Recurrence... (1 Viewer)

[timeflies]

Okay so came across this Integration question from Cambridge in class and nobody knew what we were doing after a while (even the teacher). The first and second bit of the question were fine but that last bit where you have the factorials is what we couldn't get to.

The solution in Cambridge said this:


We got up to that third line and all was good but just didn't know where to go from there. How do the last two lines work? Can someone please help explain this?

 

[RealiseNothing]

Factor out a 2 from all the brackets:

(2n-2) = 2(n-1)

Repeat this and you get the 4th line.

 

[anomalousdecay]

So from the third line we need to factor out everything that had two in it. So we get 2^(n+1) times the rest of the brackets:



Then to get to the last line, remember the definition of factorials.



And that's it.

 

[timeflies]

oh wow those steps were amazingly easy, thanks! wait can someone explain that second line to third line again too! i get the 2^n+1 n! but where does the rest come from?

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[seanieg89]

oh wow those steps were amazingly easy, thanks! wait can someone explain that second line to third line again too! i get the 2^n+1 n! but where does the rest come from?

Sent from my Nexus 5 using Tapatalk

He is multiplying both numerator and denominator by a product of even numbers which "fills in the gaps" in the factorial in the denominator.

2.4.6....(2n+2)

 

[timeflies]

He is multiplying both numerator and denominator by a product of even numbers which "fills in the gaps" in the factorial in the denominator.

2.4.6....(2n+2)

Ohhh! Makes sense now! Thanks a lot for the help!