Clara's answer to Chua Xiang Ning's Junior College 1 H2 Maths question.
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Apologies for any mistakes as I rushed this one, but hope you get the idea. Feel free to ask for more help :)
Date Posted:
7 years ago
- How do you find 1/3! is actually x^r?
- After the method of difference, how do you find the values individually for the standard expansion?
- After the method of difference, how do you find the values individually for the standard expansion?
My bad, I meant the numerator for the fractions for the series is x^r (as in x=1), since 1^r = 1 for all r
If you look at the series expansion for e^1 (i.e. x=1), up till the value r=n, you'll get 1 + 1 + 1/2! + 1/3! + 1/4! + ... + 1/n!
The series expansion for 1/(r+2)!, also up till the value r=n, is similar, as you'll get 1/3! + 1/4! + ... + 1/r! + 1/(n+1)! + 1/(n+2)!, And so you're missing 1 + 1 + 1/2!, and have and additional 1/(n+1)! + 1/(n+2)!
So 2/(r+2)! = 2e - 2(1 + 1 + 1/2!) + 2(1/(n+1)! + 1/(n+2)!)
The fractions with n don't matter much in the end anyway since they tend to zero as n tends to infinity
The rest is derived from your previous answer
If you look at the series expansion for e^1 (i.e. x=1), up till the value r=n, you'll get 1 + 1 + 1/2! + 1/3! + 1/4! + ... + 1/n!
The series expansion for 1/(r+2)!, also up till the value r=n, is similar, as you'll get 1/3! + 1/4! + ... + 1/r! + 1/(n+1)! + 1/(n+2)!, And so you're missing 1 + 1 + 1/2!, and have and additional 1/(n+1)! + 1/(n+2)!
So 2/(r+2)! = 2e - 2(1 + 1 + 1/2!) + 2(1/(n+1)! + 1/(n+2)!)
The fractions with n don't matter much in the end anyway since they tend to zero as n tends to infinity
The rest is derived from your previous answer
I understand now. Thanks for the explanation
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