How to calculate binomial theorem (x+∆x)ⁿ?

I

Indranil

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(x+∆x)ⁿ = (ⁿC₀.xⁿ.∆x⁰ + ⁿC₁.xⁿ⁻¹.∆x¹ + ⁿC₂.xⁿ⁻².∆x².................xⁿ) = xⁿ + n.x⁻ⁿ.∆x + 0.
Now my first question is why ⁿC₀ = 1, ∆x⁰=1 and ⁿC₁ = n? and my second question is why the values after ( ⁿC₀.xⁿ.∆x⁰ + ⁿC₁.xⁿ⁻¹.∆x¹) is zero?
Could you explain and if needed simplify it( binomial theorem) please?
 
Indranil said:
(x+∆x)ⁿ = (ⁿC₀.xⁿ.∆x⁰ + ⁿC₁.xⁿ⁻¹.∆x¹ + ⁿC₂.xⁿ⁻².∆x².................xⁿ) = xⁿ + n.x⁻ⁿ.∆x + 0.
Now my first question is why ⁿC₀ = 1, ∆x⁰=1 and ⁿC₁ = n? and my second question is why the values after ( ⁿC₀.xⁿ.∆x⁰ + ⁿC₁.xⁿ⁻¹.∆x¹) is zero?
Could you explain and if needed simplify it( binomial theorem) please?
Click to expand...

1) It is what it is. Why do you think it can be simplified?
2) ∆x is generally taken to be a very small value. This makes (∆x)^2 an extremely small value. Many useful approximations are derived by simply deciding that this IS zero (0).
3) ⁿC₀ = 1 and ⁿC₁ = n -- Definition (Same definition for both)
4) (∆x)⁰=1 -- Definition of Zero Exponent (on a Real Number)
 
Indranil said:
(x+∆x)ⁿ = (ⁿC₀.xⁿ.∆x⁰ + ⁿC₁.xⁿ⁻¹.∆x¹ + ⁿC₂.xⁿ⁻².∆x².................xⁿ) = xⁿ + n.x⁻ⁿ.∆x + 0
Click to expand...
Can you post a link to the page where you saw this?
 
tkhunny said:
1) It is what it is. Why do you think it can be simplified?
2) ∆x is generally taken to be a very small value. This makes (∆x)^2 an extremely small value. Many useful approximations are derived by simply deciding that this IS zero (0).
3) ⁿC₀ = 1 and ⁿC₁ = n -- Definition (Same definition for both)
4) (∆x)⁰=1 -- Definition of Zero Exponent (on a Real Number)
Click to expand...

'3) ⁿC₀ = 1 and ⁿC₁ = n -- Definition (Same definition for both)' Can you provide with the definition?
 
Indranil said:
'3) ⁿC₀ = 1 and ⁿC₁ = n -- Definition (Same definition for both)' Can you provide with the definition?
Click to expand...

This would be the definition of the binomial coefficient (read aloud as "n choose r"). It is:

\(\displaystyle \sideset{_n}{_r} C = \displaystyle \binom{n}{r} = \dfrac{n!}{r! (n - r)!}\)

What do you get when you plug in r = 0? Why? What do you get when you plug in r = 1? Why? Does the choice of n matter in either of these two cases?
 
Indranil said:
'3) ⁿC₀ = 1 and ⁿC₁ = n -- Definition (Same definition for both)' Can you provide with the definition?
Click to expand...
ⁿC₀ = 1 since there is 1 way of choosing no items from n objects and ⁿC₁ = n since there is n ways of choosing one items from n objects (you can choose the 1st item or the 2nd item or ... or the nth item)
 
my second question is why the values after ( ⁿC₀.xⁿ.∆x⁰ + ⁿC₁.xⁿ⁻¹.∆x¹) is zero?
Click to expand...

The answer is "they aren't". What you have written, with the "+ 0" is wrong. What is right is that the limit, as ∆x goes to 0 all terms is 0. (So is ⁿC₁.xⁿ⁻¹.∆x¹ so I don't know why they have included that.)
 
ksdhart2 said:
This would be the definition of the binomial coefficient (read aloud as "n choose r"). It is:

\(\displaystyle \sideset{_n}{_r} C = \displaystyle \binom{n}{r} = \dfrac{n!}{r! (n - r)!}\)

What do you get when you plug in r = 0? Why? What do you get when you plug in r = 1? Why? Does the choice of n matter in either of these two cases?
Click to expand...
Could you explain the calculation you have put above so that I can understand?
 
Indranil said:
Could you explain the calculation you have put above so that I can understand?
Click to expand...

Please tell us what part of it you don't understand. Do you not know what "n!" means, for example?

But there is something else that needs to be said: You copied the work incorrectly in your question!

What is written on the board (ignoring the denominator and the limit for now) is this:

(x+∆x)ⁿ - xn = (ⁿC₀.xⁿ.∆x⁰ + ⁿC₁.xⁿ⁻¹.∆x¹ + ⁿC₂.xⁿ⁻².∆x² + ... - xⁿ) = xⁿ + n.x⁻ⁿ.∆x - xn
​

and it is stated that the higher powers of ∆x are "neglected", which can be done because they will go to zero after the next step. That is, we actually have

xⁿ + n.x⁻ⁿ.∆x + ∆x2.{...} - xn = n.x⁻ⁿ.∆x + ∆x2.{...}
​

and divide by ∆x to get

n.x⁻ⁿ + ∆x.{...}
​

Letting ∆x approach 0, we are left with n.x⁻ⁿ. This is why powers higher than 1 can be ignored. In an introductory lesson, they should be shown, so that students unfamiliar with the idea will not be confused.

