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Why is the Binomial Formula a Derivation?

by Tenshou
Tags: binomial, derivation, formula
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Tenshou
#1
Feb7-13, 03:15 AM
P: 150
I mean the binomial formula is something of the form

##\left(a+b\right)^n## = ##\sum_{i=1}^{n}\dbinom{n}{k}a^{n}b^{n-k}##

and then you have the linear map ##\psi : A \rightarrow A## which is a derivation when;
##\theta(xy) = y\theta(x) + x\theta(y)## for all x,y in A

so the leibniz formula for a ##\psi## derivation is
##\theta^n(xy)## = ##\sum_{k=1}^{n}\dbinom{n}{k}\theta^{n}(x)\theta^{n-k}(y)##

so does anyone have any idea why these two are very very similar if not the same? Also I am trying to understand how a Lie Algebra is a "Derivation"
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#2
Feb7-13, 04:52 AM
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You are, I think, confused by the wonderful ambiguity of natural languages. The word "derive" and the words based on it such as derivation and derivative have multiple meanings. "Derive" in some contexts means to "to obtain by reasoning". That's the meaning implied by mathematical theorems. Every mathematical theorem is a tautology. The trick is to show that this is the case.

"Derive" also has a derivative meaning, "derivative". That word, too, is overloaded. For example, ask an economist or a mathematician what a "derivative" is and you will get very different answers. To economists, derivatives were the things that some nefarious operators took advantage of and indirectly caused our current economic malaise. To mathematicians, a derivative what results from differentiation.

That latter meaning is the sense of the word with regard to Lie algebras. The concept of differentiation requires two operators, subtraction (addition) and division (multiplication). Nominally, a group in mathematics is something with one operator, so nominally the concept of differentiation does not make sense with respect to a mathematical group. A Lie group is a group that somehow has enough additional structure such that the concept of differentiation somehow does make sense. The derivative of a Lie group has both addition and multiplication. That's an algebra, and hence the name Lie algebra.
Tenshou
#3
Feb7-13, 11:20 AM
P: 150
Still why do the product rule" and the binomial formula look very similar if not the same?

sankalpmittal
#4
Feb7-13, 01:27 PM
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Why is the Binomial Formula a Derivation?

Quote Quote by Tenshou View Post
Still why do the product rule" and the binomial formula look very similar if not the same?
I never thought like you. Perhaps a coincidence. They just look the similar is due to coincidence in derivation.

For example: A body is revolved in a vertical circle from bottom to top. Using conservation of mechanical energy yields the same result as kinematic equations of motion,but still you know that those kinematic equations of motion hold for one dimensional motion only.
pasmith
#5
Feb7-13, 03:50 PM
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Quote Quote by Tenshou View Post
I mean the binomial formula is something of the form

##\left(a+b\right)^n## = ##\sum_{i=1}^{n}\dbinom{n}{k}a^{n}b^{n-k}##

and then you have the linear map ##\psi : A \rightarrow A## which is a derivation when;
##\theta(xy) = y\theta(x) + x\theta(y)## for all x,y in A

so the leibniz formula for a ##\psi## derivation is
##\theta^n(xy)## = ##\sum_{k=1}^{n}\dbinom{n}{k}\theta^{n}(x)\theta^{n-k}(y)##
I think you may want to take another look at the definition of a derivation; your account is somewhat confused (What is [itex]\theta[/itex]? What is [itex]\psi[/itex]?).

so does anyone have any idea why these two are very very similar if not the same?
The inductive proofs of these results have exactly the same structure: you do an operation on a single sum which gives you two sums, which you have to recombine into a single sum.
Tenshou
#6
Feb7-13, 04:08 PM
P: 150
Quote Quote by pasmith View Post
I think you may want to take another look at the definition of a derivation; your account is somewhat confused (What is [itex]\theta[/itex]? What is [itex]\psi[/itex]?).

The inductive proofs of these results have exactly the same structure: you do an operation on a single sum which gives you two sums, which you have to recombine into a single sum.
I mean for ##\theta## which is suppose to be the linear map not ##\psi## that is a typo. that is what a derivation is. a linear map. What do you mean that they have the same structure? That is what I am asking the question about! Why is the structure the same? nothing is by coincidence ;)
micromass
#7
Feb7-13, 04:35 PM
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Consider the following algebra:

[tex]A=\bigoplus_{x\in \mathbb{R}} \mathbb{R}[/tex]

Pure formally, these are function [itex]f:\mathbb{R}\rightarrow \mathbb{R}[/itex] where only finitely many values f(x) are nonzero.

Define for [itex]a\in \mathbb{R}[/itex], the function [itex]\delta_a:\mathbb{R}\rightarrow \mathbb{R}[/itex] such that [itex]\delta_a(a)=1[/itex] and [itex]\delta_a(x)=0[/itex], for [itex]x\neq a[/itex]. We can write any [itex]f\in A[/itex] as

[tex]f=\sum_{a\in \mathbb{R}} x_a\delta_a[/tex]

where [itex]x_a\in \mathbb{R}[/itex] is nonzero for only finitely many a.

We now define a product by [itex]\delta_a\cdot \delta_b=\delta_{a+b}[/itex] and by extending this linearly.This gives a associative and commutative product on A.

Define the following derivation on A:

[tex]\theta(\sum_{a\in \mathbb{R}} x_a\delta_a ) = \sum_{a\in \mathbb{R}} a x_a \delta_a[/tex]

Note that [itex]\theta^n (\delta_a)= a^n \delta_a[/itex].

