Multinomail Theorem on Wikipedia

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The discussion focuses on understanding the multinomial theorem as presented on Wikipedia, particularly in relation to the expression (1+x+x^2)^5. Participants express confusion about the variables k1, k2, and k3, which represent combinations that sum to the total exponent. They explore the application of the theorem through examples and seek clarification on the summation and product notation used in the formula. A simpler method to solve the problem is suggested by observing patterns similar to Pascal's triangle, indicating that a more straightforward approach may exist. Overall, the conversation emphasizes the desire to grasp the theorem's application for an upcoming math exam.
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Homework Statement


Hi! This is not a homework, i am just trying to understand the multinomial theorem on Wikipedia.
http://en.wikipedia.org/wiki/Multinomial_theorem

I don't understand how to apply this formula:-
4fae69bacc2204d7641df723e4654280.png


I am trying to apply this formula in this question:-
(1+x+x2)5
This question is asked by Icetray in this https://www.physicsforums.com/showthread.php?t=521248"
and i am trying to solve it by the multinomial theorem.

Firstly i don't understand what are these k1,k2, k3...:confused:

Homework Equations





The Attempt at a Solution

 
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Pranav-Arora said:
Firstly i don't understand what are these k1,k2, k3...:confused:

Your k1,k2, k3, ... are any combination of numbers that add up to n.

The capital sigma is the regular summation symbol meaning that should sum the terms for any combination of ki's that you can think of that add up to n.

The capital pi is the product symbol, meaning that you should take the product of the factor for any t that is between 1 and m.
 
As for constructing a triangle.
You can construct Pascal's triangle by building up:
1
(1+x)
(1+x)2 = 1 + 2x + x2
(1+x)3 = 1 + 3x + 3x2 + x3
(1+x)4 = ...

You can do the same for (1+x+x2)n.
The coefficients you will find are the multinomials.


Btw, my time is up. Have to run now! :wink:
 
Thanks for your reply! :smile:

I like Serena said:
The capital sigma is the regular summation symbol meaning that should sum the terms for any combination of ki's that you can think of that add up to n.

Could you please provide an example?

I like Serena said:
As for constructing a triangle.
You can construct Pascal's triangle by building up:
1
(1+x)
(1+x)2 = 1 + 2x + x2
(1+x)3 = 1 + 3x + 3x2 + x3
(1+x)4 = ...

You can do the same for (1+x+x2)n.
The coefficients you will find are the multinomials.


Btw, my time is up. Have to run now! :wink:

Yep! I can do that but that would be more time consuming. So i thought that i should learn this theorem. And also day after tomorrow, its my maths exam, so i thought it would be very useful for that. :smile:

(Ok, bye :wink:)
 
Pranav-Arora said:
Could you please provide an example?



Yep! I can do that but that would be more time consuming. So i thought that i should learn this theorem. And also day after tomorrow, its my maths exam, so i thought it would be very useful for that. :smile:

I'm very keen to learn this as well. An example would be very much appreciated. (:
 
Icetray said:
I'm very keen to learn this as well. An example would be very much appreciated. (:

Expand (x+y)n.
Then substitute x=1 and y=x.
The equation which you would get is like this:-
(1+x)^n={}^nC_0+{}^nC_1x+{}^nC_2x^2+...+{}^nC_nx^n

We have to solve this equation:- (1+x+x2)n
For that we need to substitute x=1 and y=(x+x2) in (x+y)n expansion. :smile:

But that would be very time consuming.
 
Well, here's a general example for (A+B+C)n:
http://en.wikipedia.org/wiki/Pascal's_pyramid#Trinomial_Expansion_connection
It shows all the multinomials for this one expression in a triangle.
But I'm afraid it's still pretty messy.

Here's another example:
http://en.wikipedia.org/wiki/Multinomial_theorem#Example_multinomial_coefficients
You probably already saw it.
However, I do not think I can explain it simpler.

As for the (1+x+x2)n.
There is a simple pattern.
It's simpler because it does not contain all the multinomial coefficients, but summations of them.
To find it, I think it's easiest to work out the first 3 layers of the pyramid.
You should be able to see the pattern then.
 