It is essential to look at all of what is written (and said) in order to understand it; and when you ask for help, you need to quote the whole thing.
 
This is why we don't jump ahead in the book. There is generally a reason for the sequence. If you find yourself unfamiliar with almost ALL the notation, this may be a clue to back up and take it a little more in order.
 
ksdhart2 said:
This would be the definition of the binomial coefficient (read aloud as "n choose r"). It is:

\(\displaystyle \sideset{_n}{_r} C = \displaystyle \binom{n}{r} = \dfrac{n!}{r! (n - r)!}\)

What do you get when you plug in r = 0? Why? What do you get when you plug in r = 1? Why? Does the choice of n matter in either of these two cases?
Click to expand...
'What do you get when you plug in r = 0? Why? What do you get when you plug in r = 1? Why? Does the choice of n matter in either of these two cases?'
could you explain the part which I don't understand
 
Indranil said:
'What do you get when you plug in r = 0? Why? What do you get when you plug in r = 1? Why? Does the choice of n matter in either of these two cases?'
could you explain the part which I don't understand
Click to expand...

We don't yet know what part you don't understand! Please tell us, so we can help more effectively.

Here is what "plug in r = 0" means:

\(\displaystyle \sideset{_n}{_r} C = \dfrac{n!}{r! (n - r)!} = \dfrac{n!}{0! (n - 0)!} = \dfrac{n!}{0! n!}\)

Can you finish simplifying this? What do you get?
 
It occurs to me that you have three problems here.

You don't grasp the idea of a limit.

You don't grasp the idea of a Newton quotient.

And you don't grasp the binomial theorem.

The binomial theorem says

\(\displaystyle n \in \mathbb Z_{>0} \implies (a + b)^n = \displaystyle \left ( \sum_{j=0}^n \dfrac{n!}{j! * (n - j)!} * a^{(n-j)} * b^j \right).\)

Do you understand that?

Furthermore, \(\displaystyle 0! = 1 \text { and } n \in \mathbb Z_{>0} \implies n! = n * (n - 1)!.\)

\(\displaystyle \therefore \dfrac{n!}{0! * (n - 0)!} = \dfrac{n!}{1 * n!} = \dfrac{n!}{n!} = 1 \text {, and}\)

\(\displaystyle \dfrac{n!}{1! * (n - 1)!} = \dfrac{n * (n - 1)!}{1 * (n - 1)!} = \dfrac{n}{1} = n.\)

Do you follow that?

We ignore the trivial cases of n = 0 and n = 1.

\(\displaystyle \displaystyle (a + b)^n = \left ( \sum_{j=0}^n \dfrac{n!}{j! * (n - j)!} * a^{(n-j)} * b^j \right ) =\)

\(\displaystyle \displaystyle a^n + na^{(n-1)}b + b^2 * \left ( \sum_{j=2}^n \dfrac{n!}{j! * (n - j)!} * a^{(n-j)} * b^{(j-2)} \right ).\)

\(\displaystyle \therefore \dfrac{(a + b)^n - a^n}{b} = \dfrac{a^n + na^{(n-1)}b+ b^2 * \left ( \displaystyle \sum_{j=2}^n \dfrac{n!}{j! * (n - j)!} * a^{(n-j)} * b^{(j-2)} \right ) - a^n}{b} =\)

\(\displaystyle na^{(n-1)} + b * \text {a sum.}\)

This is just algebra. Still with me?

Now we apply that algebra to the Newton Quotient, which most generally is

\(\displaystyle \dfrac{f(x + \Delta x) - f(x)}{\Delta x} \text {, where } \Delta x \ne 0.\)

But we are specifically interested in the case where f(x) = x^n and n is an integer > 1.

\(\displaystyle f(x) = x^n \implies \dfrac{f(x + \Delta x) - f(x)}{\Delta x} = \dfrac{(x + \Delta x)^n - x^n}{\Delta x}.\)

But we figured out above with a and b what this type of fraction looks like.

\(\displaystyle f(x) = x^n \implies \dfrac{f(x + \Delta x) - f(x)}{\Delta x} = nx^{(n-1)} + \Delta x * \text { a sum.}\)

Just substitute.

To find the derivative, we take the limit of the Newton Quotient as delta x approaches zero.

\(\displaystyle \displaystyle \lim_{\Delta x \rightarrow 0} (nx^{(n-1)} + \Delta x * \text { a sum } ) = nx^{(n-1)} + 0 * \text { a sum } = nx^{(n-1)}.\)

Any questions about the limit process?
 
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Dr.Peterson said:
We don't yet know what part you don't understand! Please tell us, so we can help more effectively.

Here is what "plug in r = 0" means:

\(\displaystyle \sideset{_n}{_r} C = \dfrac{n!}{r! (n - r)!} = \dfrac{n!}{0! (n - 0)!} = \dfrac{n!}{0! n!}\)

Can you finish simplifying this? What do you get?
Click to expand...
Yes. I get 1
 
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