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AQA FP3 Expansions

I'm using the FP3 book that's on the AQA Website.

I'm on question 5 in exercise 1D on page 24. Below is a link to the book but also a screen dump of the question. I've also attached my workings so far.

It's not going well. Help needed, please. I also note that the limit has changed to x --> infinity rather than 0.


http://filestore.aqa.org.uk/subjects/AQA-MFP3-TEXTBOOK.PDF
Reply 1
Original post by maggiehodgson
I'm using the FP3 book that's on the AQA Website.

I'm on question 5 in exercise 1D on page 24. Below is a link to the book but also a screen dump of the question. I've also attached my workings so far.

It's not going well. Help needed, please. I also note that the limit has changed to x --> infinity rather than 0.


http://filestore.aqa.org.uk/subjects/AQA-MFP3-TEXTBOOK.PDF

Recall that a requirement for convergence of the infinite binomial series is that the term y in (1+y)n(1+y)^n is that y is very small.
Consequently, you need to take out more than just a factor of (3x)1/2(3x)^{1/2}

EDIT: Also your solution to Q4 is a little off, recall that there is a -1 in the numerator and denominator of the fraction and use that :smile:
(edited 9 years ago)
Original post by joostan
Recall that a requirement for convergence of the infinite binomial series is that the term y in (1+y)n(1+y)^n is that y is very small.
Consequently, you need to take out more than just a factor of (3x)1/2(3x)^{1/2}

EDIT: Also your solution to Q4 is a little off, recall that there is a -1 in the numerator and denominator of the fraction and use that :smile:



Struggling to see. What factor could I take out? Won't x/3 now be small?

Also in question 4 have I just forgotten the -1 on numerator and denominator so would begin xln2 / xln3? Then I'd divide each by x and still get the same answer? Is it just a missing step?

Thanks for your time
Reply 3
Original post by maggiehodgson
Struggling to see. What factor could I take out? Won't x/3 now be small?

Also in question 4 have I just forgotten the -1 on numerator and denominator so would begin xln2 / xln3? Then I'd divide each by x and still get the same answer? Is it just a missing step?

Thanks for your time


If x is very large, and you divide it by 3 it certainly won't be very small, take 30000000000 for example.
As for your second question, yes that's right.

I have to disappear off now, but I'l give you a hint:

Spoiler

(edited 9 years ago)
Original post by joostan
If x is very large, and you divide it by 3 it certainly won't be very small, take 30000000000 for example.
As for your second question, yes that's right.

I have to disappear off now, but I'l give you a hint:

Spoiler




Ah yes. Thanks
Reply 5
Original post by maggiehodgson
Ah yes. Thanks


No worries. :smile:

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