Binomial expansion

[burgerandcheese]

If I was to expand the expression (3 + 2x)^-3 using this formula,



would it be correct to make the x in the formula as 2x + 2? So I get (1 + [2x + 2])^-3 and use the formula. But I got 19 + 42x + 20x² + ... this is not the same as the provided answer.

My teacher said she wasn't sure, so she advised me to stick with the traditional method which is to factor out the 3 to get (3(1 + 2x/3))^-3 = 3^-3(1 + 2x/3)^-3 and expand from there. I did realise that in my book it said that when n is not a positive integer, then the expansion is valid only for |x|<1.

Does that mean for the former method (of expanding from (1 + [2x + 2])^-3) to work , I have to know whether or not |2x+2|<1? If so, then how do we know that |2x/3|<1?

 

[Romsek]

it should work

[MATH]u=2x+2\\ 1+u = 3+2x (1+u)^3 = 1 + 3u+3u^2 + u^3 = \\ 1 + 3(2x+2) + 3(2x+2)^2 + (2x+2)^3 = \\ 8 x^3+36 x^2+54 x+27\\ (3+2x)^{-3} = \dfrac{1}{8 x^3+36 x^2+54 x+27}[/MATH]
Now if we just expand it the usual way

[MATH](2x+3)^3 = 8 x^3+36 x^2+54 x+27\\ (2x+3)^{-3} = \dfrac{1}{8 x^3+36 x^2+54 x+27}[/MATH]
Same answer.

A better way of doing this is to use the source of the formula for [MATH](1+x)^n[/MATH]
[MATH](a+b)^n = \sum \limits_{k=0}^n \dbinom{n}{k}a^k b^{n-k}[/MATH]
[MATH]a = 2x\\ b=3\\ (2x+3)^3 = 1\cdot 1 \cdot 3^3 + 3 \cdot 2x \cdot 3^2 + 3 \cdot (2x)^2 \cdot 3 + 1 \cdot (2x)^3 \cdot 1 =\\ 27+54x + 36x^2 + 8x^3[/MATH]

 

[Dr.Peterson]

If I was to expand the expression (3 + 2x)^-3 using this formula,

View attachment 16381

would it be correct to make the x in the formula as 2x + 2? So I get (1 + [2x + 2])^-3 and use the formula. But I got 19 + 42x + 20x² + ... this is not the same as the provided answer.

My teacher said she wasn't sure, so she advised me to stick with the traditional method which is to factor out the 3 to get (3(1 + 2x/3))^-3 = 3^-3(1 + 2x/3)^-3 and expand from there. I did realise that in my book it said that when n is not a positive integer, then the expansion is valid only for |x|<1.

Does that mean for the former method (of expanding from (1 + [2x + 2])^-3) to work , I have to know whether or not |2x+2|<1? If so, then how do we know that |2x/3|<1?

As I understand it, you are applying the formula directly, with n = -3, not with n = 3 as Romsek did, so that you get an infinite series.

Can you show us the gist of your work, and what the provided answer is? I want to see how you handled (2x+2) in your expansion.

But you are correct that there is no promise that the answers obtained the two ways will converge for the same values of x, which might explain a difference in the result. The answer by the "traditional" way can only be applied for |x| < 3/2, while yours applies only to x in (-3/2, -1/2), which turns out to be a subset of the other; in each case, the domain must be stated as part of the answer -- that's how you "know" that the condition holds!

 

[burgerandcheese]

this was my working.

 

[Dr.Peterson]

Aha!

In method (1), do you notice that the terms you ignored contain x, x^2, ... terms when expanded? Doing it this way, you can never get a final coefficient for each term (in fact, they might be infinite ...). If you don't see this, add in the next term, with (2x+2)^3.

Method (2) is correct; each term is final as you write it.

Method (3), of course, does not produce a series. But it is exact, since the series in the denominator terminates.

 

[burgerandcheese]

In method (1), do you notice that the terms you ignored contain x, x^2, ... terms when expanded? Doing it this way, you can never get a final coefficient for each term (in fact, they might be infinite ...). If you don't see this, add in the next term, with (2x+2)^3.

Continuing the series for method (1),
((-3)(-3-1)(-3-2)/3!)(2x + 2)³ = -80 - 240x - 240x² - 80x³
and ((-3)(-3-1)(-3-2)(-3-3)/4!)(2x+2)⁴ = 240 + 960x + 1440x² + 960x³ + 240x⁴

So basically method (1) is correct, but since the coefficients for x and x² and x³ etc. keeps getting larger then that means the series for (1) diverges instead of converges and hence it can't be used for approximation?

 

[Maths_Mike]

you need to remember that when using method 1 mod x has to be less than one.

You have a expansion that only works for mod (2x+2) < 1

eg if x=1 we have 5^-3 which is definitely not = 19 + 42 + 24 +

So method 1 is not correct. The two expressions are not the same and even if we restrict the possible values of x the right hand side would only be an approximation (because you can't actual sum an infinite number of terms)

 

[Dr.Peterson]

Continuing the series for method (1),
((-3)(-3-1)(-3-2)/3!)(2x + 2)³ = -80 - 240x - 240x² - 80x³
and ((-3)(-3-1)(-3-2)(-3-3)/4!)(2x+2)⁴ = 240 + 960x + 1440x² + 960x³ + 240x⁴

So basically method (1) is correct, but since the coefficients for x and x² and x³ etc. keeps getting larger then that means the series for (1) diverges instead of converges and hence it can't be used for approximation?

Worse than that: You can't even identify terms in the series you are trying to make, because each new step changes all existing terms. So we don't even know what the constant term would be. This method doesn't actually make a series (in powers of x) at all.