Binomial Theorem - determine the term with...

[stunner5000pt]

Homework Statement

Determine the coefficient of the x^2 term in the binomial expansion of
$$\frac{(1+3x)^3}{(1+2x)^2}$$

Relevant Equations

binomial theorem: https://en.wikipedia.org/wiki/Binomial_series

I'm sort of stumped here , do i do this?

$$(1+3x) \left( \frac{1+3x}{1+2x} \right)^2 = (1+3x) \left( \frac{3}{2} - \frac{1}{2(2x+1)} \right)^2$$

$$(1+3x) \left( \frac{3}{2} \right)^2 \left( 1 + \frac{-1}{3(2x+1)} \right)^2$$

and then apply the binomial theorem formula on the squared term above? But this gives way to a series that looks like this

$$\frac{9}{4} (1+3x) \left( 1 + 2 \frac{-1}{3(2x+1)} + \frac{2(2-1)}{2!} \left( \frac{-1}{3(2x+1)} \right)^2 + ... \right)$$

or is there a different (or more efficient way) to go about this

your help is greatly appreciated

 

[PeroK]

or is there a different (or more efficient way) to go about this

A better approach is:
$$\frac{(1+3x)^3}{(1+2x)^2} = (1+3x)^3(1+2x)^{-2} $$And apply the Binomial Theorem to both.

 

[Cutter Ketch]

Hmmm. I couldn’t do this using the binomial theorem alone. Maybe I did it the hard way, but I, personally, liked what you did in your first step separating out that squared term. Somehow you have to get x out of the denominator. I’m thinking if you can’t do it with that simple thing in the parenthesis, it isn’t going to be any easier with more complicated expressions. The trick here is to remember they only want one coefficient, not a whole solution. Assume you can write that simple expression as a polynomial and see how far you can get determining the coefficients up to x^2.

And you’ll still get to use the binomial expansion a couple of times after that, so it still fits the problem description.

 

[mathwonk]

here is a reference for the binomial theorem for negative exponents:
https://www.whitman.edu/mathematics/cgt_online/book/section03.01.html

 

[PeroK]

Hmmm. I couldn’t do this using the binomial theorem alone.

Whyever not?

 

[martinbn]

You can also do the following. The binomial theorem tells you that you can write
$$\frac{(1+3x)^3}{(1+2x)^2} = a + bx + cx^2+\cdots$$
and you need $c$. If you evaluate both side at $x=0$ you get $1=a$. If you first differentiate both sides and then evaluate you will get $b$. For $c$ you need the second derivative. Of course this needs derivatives and if it is precalculus may be you want to avoid that. Then with a bit more work you can do the following. You have found $a$, so you can write
$$\frac{(1+3x)^3}{(1+2x)^2} -1 = bx + cx^2+\cdots$$
Do the algebra on the left, and it should simplify to an expression with a factor of $x$. Then divide by $x$ both sides and evaluate at zero to get $b$. The repeat the process one more time to find $c$.

 

[pasmith]

You can also do the following. The binomial theorem tells you that you can write
$$\frac{(1+3x)^3}{(1+2x)^2} = a + bx + cx^2+\cdots$$
and you need $c$. If you evaluate both side at $x=0$ you get $1=a$. If you first differentiate both sides and then evaluate you will get $b$.

Alternatiely, you can expand $$\begin{split} (1 + 3x)^2 &= (1 + 2x)^2\sum_{n=0}^\infty a_nx^n \\ &= \sum_{n=0}^\infty (a_nx^n + 4a_nx^{n+1} + 4a_nx^{n+2}) \\ 1 + 9x + 27x^2 + 27x^3 &= a_0 + (a_1 + 4a_0)x + \sum_{n=2}^\infty (a_n + 4a_{n-1} + 4a_{n-2})x^n \end{split}$$ and compare coefficients of $x^n$.

 

[Cutter Ketch]

Whyever not?

Because the negative binomial series is not the binomial theorem.

 

[PeroK]

Because the negative binomial series is not the binomial theorem.

That's stretching a point. The binomial theorem was generalised to negative integers by Newton, I believe. Given the context of the problem, I wouldn't give it a second thought.

