Question

What is the coefficient on the x^n term?

Answer 1

1

Answer 2

\[f(x)=x+x^2(x+1)+x^3(x+1)^2+x^4(x+1)^3+...\]

Answer 3

lol

Answer 4

:P:LOL @mrdoldum good job :)

Answer 5

well coe(x)=1
and coe(x^2)=1
and coe(x^3)=2 I believe
just trying to see a pattern

Answer 6

no

Answer 7

coe(x^4)=2
I think for n>1 it might be 2

Answer 8

The coefficient for all the polynomials you posted is still 1.

Answer 9

hmm so far @freckles you are doing good

Answer 10

wait that is wrong
I take back for n>1 it might be 2

Answer 11

coe(x^5)=
is harder

Answer 12

Well, the coefficient for x in all those polynomials is 1.

Answer 13

Are you sure coe(x^4)=2? (or 1... lol)

Answer 14

one of the ways to look is to find a pattern

Answer 15

The coefficient of a variable is the number in front. For 2x the coefficient of x is 2. For \[x^{2}\] the coefficient of x is 1. If there is not a number in front of a variable the coefficient is 1 as \[1*x=x\]For \[a(b+x)\] the coefficient of x is a.

Answer 16

\[f(x)=x+x^3+x^2+x^3(x^2+2x+1)+x^4(x^3+3x^2+x+1)+\cdots \\ f(x)=x+x^2+(1+1)x^3+(2+1+\cdots)x^4+\cdots \]

Answer 17

@mrdoldum To clarify the problem, I mean after you multiply the polynomial out completely.

Answer 18

so coe(x^4)>2

Answer 19

the coe(4)=3

Answer 20

is stops there i believe
coe(x^4)=3

Answer 21

Anyone want to guess the 5th one? =P

Answer 22

so far we have 1,1,2,3,...

Answer 23

is it 4?

Answer 24

The \[x^{n}\] term would be \[x^{n}(x+1)^{n-1}\] So, yes, I read the problem wrong, sorry. For this the coefficient of \[x^{n}\] is 1, but \[x^{n}\] is the coefficient of \[(x+1)^{n-1}\]

Answer 25

looks like a familiar sequence

Answer 26

or the fibnoccia sequence

Answer 27

I'm gonna go get some hot tea, it's cold and rainy outside. ;)

Answer 28

or however you spell it

Answer 29

i say the coe(5)=5

Answer 30

1,1,2,3,5,8,13,...

Answer 31

\(\huge\color{white}{{\rm coe }~\rm = chocolate~ or ~eating}\)
my guess is 5 :D

Answer 32

so Fibonacci sequence

Answer 33

fibonacci sequence is the best :D

Answer 34

^^^ ????

Answer 35

http://www.wolframalpha.com/input/?i=power+series+x%2F%281-x%28x%2B1%29%29

Answer 36

@ganeshie8 with his cute power series thingy

Answer 37

thats a generating function for fibonacci series :D

Answer 38

wow, that's the same sequence :)

Answer 39

ganeshie is generating

Answer 40

\[\begin{align}f(x)&=x+x^2(x+1)+x^3(x+1)^2+x^4(x+1)^3+...\\~\\
x(x+1)f(x)&=~~~+x^2(x+1)+x^3(x+1)^2+x^4(x+1)^3+...\\~\\

\end{align}\]

subtracting both we get

\[f(x) - x(x+1)f(x) = x \]

\[\implies f(x) = \dfrac{x}{1-x(x+1)}\]

Answer 41

partial fractions and power series... and all that cool stuff xD

Answer 42

hey @ganeshie8 since you like power series questions so much
i have a question i have been eyeballing for awhile and have felt too intimidated to try

Answer 43

lol eyeballing

Answer 44

i will post a separate thread for it

Answer 45

That's pretty much the reverse of how I created this, but slightly different: \[\Large f(x)=\frac{x}{1-x-x^2} \\ \Large \frac{f(x)}{x} = \frac{1}{1-[x(x+1)]} \\ \Large f(x) =x \sum_{n=0}^\infty x^n(x+1)^n\]

Answer 46

okie power series is the king of all series xD

Answer 47

*queen of all series (for the sake of lady audience in this thread :P)

Answer 48

just cause it called ''POWER''

Answer 49

A sequence as coefficients in front the geometric series is considered its generating function, does that mean a function multiplied by an integral x^n dn from 0 to infinity is considered its generating function?

Answer 50

It is 1, did you notice a pattern to the polynomial here?

Answer 51

I mean, isn't a function just a sequence of numbers except there's like a bunch more closer together lol

Answer 52

You have me confuzzled

Answer 53

looks interesting \(\int_0^{\infty} x^n dn\) assuming |x| < 1 is it

Answer 54

you want to multiply this with functions wow

Answer 55

Yeah makes sense yeah? Haha this is really just the laplace transform isn't it?

Answer 56

Actually we could multiply a laplace transform by ln(1/x) I think to get the "generating polynomial" of a function I think.