, n \[Element] Integers && n >= 0 &&
L \[Element] Reals && L > 0 && z \[Element] Reals]

[dh]

Hi,
an assumption of e.g.

Element[(n+1)/2,Integers]

will do the trick,
sincerely, Daniel

Post by ***@sd.com
Is there a way in Mathematica to evaluate an expression assuming some
varialbe is Odd integer, for example in below integral how can i tell it
that n is odd integer in the Assumptions section of FullSimplify command or
In[35]:=
Integrate[Sin[n*Pi*(z/L)], {z, 0, L}]
FullSimplify[%, n \[Element] Integers && n >= 0 &&
L \[Element] Reals && L > 0 && z \[Element] Reals]

[Andrzej Kozlowski]

Post by ***@sd.com
Is there a way in Mathematica to evaluate an expression assuming some
varialbe is Odd integer, for example in below integral how can i
tell it
that n is odd integer in the Assumptions section of FullSimplify
command or
In[35]:=
Integrate[Sin[n*Pi*(z/L)], {z, 0, L}]
FullSimplify[%, n \[Element] Integers && n >= 0 &&
L \[Element] Reals && L > 0 && z \[Element] Reals]

All you *actually* need is:



Simplify[Integrate[Sin[(Pi*n*z)/L], {z, 0, L}],
Mod[n, 2] == 1]


(2*L)/(n*Pi)

Andrzej Kozlowski

[Bob Hanlon]

Clear[n];

int=Integrate[Sin[n*Pi*(z/L)],{z,0,L}];

Simplify[
Simplify[
int/.n->2*m+1,
Element[m, Integers]]/.
m->(n-1)/2]

(2*L)/(n*Pi)


Bob Hanlon

Date: 2005/07/25 Mon AM 01:12:15 EDT
Subject: evaluate assuming odd integers?
Is there a way in Mathematica to evaluate an expression assuming some
varialbe is Odd integer, for example in below integral how can i tell it
that n is odd integer in the Assumptions section of FullSimplify command

or

In[35]:=
Integrate[Sin[n*Pi*(z/L)], {z, 0, L}]
FullSimplify[%, n \[Element] Integers && n >= 0 &&
L \[Element] Reals && L > 0 && z \[Element] Reals]

[Jean-Marc Gulliet]

Post by ***@sd.com
Is there a way in Mathematica to evaluate an expression assuming some
varialbe is Odd integer, for example in below integral how can i tell it
that n is odd integer in the Assumptions section of FullSimplify command or
In[35]:=
Integrate[Sin[n*Pi*(z/L)], {z, 0, L}]
FullSimplify[%, n \[Element] Integers && n >= 0 &&
L \[Element] Reals && L > 0 && z \[Element] Reals]

Replace n by 2*k+1, k being any non-negative integer:

In[1]:=
Integrate[Sin[(2*k + 1)*Pi*(z/L)], {z, 0, L}, Assumptions -> {k ∈
Integers && k >= 0 && L > 0 && z ∈ Reals}]

Out[1]=
(2*L)/(Pi + 2*k*Pi)

Regards,
/J.M.

[Pratik Desai]

Post by ***@sd.com
Is there a way in Mathematica to evaluate an expression assuming some
varialbe is Odd integer, for example in below integral how can i tell it
that n is odd integer in the Assumptions section of FullSimplify command or
In[35]:=
Integrate[Sin[n*Pi*(z/L)], {z, 0, L}]
FullSimplify[%, n \[Element] Integers && n >= 0 &&
L \[Element] Reals && L > 0 && z \[Element] Reals]

I guess the two simplest approaches that come to mind are


int[n_?OddQ] = Integrate[Sin[n*Pi*(z/L)], {z, 0, L}]
int2 = Table[Integrate[Sin[(2*n - 1)*Pi*(z/L)], {z, 0, L}], {n, 1, 5}]

Best regards
Pratik

[David Park]

Perhaps there is a simpler method, but the following works.

Integrate[Sin[n*Pi*(z/L)], {z, 0, L}]
% /. n -> 2*k + 1
Simplify[%, k \[Element] Integers]
Simplify[% /. {k -> (1/2)*(-1 + n)}]

David Park
***@earthlink.net
http://home.earthlink.net/~djmp/



From: ***@sd.com [mailto:***@ads.com]


Is there a way in Mathematica to evaluate an expression assuming some
varialbe is Odd integer, for example in below integral how can i tell it
that n is odd integer in the Assumptions section of FullSimplify command or
the integral itself??:
In[35]:=
Integrate[Sin[n*Pi*(z/L)], {z, 0, L}]
FullSimplify[%, n \[Element] Integers && n >= 0 &&
L \[Element] Reals && L > 0 && z \[Element] Reals]

[Yasvir A. Tesiram]

OddQ[expr]

Post by ***@sd.com
Is there a way in Mathematica to evaluate an expression assuming some
varialbe is Odd integer, for example in below integral how can i tell it
that n is odd integer in the Assumptions section of FullSimplify command or
In[35]:=
Integrate[Sin[n*Pi*(z/L)], {z, 0, L}]
FullSimplify[%, n \[Element] Integers && n >= 0 &&
L \[Element] Reals && L > 0 && z \[Element] Reals]

[stephen layland]

Post by ***@sd.com
Is there a way in Mathematica to evaluate an expression assuming some
varialbe is Odd integer, for example in below integral how can i tell it
that n is odd integer in the Assumptions section of FullSimplify command or
In[35]:=
Integrate[Sin[n*Pi*(z/L)], {z, 0, L}]
FullSimplify[%, n \[Element] Integers && n >= 0 &&
L \[Element] Reals && L > 0 && z \[Element] Reals]

OddQ[]

--
/*------------------------------*\
| stephen layland |
| Documentation Programmer |
| http://members.wri.com/layland |
\*------------------------------*/

[Steve Luttrell]

Simplify[Integrate[Sin[(2m+1)*Pi*(z/L)], {z, 0,
L}],Assumptions->m\[Element]Integers]

(2*L)/(Pi + 2*m*Pi)

Steve Luttrell

Post by ***@sd.com
Is there a way in Mathematica to evaluate an expression assuming some
varialbe is Odd integer, for example in below integral how can i tell it
that n is odd integer in the Assumptions section of FullSimplify command or
In[35]:=
Integrate[Sin[n*Pi*(z/L)], {z, 0, L}]
FullSimplify[%, n \[Element] Integers && n >= 0 &&
L \[Element] Reals && L > 0 && z \[Element] Reals]