What are the constants in the binomial series expansion for (1+mx)^-n?

[sooyong94]

Homework Statement

For $n>0$, the expansion of $(1+mx)^{-n}$ in ascending powers of $x$ is $1+8x+48x^{2}+...$

(a) Find the constants $m$ and $n$
(b) Show that the coefficient of $x^{400}$ is in the form of $a(4)^{k}$, where $a$ and $k$ are real constants.

Homework Equations

Binomial expansion

The Attempt at a Solution

I have expanded the series using binomial theorem, and compared each coefficient. Then I solved them using simultaneous equation. That works out to be $m=-4, n=2$.

(b) I know the binomial coefficient is $\frac{n(n-1)(n-2)...(n-r+1)}{r!}$, though I'm stuck on that... :(

 

[Simon Bridge]

I have expanded the series using binomial theorem, and compared each coefficient. Then I solved them using simultaneous equation. That works out to be $m=-4, n=2$.

if $n=2$, then how can there be an $x^{400}$ term? Why does the provided expansion not terminate?
if m=-4, as well, then wouldn't that mean that the second term in the expansion is -8x instead of the +8x given?

 

[sooyong94]

Oops... It should be $(1+mx)^{-n}$... Sorry about that. :P

 

[Simon Bridge]

OK - so you have the expansion, in principle, of form:
$$(1+mx)^{-n}=\sum_{i=1}^N a_ix^i$$ ... the coefficient of $x^{400}$ is $a_{400}$ ... and you know that $a_i$ depends on n,m and r from the binomial expansion formula.

You have a formula for the binomial coefficient - but $a_{400}$ has an extra term due to the value of m.
The trick is to match up the values you have with the formulas ... i.e. while n=2 in the equation above, the "n" in the formula for the binomial coefficient is different ... it may help to rewrite the formula taking this into account.
If $a_{400}$ corresponds to the r'th binomial coefficient then what is r?

 

[sooyong94]

I don't get it... :/
I know the coefficient of $x^{400}$ would look something like this:
$\frac{-2(-3)(-4)...(-400-2+1)}{400!}$
Though it doesn't look like what the question needs... :S

 

[Simon Bridge]

You forgot about the value of m ;)

Anyway - you have to show that the series is the same as what they gave you.
Find the value of a and k.

Just to get you thinking the right way: is the product a positive or negative number?
Maybe it is -2x3x4x...x401 or +2x3x4x...x401

... can you represent it in some kind of shorthand?
Say using factorial notation?

 

[sooyong94]

Where's the value of m?

That actually looks like (401!) ?

 

[Simon Bridge]

Where's the value of m?

You said that m=-4 remember?

Notice that the question does not ask for the binomial coefficient.
The coefficient of x^400 includes the binomial coefficient but is not equal to it... just like the coefficient of x^2 is not equal to it's binomial coefficient. Compare.

That actually looks like (401!) ?

That's right! So to ratio comes to 401!/400! doesn't it? ... what is that equal to?

 

[sooyong94]

Still don't get it on the binomial coefficient though...

I know that 401!/400!=(401x400!)/(400!)=401

 

[Simon Bridge]

I know that 401!/400!=(401x400!)/(400!)=401

That's all there is to it! Well done ;)
How did you work out that it should be positive rather than negative?

You need the part where "m=4" fits into make the connection.
Remember, you need to show that $a_{400}=b(4)^k$ where #a_n$is the coefficient of$x^n## in the polynomial, with b and k both positive real numbers.

 

[sooyong94]

That's all there is to it! Well done ;)
How did you work out that it should be positive rather than negative?

You need the part where "m=4" fits into make the connection.
Remember, you need to show that $a_{400}=b(4)^k$ where $a_n$ is the coefficient of $x^n$ in the polynomial, with b and k both positive real numbers.

I don't know about that though... :/

But how do I use that m=4?

 

[Simon Bridge]

Well consider: what is the coefficient of x and what is the coefficient of x^2?
How does that m=-4 contribute to those coefficients?
How does the binomial coefficient contribute?

You calculated it - you should know!
Write ut the coefficient of x and x^2 in terms of n and m again - instead of actual numbers.

 

[sooyong94]

Coefficient of x would look something like this: -nm
Coefficient of x^2 would look something like this: $\frac{(-n)(-n-1))}{2!} m^{2}$

 

[haruspex]

Coefficient of x would look something like this: -nm
Coefficient of x^2 would look something like this: $\frac{(-n)(-n-1))}{2!} m^{2}$

Yes. m is just a multiplier on x, so you can start with the expansion for m=1 and just replace x with (mx) in the expansion.

 

[sooyong94]

So that means... the coefficient of $x^{400}$ is
$\frac{(-2)(-3)(-4)...(-400-2+1)}{400!} (-4)^{400}$ ?

 

[haruspex]

So that means... the coefficient of $x^{400}$ is
$\frac{(-2)(-3)(-4)...(-400-2+1)}{400!} (-4)^{400}$ ?

Yes, but you can get rid of all of those awkward minus signs. Each extra factor like (-400-2+1) inverts the sign, but so does the extra -4 term, so they all cancel.

 

[sooyong94]

Yes, but you can get rid of all of those awkward minus signs. Each extra factor like (-400-2+1) inverts the sign, but so does the extra -4 term, so they all cancel.

Erm... why? How does it invert the minus sign?
I know there are 400 terms, so the signs cancel each other, because 400 is an even number?

 

[haruspex]

Erm... why? How does it invert the minus sign?
I know there are 400 terms, so the signs cancel each other, because 400 is an even number?

