1. ## Binomial Expansion

I am not a student, but I am trying to figure this out.

The text book states:
"We expand our previous forma(using a binomial expansion) and drop 'higher order' terms"

The formula the text is referring to is
$\displaystyle \frac{F}{1-dF}$

And after binomial expansion reduces to:

$\displaystyle F^2 \time d$

I appreciate any help.

2. ## Re: Binomial Expansion

just do the division and truncate terms where the order of $F$ is greater than 1.

I get $F + d F^2$ though

are you sure the answer is just $d F^2$ ?

3. ## Re: Binomial Expansion

Yes, I just double checked.

4. ## Re: Binomial Expansion

well...

a) the binomial expansion is $(a + b)^n = \displaystyle{\sum_{k=0}^n}~\binom{n}{k} a^k b^{n-k}$

and I confess I don't see the relevance of this for this problem.

b)

$\dfrac{1}{1 - d F} = \displaystyle{\sum_{k=0}^\infty}~(dF)^k = 1 + d F + d^2 F^2 + \dots,~\text{for }|d F|<1$

so

$\dfrac{F}{1- d F} = F + d F^2 + d^2 F^3 \dots$

and we truncate this to

$\dfrac{F}{1- d F} = F + d F^2$

5. ## Re: Binomial Expansion

Originally Posted by romsek
well...

a) the binomial expansion is $(a + b)^n = \displaystyle{\sum_{k=0}^n}~\binom{n}{k} a^k b^{n-k}$

and I confess I don't see the relevance of this for this problem.
$\dfrac{F}{1 - dF} =$

$F(1 - dF)^{-1} \approx$

$F[(1)^{-1} - 1(1)^{-2}(-dF)] =$

$F[1 + dF] =$

$F + dF^2$

6. ## Re: Binomial Expansion

Originally Posted by romsek
well...

a) the binomial expansion is $(a + b)^n = \displaystyle{\sum_{k=0}^n}~\binom{n}{k} a^k b^{n-k}$

and I confess I don't see the relevance of this for this problem.

b)

$\dfrac{1}{1 - d F} = \displaystyle{\sum_{k=0}^\infty}~(dF)^k = 1 + d F + d^2 F^2 + \dots,~\text{for }|d F|<1$

so

$\dfrac{F}{1- d F} = F + d F^2 + d^2 F^3 \dots$

and we truncate this to

$\dfrac{F}{1- d F} = F + d F^2$
I am not sure what the relevance is. I just want to fully understand the text. Honestly the approximation is more work. I appreciate you help. Is the binomial theorem just for variables?

Originally Posted by greg1313
$\dfrac{F}{1 - dF} =$

$F(1 - dF)^{-1} \approx$

$F[(1)^{-1} - 1(1)^{-2}(-dF)] =$

$F[1 + dF] =$

$F + dF^2$
How did you go from
$F[(1)^{-1} - 1(1)^{-2}(-dF)] =$

to

$F[1 + dF] =$

This is what I came up with

\begin{aligned}\sum_{k=-1}^{n} 1^{n-k}dF^k\\ F(1-dF)^{-1} &\approx \sum_{k=-1}^{2} 1^{n-(-1)}dF^{-1}\\ &\approx 1^{2 -(-1)}dF^{-1} + 1^{2-0}dF^{0} + 1^{2-1}dF^{1} + 1^{2-2}dF^{2}\\ &\approx dF^{-1} + dF + dF^2\\ &\approx \frac{1}{dF} + dF + dF^2\\ \end{aligned}

Thanks again

7. ## Re: Binomial Expansion

Originally Posted by bmccardle
Is the binomial theorem just for variables? No. For example,

it can be used on $\displaystyle \ \ (9 + 1)^{\frac{1}{2}}$.See below.

How did you go from

$F[(1)^{-1} - 1(1)^{-2}(-dF)] =$

to

$F[1 + dF] =$
$F[(1)^{-1} - 1(1)^{-2}(-dF)] =$

$F[1 - 1(1)(-dF)] =$

$F[1 + dF)] =$

$F + dF^2$

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

This is what I came up with

\begin{aligned}\sum_{k=-1}^{n} 1^{n-k}dF^k\\ F(1-dF)^{-1} &\approx \sum_{k=-1}^{2} 1^{n-(-1)}dF^{-1}\\ &\approx 1^{2 -(-1)}dF^{-1} + 1^{2-0}dF^{0} + 1^{2-1}dF^{1} + 1^{2-2}dF^{2}\\ &\approx dF^{-1} + dF + dF^2\\ &\approx \frac{1}{dF} + dF + dF^2\\ \end{aligned}

\begin{aligned}F*\sum_{k=0}^{n} 1^{n-k}(dF)^k\\ F(1-dF)^{-1} &\approx \sum_{k=0}^{2} 1^{n-(0)}(dF)^{-1}\\ &\approx F[ 1^{2-0}(dF)^{0} + 1^{2-1}(dF)^{1} + 1^{2-2}(dF)^{2}]\\ &\approx F(1 + dF + d^2F^2)\\ &\approx F + dF^2 + d^2F^3 \\ \end{aligned}

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

$\displaystyle \ \ (9 + 1)^{\frac{1}{2}} \ \approx$

$\displaystyle 9^{\frac{1}{2}} \ + \ \bigg(\frac{1}{2}\bigg)9^{\frac{-1}{2}}(1)^1 \ =$

$\displaystyle 3 + \bigg(\dfrac{1}{2}\bigg)\bigg(\dfrac{1}{3}\bigg) =$

$\displaystyle 3 + \dfrac{1}{6} \approx$

$\displaystyle 3.167$

8. ## Re: Binomial Expansion

* * * This is an edit (correction) for the previous post that did not make it,
because the time limit for edits ran out. * * *

\begin{aligned}\sum_{k=0}^{n} 1^{n-k}(dF)^k\\ F(1-dF)^{-1} &\approx F*\sum_{k=0}^{2} 1^{n-(0)}(dF)^{-1}\\ &\approx F[ 1^{2-0}(dF)^{0} + 1^{2-1}(dF)^{1} + 1^{2-2}(dF)^{2}]\\ &\approx F(1 + dF + d^2F^2)\\ &\approx F + dF^2 + d^2F^3 \\ \end{aligned}