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binomial expansion question

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Original post by EEngWillow
I assume by "rational" you actually mean "fractions where the denominator isn't equal to 1 (...or 0)", since the integers are rational numbers :tongue:

Yes, I do. Sorry. If you're being picky, all numbers are complex too! :tongue:

Original post by EEngWillow

One point I should have added is that from the infinite expansion you can hopefully see the need for |x| < 1: you need a value between 0 and 1 for the terms to get smaller in value each time (take the example x = 0.5, so x^2 = 0.25, x^3 = 0.125 etc) -> under these circumstances the series converges. If you had instead x = 2, then each term would get larger (x = 2, x^2 = 4, x^3 = 8) and it diverges.

Yes, I see that you need |x| < 1 for convergence.
Original post by Plato's Trousers
Yes, I do. Sorry. If you're being picky, all numbers are complex too! :tongue:


Yes, I see that you need |x| < 1 for convergence.


...And all is good in the world again. For now.
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kayla:)
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Hey

I'm looking at a past question on a WJEC C4 paper and I'm wondering about the validity of the expansion

As I've got 2 expansions in one, do I find the validity of both of them, then use the one that's closer to 0?
Say if it was 1<|4x| and 1<|3x| would I go with the 1<|4x| ??

Thanks!!
binomial question.png8.1KB
Original post by kayla:)
Hey

I'm looking at a past question on a WJEC C4 paper and I'm wondering about the validity of the expansion

As I've got 2 expansions in one, do I find the validity of both of them, then use the one that's closer to 0?
Say if it was 1<|4x| and 1<|3x| would I go with the 1<|4x| ??

Thanks!!

Yes, for the expansion to be valid it must satisfy both conditions (otherwise one expansion exists and the other doesn't, or both expansions don't exist). In other words, you need ∣x∣<13 |x|< \frac{1}{3} and ∣x∣<14 |x| < \frac{1}{4} . Which of those is harder to achieve?

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