 

[Cutter Ketch]

You can also do the following. The binomial theorem tells you that you can write
$$\frac{(1+3x)^3}{(1+2x)^2} = a + bx + cx^2+\cdots$$
and you need $c$. If you evaluate both side at $x=0$ you get $1=a$. If you first differentiate both sides and then evaluate you will get $b$. For $c$ you need the second derivative. Of course this needs derivatives and if it is precalculus may be you want to avoid that. Then with a bit more work you can do the following. You have found $a$, so you can write
$$\frac{(1+3x)^3}{(1+2x)^2} -1 = bx + cx^2+\cdots$$
Do the algebra on the left, and it should simplify to an expression with a factor of $x$. Then divide by $x$ both sides and evaluate at zero to get $b$. The repeat the process one more time to find $c$.

As I mentioned, I liked the OP’s first step of pulling out one power of the numerator and making that squared term
$$ \left( \frac{1+3x}{1+2x} \right) ^2 $$
Because it seems much simpler to me to expand
$$ \frac{1+3x}{1+2x} $$
Writing
$$ \frac{1+3x}{1+2x} = \sum_{k=0}^\infty C_k x^k $$
Becomes
$$ 1+3x = \sum_{k=0}^\infty C_k x^k + 2x \sum_{k=0}^\infty C_k x^k $$
And then
$$ 1+3x = C_0 + \sum_{k=1}^\infty (2C_{k-1}+ C_k) x^k $$
Comparing the two sides gives the coefficients and I get
$$ \frac{1+3x}{1+2x} = 1 + x - 2 x^2 + O(x^3) $$
That gets the x’s out of the denominator and the rest is all the usual expansions

 

[PeroK]

Writing
$$ \frac{1+3x}{1+2x} = \sum_{k=0}^\infty C_k x^k $$

How do you justify that? If the binomial theorem applies only to positive integers?

 

[PeroK]

@Cutter Ketch The link in the original post explicitly includes negative integers:

Homework Statement: Determine the coefficient of the x^2 term in the binomial expansion of
$$\frac{(1+3x)^3}{(1+2x)^2}$$
Relevant Equations: binomial theorem: https://en.wikipedia.org/wiki/Binomial_series

 

[Cutter Ketch]

How do you justify that? If the binomial theorem applies only to positive integers?

You don’t have to justify it. Assume it can be written as a polynomial and then see if you can derive coefficients that make it true.

 

[PeroK]

You don’t have to justify it. Assume it can be written as a polynomial and then see if you can derive coefficients that make it true.

This is in the mathematics forum. You can't just assume that an arbitrary function can be written as an infinite power series. Not while inisting that the binomial theorem for negative integers cannot be used. If that function can be written as a power series, then by implication so can the denominator.

 

[Cutter Ketch]

@Cutter Ketch The link in the original post explicitly includes negative integers:

I saw that. I also saw your first post (the very first reply on the thread) strongly suggesting using that. I also saw that in that post it is described as a different expansion. I saw I could get there without it.

 

[Cutter Ketch]

This is in the mathematics forum. You can't just assume that an arbitrary function can be written as an infinite power series. Not while inisting that the binomial theorem for negative integers cannot be used. If that function can be written as a power series, then by implication so can the denominator.

In mathematics you can test for equality. I show a solution where each coefficient can be constructed from the previous coefficient. The polynomial exists. Proof by induction.

 

[Cutter Ketch]

Sad !?!? Why are you busting my chops? Go look at my posts. I never said don’t use the negative binomial expansion. I never said you can’t use the negative binomial expansion. I only said I got there a different way. I haven’t said one thing wrong and yet you keep needling me like I’m doing something wrong. Well, congratulations, you’ve successfully driven me from the forum. You won’t have to deal with my particular brand of mental inferiority again.

 

[FactChecker]

1) It seems wrong to me to talk about the "binomial expansion of $\frac {(1+3x)^3}{(1+2x)^2}$". There is the binomial expansion of the numerator and of the denominator, but not of the fraction.
2) A couple of simple approaches:

2A) Multiply out the numerator and the denominator (using the binomial expansion if desired) and then use simple long division on the fraction.
$\frac {(1+3x)^3}{(1+2x)^2} = \frac {27 x^3 + 27 x^2 + 9 x + 1}{4 x^2 + 4 x + 1}$
$= (27/4) x+lowerPowersOfX$

2B) Notice that the numerator grows (for large x) like $(3x)^3$ and the denominator grows like $(2x)^2$.
For very large $x$ values, all the rest can be ignored.
So what does the ratio grow like?
If the ratio has a non-zero coefficient of $x^2$, what would it grow like?
What can you conclude?

 

[WWGD]

Well, you can always divide polynomials , as in the ring $\mathbb R[x]$and get a polynomial as a remainder. It is a Euclidean ring after all.