So [itex]\theta^n (\delta_a\delta_b) = (a+b)^n \delta_{a+b}[/itex]. And

[tex]\sum_{k=0}^n \binom{n}{k} \theta^k(\delta_a)\theta^{n-k}(\delta_b) = \sum_{k=0}^n \binom{n}{k} a^kb^{n-k}\delta_{a+b}[/tex]

So by the Leibniz rule, we get that

[tex](a+b)^n \delta_{a+b} = \sum_{k=0}^n \binom{n}{k} a^kb^{n-k}\delta_{a+b}[/tex]

and in particular, the coefficients of [itex]\delta_{a+b}[/itex] must equal so we have proven that

[tex](a+b)^n = \sum_{k=0}^n \binom{n}{k} a^kb^{n-k}[/tex]

which is the binomial formula. So this is a way to prove the binomial formula from the Leibniz rule.
micromass
#8
Feb7-13, 04:37 PM
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Quote Quote by sankalpmittal View Post
I never thought like you. Perhaps a coincidence. They just look the similar is due to coincidence in derivation.

For example: A body is revolved in a vertical circle from bottom to top. Using conservation of mechanical energy yields the same result as kinematic equations of motion,but still you know that those kinematic equations of motion hold for one dimensional motion only.
Nothing in math is a coincidence In physics maybe, but not in math.
WannabeNewton
#9
Feb7-13, 04:44 PM
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Quote Quote by sankalpmittal View Post
For example: A body is revolved in a vertical circle from bottom to top. Using conservation of mechanical energy yields the same result as kinematic equations of motion,but still you know that those kinematic equations of motion hold for one dimensional motion only.
That's because you have constrained the body to move in a circle. You eliminated one of the two degrees of freedom so you have reduced the system to a one dimensional problem. This isn't a coincidence =D.
Tenshou
#10
Feb7-13, 04:53 PM
P: 150
I cannot spot any flaws, and thank you! the ##\delta## looks kind of like an exponential function... Can I say that this helped me understand possibly how the Lie Algebra is a Derivative, I mean it is if you can tack on the structure of a commutator. Thank you micro mass, you work seems elegant (/•ิ_•ิ)/ I salute you, sir.
Tenshou
#11
Jul2-13, 05:06 AM
P: 150
I think I have found a flaw, after several weeks of looking at this in awe, I decided to put this to a test. And this proof fails for all numbers except when ##a=b=1## look...

Since ##a## must be equal to ##b## such that the coefficients of ##\delta_{a+b} \neq 0## me have ## (a+a)^n=2^{n}a^{n} \ge \sum\limits_{i=0}^{n} \binom{n}{k} a^{n}a^{n-k}=\sum \limits_{i=0}^{n} \binom{n}{k}a^{n}## The equality comes about when a = 1, To prove that it only happens at a=1 where it doesn't fail then... lets go :D

##\begin{align*}(1+1)^n=2^n= \sum \limits_{i=0}^{n} \binom{n}{k} 1^{k}1^{n-k} \\ = \sum\limits_{i=0}^{n} \binom{n}{k} 1^{n}\end{align*}##

you can discard the one and there you go... I think the error is in how you defined the derivative. for example, it isn't clear if you took the derivative of ##x_a## or ##\delta_{a+b}## if you could clear that up it would fantastic :3
micromass
#12
Jul2-13, 06:38 AM
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Quote Quote by Tenshou View Post
I think I have found a flaw, after several weeks of looking at this in awe, I decided to put this to a test. And this proof fails for all numbers except when ##a=b=1## look...

Since ##a## must be equal to ##b## such that the coefficients of ##\delta_{a+b} \neq 0## me have ## (a+a)^n=2^{n}a^{n} \ge \sum\limits_{i=0}^{n} \binom{n}{k} a^{n}a^{n-k}=\sum \limits_{i=0}^{n} \binom{n}{k}a^{n}## The equality comes about when a = 1, To prove that it only happens at a=1 where it doesn't fail then... lets go :D
Why is the equality only true for a=1?

What you're saying now is not that the proof is flawed, but that the entire binomial theorem is flawed!
Tenshou
#13
Jul2-13, 07:27 AM
P: 150
I actually have the inequality going the wrong way, I am sorry I have been up for the past 24 hours I think I need some sleep... the sum to n should be greater than the other side of ##2^{n}a^n## It may be true that the proof is flawed, or the fact that math is either consistent and not complete or complete and not consistent, but for the sake of this I will take the latter. I mean can you just define the what it means to take ##\theta(f)## I am not sure what the derivation is on it I mean it is ##af##, it is that right?
PSarkar
#14
Nov18-13, 07:45 PM
P: 43
Quote Quote by Tenshou View Post
Since ##a## must be equal to ##b## such that the coefficients of ##\delta_{a+b} \neq 0## me have ## (a+a)^n=2^{n}a^{n} \ge \sum\limits_{i=0}^{n} \binom{n}{k} a^{n}a^{n-k}=\sum \limits_{i=0}^{n} \binom{n}{k}a^{n}##
How did you get the inequality? It should actually be
[tex]
(a+a)^n=2^{n}a^{n} = \sum\limits_{i=0}^{n} \binom{n}{k} a^{k}a^{n-k} = \sum\limits_{i=0}^{n} \binom{n}{k} a^n = a^n\sum\limits_{i=0}^{n} \binom{n}{k}.
[/tex]
This only proves that
[tex]
\sum\limits_{i=0}^{n} \binom{n}{k} = 2^{n}
[/tex]
which you can check is indeed true. So there is nothing wrong with the proof by micromass.


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