Binomial Expansion Coefficient of x^n

[erisedk]

Homework Statement

Find the coefficient of x^n in the expansion of ( 1 + x/1! + x^2/2! + x^3/3! + ... + x^n/n! )^2 .

Homework Equations

The Attempt at a Solution

At first glance, this looks like the polynomial form of e^x, but the expansion of e^x goes to infinity, so any use of that seems ruled out. Furthermore, I can't try expanding this or writing it in terms of the rth term and equating the power of x to n and doing something with that. Basically, I don't know how to begin.

 

[Dick]

Homework Statement

Find the coefficient of x^n in the expansion of ( 1 + x/1! + x^2/2! + x^3/3! + ... + x^n/n! )^2 .

Homework Equations

The Attempt at a Solution

At first glance, this looks like the polynomial form of e^x, but the expansion of e^x goes to infinity, so any use of that seems ruled out. Furthermore, I can't try expanding this or writing it in terms of the rth term and equating the power of x to n and doing something with that. Basically, I don't know how to begin.

Ask yourself whether the terms that are missing from the expansion of e^x would make any difference to the coefficient of the x^n term.

 

[PeroK]

Homework Statement

Find the coefficient of x^n in the expansion of ( 1 + x/1! + x^2/2! + x^3/3! + ... + x^n/n! )^2 .

Homework Equations

The Attempt at a Solution

At first glance, this looks like the polynomial form of e^x, but the expansion of e^x goes to infinity, so any use of that seems ruled out. Furthermore, I can't try expanding this or writing it in terms of the rth term and equating the power of x to n and doing something with that. Basically, I don't know how to begin.

Why not work it out for n = 2, n = 3 and possibility n =4 and see if you can work out what's happening.

 

[erisedk]

I tried for n=2, I got 2. However, looking at it, I don't see any pattern that might be of much use for figuring out x^n.

 

[Dick]

I tried for n=2, I got 2. However, looking at it, I don't see any pattern that might be of much use for figuring out x^n.

Try it for the other ones and compare with the coefficient of x^n in (e^x)^2.

 

[Ray Vickson]

Homework Statement

Find the coefficient of x^n in the expansion of ( 1 + x/1! + x^2/2! + x^3/3! + ... + x^n/n! )^2 .

Homework Equations

The Attempt at a Solution

At first glance, this looks like the polynomial form of e^x, but the expansion of e^x goes to infinity, so any use of that seems ruled out. Furthermore, I can't try expanding this or writing it in terms of the rth term and equating the power of x to n and doing something with that. Basically, I don't know how to begin.

Think of putting two copies of your polynomial side-by-side; these are the two factors in the squared expression. The coefficient of x^k in the product consists of a sum of coefficient products: 1= 1/0! from factor 1 and 1/k! from factor 2, or 1/1! from factor 1 and 1/(k-1)! from factor 2, or 1/2! from factor 1 and 1/(k-2)! from factor 2, or ... or 1/k! from factor 1 and 1/0! from factor 2. You want the sum of all the two-by-two products of these numbers.

 

[Joffan]

If we write the squaring multiplication with one of the terms reversed, it's may make it easier to see the components of the x^n coefficient in the answer.

##
\begin{matrix}
( & 1 & + &x/1! & + &x^2/2! & + & \dots & + & x^{n-2}/(n-2)! & + &x^{n-1}/(n-1)! & + & x^n/n! & ) & \times \\
( & x^n/n! & + & x^{n-1}/(n-1)! & + & x^{n-2}/(n-2)! & + & \dots & + & x^2/2! & + & x/1! & + & 1 & ) &
\end{matrix}
##

 

[Dick]

If we write the squaring multiplication with one of the terms reversed, it's may make it easier to see the components of the x^n coefficient in the answer.

##
\begin{matrix}
( & 1 & + &x/1! & + &x^2/2! & + & \dots & + & x^{n-2}/(n-2)! & + &x^{n-1}/(n-1)! & + & x^n/n! & ) & \times \\
( & x^n/n! & + & x^{n-1}/(n-1)! & + & x^{n-2}/(n-2)! & + & \dots & + & x^2/2! & + & x/1! & + & 1 & ) &
\end{matrix}
##

Sure, you can write down the series. But it's also not terribly hard just to write down the number that it must sum to.

 

[erisedk]

Think of putting two copies of your polynomial side-by-side; these are the two factors in the squared expression. The coefficient of x^k in the product consists of a sum of coefficient products: 1= 1/0! from factor 1 and 1/k! from factor 2, or 1/1! from factor 1 and 1/(k-1)! from factor 2, or 1/2! from factor 1 and 1/(k-2)! from factor 2, or ... or 1/k! from factor 1 and 1/0! from factor 2. You want the sum of all the two-by-two products of these numbers.

So I want to sum the following-
( 1/0! * 1/n! ) + ( 1/1! * 1/(n-1)! ) + ... + ( 1/n! * 1/0! )
How do I sum that up to get the answer 2^n/n! ?

 

[Dick]

So I want to sum the following-
( 1/0! * 1/n! ) + ( 1/1! * 1/(n-1)! ) + ... + ( 1/n! * 1/0! )
How do I sum that up to get the answer 2^n/n! ?

What powers of x are in the difference between your product and the power series expansion of e^(2x)?

 

[Joffan]

So I want to sum the following-
( 1/0! * 1/n! ) + ( 1/1! * 1/(n-1)! ) + ... + ( 1/n! * 1/0! )
How do I sum that up to get the answer 2^n/n! ?

Does 1/(k!(n-k)!) look familiar to you at all? How about n!/(k!(n-k)!) ?

 

[erisedk]

Does 1/(k!(n-k)!) look familiar to you at all? How about n!/(k!(n-k)!) ?

.
Yes of course. So I multiply and divide by n!, and I get ( C0+C1+...Cn )/n! Which is just 2^n/n!

Thank you everyone :D

 

[Dick]

.
Yes of course. So I multiply and divide by n!, and I get ( C0+C1+...Cn )/n! Which is just 2^n/n!

Thank you everyone :D

Yes, but you didn't even have to sum the series. Your product is identical to the power series for e^(2x) up to order n. The nth term of that is 2^n/n!.

 

[erisedk]

Yes, but you didn't even have to sum the series. Your product is identical to the power series for e^(2x) up to order n. The nth term of that is 2^n/n!.

That's pretty neat. I haven't formally learned power series and stuff, so I didn't know that.

 

[erisedk]

Yes, but you didn't even have to sum the series. Your product is identical to the power series for e^(2x) up to order n. The nth term of that is 2^n/n!.

I just googled operations on power series, and that's definitely a much better way of doing this problem.