Binomial Expansion Coefficient of x^n

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Homework Help Overview

The discussion revolves around finding the coefficient of x^n in the expansion of the expression (1 + x/1! + x^2/2! + x^3/3! + ... + x^n/n!)^2, which is related to binomial expansion and series. Participants explore the implications of using the exponential function e^x and its polynomial form.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the initial challenge of relating the expression to the exponential function and question the significance of missing terms from e^x. Some suggest testing specific values of n to identify patterns, while others consider the structure of the polynomial when squared.

Discussion Status

The conversation has progressed with various attempts to analyze the problem, including specific calculations for small values of n. Some participants have offered insights into the summation of coefficients and the relationship to the power series of e^(2x), although no consensus has been reached on a definitive method.

Contextual Notes

Participants note the complexity of the problem and the potential constraints of not having formally learned about power series, which affects their approach to the solution.

erisedk
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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.
 
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erisedk said:

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.
 
erisedk said:

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.
 
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.
 
erisedk said:
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.
 
erisedk said:

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.
 
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}
##
 
Joffan said:
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.
 
Ray Vickson said:
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! ?
 
  • #10
erisedk said:
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)?
 
  • #11
erisedk said:
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)!) ?
 
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  • #12
Joffan said:
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
 
  • #13
erisedk said:
.
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!.
 
  • #14
Dick said:
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.
 
  • #15
Dick said:
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.
 

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