All the terms in the expansion are positive. (−2)(−3)(−4)...(−n−2+1) has n terms, so has the same sign as (-4)ⁿ.

 

[sooyong94]

All the terms in the expansion are positive. (−2)(−3)(−4)...(−n−2+1) has n terms, so has the same sign as (-4)ⁿ.

How does all terms in the expansion are positive?

 

[haruspex]

How does all terms in the expansion are positive?

The first term (1) is positive. To get the second term you multiply by (-2)(-4) = 8. To get the next you multiply be (-3)(-4) = 12. At every step you multiply by two factors, both negative, so the minus signs cancel.

 

[Simon Bridge]

It may be easier to think of it like this ...
the product of N negative numbers is positive if N is even and negative if N is odd right?
i.e. if $y=(-x_1)(-x_2)(-x_3)\cdots(-x_{N-1})(-x_N): x_i>0$
then y > 0 if N is even and y < 0 if N is odd.
Can you see why this is the case?

...similarly $(-4)^N=4^N$ if N is even and $-4^N$ if N is odd.

 

[Simon Bridge]

Back to this:

Coefficient of x would look something like this: -nm
Coefficient of x^2 would look something like this: $\frac{(-n)(-n-1))}{2!} m^{2}$

Good - so what would the coefficient of x^3 look like?
How about x^4?

Follow the pattern to find the expression for the coefficient of x^400 - put in the values of n and m you found and remember posts 4 and 5.

 

[sooyong94]

It may be easier to think of it like this ...
the product of N negative numbers is positive if N is even and negative if N is odd right?
i.e. if $y=(-x_1)(-x_2)(-x_3)\cdots(-x_{N-1})(-x_N): x_i>0$
then y > 0 if N is even and y < 0 if N is odd.
Can you see why this is the case?

...similarly $(-4)^N=4^N$ if N is even and $-4^N$ if N is odd.

I get it now... :D
Because when you multiply N negative number, and if N is an even number (in this case 400), the signs of each number cancel. :P

 

[sooyong94]

Back to this:

Good - so what would the coefficient of x^3 look like?
How about x^4?

Follow the pattern to find the expression for the coefficient of x^400 - put in the values of n and m you found and remember posts 4 and 5.

So... that would look something like this...
the coefficient of $x^{400}$ is
$\frac{(-2)(-3)(-4)...(-400-2+1)}{400!} (-4)^{400}$

 

[haruspex]

So... that would look something like this...
the coefficient of $x^{400}$ is
$\frac{(-2)(-3)(-4)...(-400-2+1)}{400!} (-4)^{400}$

Yes, but you should simplify the signs, as you've been shown.

 

[sooyong94]

Yes, but you should simplify the signs, as you've been shown.

the coefficient of $x^{400}$ is
Does it sound right to you?
$\frac{(2)(3)(4)...(399}{400!} (4)^{400}$

 

[haruspex]

the coefficient of $x^{400}$ is
Does it sound right to you?
$\frac{(2)(3)(4)...(399)}{400!} (4)^{400}$

No, you've made a mistake in the arithmetic (the 399).
Having fixed that, there's another simplification to do (which you did much earlier in the thread, post #9).

 

[sooyong94]

No, you've made a mistake in the arithmetic (the 399).
Having fixed that, there's another simplification to do (which you did much earlier in the thread, post #9).

Why?

 

[haruspex]

why?

-400-2+1 = ?

 

[Simon Bridge]

I have pretty much the same notes for you:

So... that would look something like this...
the coefficient of $x^{400}$ is
$\frac{(-2)(-3)(-4)...(-400-2+1)}{400!} (-4)^{400}$

Therefore:

the coefficient of $x^{400}$ is
Does it sound right to you?
$\frac{(2)(3)(4)...(399}{400!} (4)^{400}$

... $-400-2+1\neq -399$ check arithmetic.

You have to remember to use what you learned in previous posts when you do future ones.
These lessons all fit together.

i.e. the numerator in the last calculation can be written in a shorter way remember?
This leads to a simpler way to write the whole fraction part.

Once you've done that - you need to compare what you got with what you have to show way back in post #1. If you go reread that but you should see the pattern already.

 

[sooyong94]

I have pretty much the same notes for you:
Therefore:
... $-400-2+1\neq -399$ check arithmetic.

You have to remember to use what you learned in previous posts when you do future ones.
These lessons all fit together.

i.e. the numerator in the last calculation can be written in a shorter way remember?
This leads to a simpler way to write the whole fraction part.

Once you've done that - you need to compare what you got with what you have to show way back in post #1. If you go reread that but you should see the pattern already.

-401... Ok.. I get it now. :P

 

[Simon Bridge]

Great - so now all you have to do is put it all together!
Remember what you are trying to show?

 

[sooyong94]

Great - so now all you have to do is put it all together!
Remember what you are trying to show?

$\frac{(2)(3)(4)...(401)}{400!} (4)^{400}$

$=\frac{(1)(2)(3)(4)...(401)}{400!} 4^{400}$
$=\frac{(401!)}{400!} 4^{400}$
$=\frac{(401)(400!)}{400!} 4^{400}$
$=401 (4)^{400}$

$a=401, k=400$

 

[haruspex]

$\frac{(2)(3)(4)...(401)}{400!} (4)^{400}$

$=\frac{(1)(2)(3)(4)...(401)}{400!} 4^{400}$
$=\frac{(401!)}{400!} 4^{400}$
$=\frac{(401)(400!)}{400!} 4^{400}$
$=401 (4)^{400}$

$a=401, k=400$

That's it.

 

[Simon Bridge]

Beautiful - pour yourself a drink!