Multinomail Theorem on Wikipedia

Then here, n is the number of terms, a1, a2, a3, ... an are the terms.For example, if n = 3, then:\Pi_{i=1}^{3}a_i=a_1 \cdot a_2 \cdot a_3In the context of the multinomial theorem, the notation:\Pi_{1 \leq t \leq m} x^{k_t}_tmeans that you should take the product of all the terms in which k1, k2, k3,... are used as exponents for the variables x1, x2, x3,... respectively.In the case of (1+x+x2)5
  • #1
Saitama
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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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  • #2
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.
 
  • #3
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:
 
  • #4
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:)
 
  • #5
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. (:
 
  • #6
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:-
[tex](1+x)^n={}^nC_0+{}^nC_1x+{}^nC_2x^2+...+{}^nC_nx^n[/tex]

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.
 
  • #7
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.
 
  • #8
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:-
[tex](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[/tex]

Now what should i do next? :confused:
 
  • #9
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:-
[tex](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[/tex]

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:-

[tex](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[/tex]
Is it ok now?
 
  • #11
Pranav-Arora said:
Ok, i am trying it out again:-

[tex](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[/tex]
Is it ok now?

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

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

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 [tex]\sum_{2+1+2=5}[/tex] :confused:
 
  • #15
Pranav-Arora said:
I am still not able to get what happens to [tex]\sum_{2+1+2=5}[/tex] :confused:

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

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

Does that help?

Yes that helped. :smile:
But then what we have to do with this:-
[tex]\Pi_{1 \leq t \leq m} x^{k_t}_t[/tex]
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:-
[tex]\Pi_{i=1}^{n}a_i=a_1 \cdot a_2 \cdot a_3... \cdot a_n[/tex]
 
  • #17
The symbol is a capital pi. It is the product symbol.

As an example:
[tex]\prod_{1 \leq t \leq 3} x^{k_t}_t = x_1^{k_1} \cdot x_2^{k_2} \cdot x_3^{k_3}[/tex]
 
  • #18
Thanks fro helping me to understand the Multinomial theorem but that is still a clumsy method to solve this question:-
[tex](1+x+x^2)^5[/tex]
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:
 

1. What is the Multinomial Theorem?

The Multinomial Theorem is a mathematical formula that expands the power of a multinomial expression. It is an extension of the Binomial Theorem, which is used to expand the power of a binomial expression. The Multinomial Theorem is used to simplify calculations involving multinomial expressions, making them easier to solve.

2. What is a multinomial expression?

A multinomial expression is a mathematical expression that contains multiple terms with different variables and exponents. It is a generalization of a polynomial expression, which contains only one variable. An example of a multinomial expression is (x + y + z)^2.

3. How is the Multinomial Theorem applied?

The Multinomial Theorem is applied by using the formula (x1 + x2 + ... + xn)^m = Σ(n choose k) x1^k * x2^(m-k) * ... * xn^(m-k), where n is the number of terms in the expression, m is the power, and k ranges from 0 to m. This formula allows us to expand the power of the multinomial expression and simplify the calculations.

4. What is the difference between the Multinomial Theorem and Binomial Theorem?

The main difference between the Multinomial Theorem and Binomial Theorem is the number of terms they can handle. The Binomial Theorem is used for expanding the power of a binomial expression, which contains two terms, while the Multinomial Theorem is used for expanding the power of a multinomial expression, which contains more than two terms.

5. What are some real-life applications of the Multinomial Theorem?

The Multinomial Theorem has many real-life applications, especially in statistics and probability. It is used to calculate the probabilities of different outcomes in a multinomial experiment, such as rolling a die multiple times. It is also used in economics and finance, particularly in portfolio optimization and risk management. Additionally, the Multinomial Theorem has applications in physics, chemistry, and engineering, where it is used to simplify complex calculations involving multiple variables and exponents.

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