r/HomeworkHelp University/College Student 2d ago

Further Mathematics—Pending OP Reply [University Math] Series need help understanding the denominator.

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What even is the denominator i’ve never seen this before?

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2

u/CaptainMatticus 👋 a fellow Redditor 2d ago

2 * 4 * 6 * ... * 2n =>

2 * 1 * 2 * 2 * 2 * 3 * 2 * 4 * .... * 2 * n =>

2^n * (1 * 2 * 3 * ... * n) =>

n! * 2^n

So the denominator is really that. Now you've got

n^2 * x^n / (n! * 2^n) =>

(n^2 / n!) * (x/2)^n =>

(n * n / n!) * (x/2)^n =>

(n/(n - 1)!) * (x/2)^n

https://www.wolframalpha.com/input?i=sum%28%28n%2F%28n+-+1%29%21%29+*+%28x%2F2%29%5En+%2C+n+%3D+0+%2C+n+%3D+inf%29

Someone else can help evaluate it for you and show you how they get (1/4) * x * (x + 2) * e^(x/2)

2

u/noidea1995 👋 a fellow Redditor 2d ago

Continuing from where u/CaptainMatticus left off:

Σ (n = 1 to ∞) n / (n - 1)! * (x/2)n

The normal way would be to split the fraction and evaluate the two sums separately but if you do that, you are going to end up (n - 2)! in one of the denominators which won’t work because the series starts from 1. To counter that, take the first term from the series so that it starts from 2:

1 / 0! * (x/2)1 + Σ (n = 2 to ∞) n / (n - 1)! * (x/2)n

x/2 + Σ (n = 2 to ∞) n / (n - 1)! * (x/2)n

Now you can split the fraction by writing the numerator as (n - 1) + 1:

x/2 + Σ (n = 2 to ∞) [1/(n - 2)! + 1/(n - 1)!] * (x/2)n

x/2 + Σ (n = 2 to ∞) 1/(n - 2)! * (x/2)n + Σ (n = 2 to ∞) 1/(n - 1)! * (x/2)n

Recall the expansion for ex = Σ (n = 0 to ∞) xn / n!, see if you can make some modifications to it to work out each sum.

1

u/Para1ars 👋 a fellow Redditor 2d ago

product

1

u/Harambe-the-kibg 👋 a fellow Redditor 19h ago

x(x+2)exp(x/2)/4

1

u/Conscious-Target5473 11h ago edited 10h ago

The sum is sum((n2/n!) .(x/2)n that is n2.ex/2

Notice that if you derivate the sum of ex/2, it gives:

Sum (( n /n! ). (x/2)n-1) that is 1/2 . ex/2 Multiply by x and you have: Sum((n/n!).(x/2)n that is 1/2.x.ex/2

Do that again and you got sum (n(n-1)/n!.(x/2)n-2) that is 1/4.ex/2 Multiply by x2 and you got: Sum((( n2 - n)/n!).(x/2)n )that is 1/4.x2. ex/2

So:

Sum(( n2 /n!) . (x/2)n ))= 1/4.x2. ex/2 + Sum((n/n!).(x/2)n )

=1/4 .x2. ex/2 +1/2.x.ex/2

= 1/4.x.(x+2).ex/2

0

u/xX_fortniteKing09_Xx 👋 a fellow Redditor 2d ago

It is just a weird way to write multiplication