Question

Find the coefficient of x^10 in the power series (x^2 +x^4+x^6+....)(x^3+x^6+x^9+.....)(x^4+x^8+x^12+....).Can someone help me?

Answer 1

the ans is 0
since no solution for 2a+3b+4c=10, for any natural number a, b, c

Answer 2

(x^2 +x^4+x^6+....)(x^3+x^6+x^9+.....)(x^4+x^8+x^12+....)
= x^2(1+x^2+x^4+...)x^3(1+x^3+x^6+...)x^4(1+x^4+x^8+...)
= x^9 (1+x^2+x^4+...)(1+x^3+x^6+...)(1+x^4+x^8+...)
the multiply of terms in brackets cant getting the x term, so there is not a x^10, in other words @enka is right :)

Answer 3

I don't understand. how can you lead to that equation? and even though if so, you still have so many cases to get that for example 6 [3*2] +0{3*0] +4 [4*1]

Answer 4

ok . I got it now thanks alot. what if every terms has 1+.... in the front of the sequence?

Answer 5

then, you'll get the ans

Answer 6

I rewrite the whole thing (1+x^2+x^4+x^6+.....)(1+x^4+x^8+x^12+....)(1+x^6+x^12+x^18+....)

Answer 7

the question is the same, coefficient of x^10

Answer 8

These can be a little tedious, but you can simplify a bit, depending on the exct question.

It is sufficient to track this:
(x^2 +x^4+x^6+x^8+x^10)(x^3+x^6+x^9)(x^4+x^8)
Everything else will be over 10

Multiply the first two:

x^5 + x^7 + x^8 + x ^9 + x^10 + a bunch of other stuff that we can discard.

(x^5 + x^7 + x^8 + x ^9 + x^10)(x^4+x^8)

Recognizing that the lowest exponent we brought with us in the last factor is 4, we can discard from the larger factor all but the first term

(x^5)(x^4+x^8)

x^9 + x^13 -- There is no x^10

Answer 9

now the problem count back 3 post from this message

Answer 10

Better:

It is sufficient to track this:
(x^2 +x^4+x^6+x^8+x^10)(x^3+x^6+x^9)(x^4+x^8)
Everything else will be over 10

Factor

(x^9)(1 + x^2 + x^4 + x^6 + x^8)(1 + x^3 + x^6)(1 + x^4)

And the question becomes, what is the coefficient of x^1 in the following expression.

(1 + x^2 + x^4 + x^6 + x^8)(1 + x^3 + x^6)(1 + x^4)

It should be relatively obvious that the desired coefficient is zero (0).

Answer 11

It is not that. they are the infinite polynomial . they don't stop at any where

Answer 12

the first ( ) =\[\frac{ 1 }{ 1-x^2}\]

Answer 13

the second( ) =\[\frac{ 1 }{1-x^4 }\]

Answer 14

the third ( ) =

Answer 15

\[\frac{ 1 }{ 1-x^6 }\]

Answer 16

Okay, then

(1 + x^2 + ...)
*(1 + x^3 + ...)
*(1 + x^4 + ...)

Again, it is sufficient to track these:

(1 + x^2 + x^4 + x^6 + x^8 + x^10)
*(1 + x^3 + x^6 + x^9)
*(1 + x^4 + x^8)

Multiply the 1st 2

(1 + x^2 + x^3 + x^4 + x^5 + 2x^6 + x^7 + 2x^8 + 2x^9 + 2x^10 + other stuff that we can ignore)

(1 + x^2 + x^3 + x^4 + x^5 + 2x^6 + x^7 + 2x^8 + 2x^9 + 2x^10)

Piecemeal the last factor

1 * 2x^10 = 2x^10

x^4 * 2x^6 = 2x^10

x^8 * x^2 = x^10

and we get 5x^10

Like I said, these can be a little tedious.

Answer 17

yes you are right. Thanks a lot. I have to copy it to figure out the new way to have the answer

Answer 18

I am sorry when I cannot give you the medal for your answer since it is not an independent question and from the original problem I gave out the all of them . Terribly sorry and thanks a looooot

Answer 19

Infinite though they are, if you ask me about the coefficient of x^10, there is quite a bit of the infinite polynomial that will not have any impact and it can be simply ignored.

Answer 20

now I learn a new way to solve the problem

Answer 21

I will post another question in Algebra. are you interesting in?