I just want to understand this formula:-
4fae69bacc2204d7641df723e4654280.png


If i try to solve this question (1+x+x2)5 using the multinomial theorem, should i go like this:-
(1+x+x^2)^5=\sum_{0+1+2=3}\frac{5!}{0!1!2!}\Pi_{1 \leq t \leq m} x^{k_t}_t

Now what should i do next? :confused:
 
Pranav-Arora said:
I just want to understand this formula:-
4fae69bacc2204d7641df723e4654280.png


If i try to solve this question (1+x+x2)5 using the multinomial theorem, should i go like this:-
(1+x+x^2)^5=\sum_{0+1+2=3}\frac{5!}{0!1!2!}\Pi_{1 \leq t \leq m} x^{k_t}_t

Now what should i do next? :confused:

You need k1 + k2 + k3 = 5, NOT 3. In other words, once you pick k1 & k2, then you must have k3 = 5 - (k1 + k2) .

You can think of the sum as having the following:
k1 = 0 to n
k2 = 0 to n - k1
k3 = 0 to n - (k1 + k2)


 
  • #10
Hi SammyS!:smile:

SammyS said:
You need k1 + k2 + k3 = 5, NOT 3. In other words, once you pick k1 & k2, then you must have k3 = 5 - (k1 + k2) .

You can think of the sum as having the following:
k1 = 0 to n
k2 = 0 to n - k1
k3 = 0 to n - (k1 + k2)​

Ok, i am trying it out again:-

(1+x+x^2)^5=\sum_{2+1+2=5}\frac{5!}{2!1!2!}\Pi_{1 \leq t \leq m} x^{k_t}_t
Is it ok now?
 
  • #11
Pranav-Arora said:
Ok, i am trying it out again:-

(1+x+x^2)^5=\sum_{2+1+2=5}\frac{5!}{2!1!2!}\Pi_{1 \leq t \leq m} x^{k_t}_t
Is it ok now?

It should be:
(1+x+x^2)^5=... + \frac{5!}{2!1!2!}(1^2 x^1 (x^2)^2) + ...

The total number of terms is 21.
 
  • #12
I like Serena said:
It should be:
(1+x+x^2)^5=... + \frac{5!}{2!1!2!}(1^2 x^1 (x^2)^2) + ...

The total number of terms is 21.

Where's the sigma?
 
  • #13
Pranav-Arora said:
Where's the sigma?

The sigma is a shorthand notation to indicate a summation.
When written out in terms, there is no sigma any more.
Of course I did not write out all of the terms - too much work! :wink:
 
  • #14
I am still not able to get what happens to \sum_{2+1+2=5} :confused:
 
  • #15
Pranav-Arora said:
I am still not able to get what happens to \sum_{2+1+2=5} :confused:

Let me give a simpler example.
\sum_{k+m=2} k \cdot m = 0 \cdot 2 + 1 \cdot 1 + 2 \cdot 0

Does that help?
 
  • #16
I like Serena said:
Let me give a simpler example.
\sum_{k+m=2} k \cdot m = 0 \cdot 2 + 1 \cdot 1 + 2 \cdot 0

Does that help?

Yes that helped. :smile:
But then what we have to do with this:-
\Pi_{1 \leq t \leq m} x^{k_t}_t
I don't know what's this symbol called and i have never dealt with it. I only know what is its function.
If it is given like this:-
\Pi_{i=1}^{n}a_i=a_1 \cdot a_2 \cdot a_3... \cdot a_n
 
  • #17
The symbol is a capital pi. It is the product symbol.

As an example:
\prod_{1 \leq t \leq 3} x^{k_t}_t = x_1^{k_1} \cdot x_2^{k_2} \cdot x_3^{k_3}
 
  • #18
Thanks fro helping me to understand the Multinomial theorem but that is still a clumsy method to solve this question:-
(1+x+x^2)^5
Isn't there any easy method? :smile:
 
  • #19
In a previous post I wrote:

ILSe said:
As for the (1+x+x2)n.
There is a simple pattern.
It's simpler because it does not contain all the multinomial coefficients, but summations of them.
To find it, I think it's easiest to work out the first 3 layers of the pyramid.
You should be able to see the pattern then.
I haven't worked out a formula yet, but I can see a simple pattern, similar to the regular Pascal's triangle.
Based on this pattern I can probably work out a formula, but so could you! :wink:
 
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