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C4 really hard question need help :)

Screen Shot 2016-06-06 at 21.42.03.png
the answer to part c is x<1/2

But I have no idea how to work that out.
Reply 1
Avatar for Yua
Yua
14
Original post by suibster
Screen Shot 2016-06-06 at 21.42.03.png
the answer to part c is x<1/2

But I have no idea how to work that out.


when you get the partial fractions, say if one of them is (1-2x)^-1 then you get 1 + (-1)(-2x) + (-1)(-2)(-2x)/2

the 2x bit is the key

|2x| <1

so |x| <1/2
Reply 2
Avatar for suibster
suibster
OP
3
Thanks but I don't really understand the thought process into this.
why (1-2x)^-1 in particular ?
There's also a bracket for (1-x)^-1 and (1-x)^-2, do you not take them into consideration?
Reply 3
Avatar for Zacken
Zacken
22
Original post by suibster
Thanks but I don't really understand the thought process into this.
why (1-2x)^-1 in particular ?
There's also a bracket for (1-x)^-1 and (1-x)^-2, do you not take them into consideration?


You do, so you need all of

|-x| < 1 AND |-x| < 1 AND |2x| < 1.

But this simplifies down to just

|x| < 1/2.

Since if that's true then all three are true that the same time.

Your thought process should be to always take the smallest bound you can. So if there was a (1-3x)^n in there, you'd also do |3x| < 1.
Reply 4
Avatar for suibster
suibster
OP
3
Original post by Zacken
You do, so you need all of

|-x| < 1 AND |-x| < 1 AND |2x| < 1.

But this simplifies down to just

|x| < 1/2.

Since if that's true then all three are true that the same time.

Your thought process should be to always take the smallest bound you can. So if there was a (1-3x)^n in there, you'd also do |3x| < 1.



Thank you! Very well explained.
Reply 5
Avatar for Zacken
Zacken
22
Original post by suibster
Thank you! Very well explained.


No worries, thank you. Appreciated. :smile:

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