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Another question needed btw Thanks Enigma for the reply (1 Viewer)

ssukarieh1

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Hello, limiting sums question from 3 Unit Cambridge, last question in the exercise.
17. Find the condition for each GP to have a limiting sum, then find that limiting sum:
a) 1+(x^2-1)+(x^2-1)^2+...

g) 1+2x/1+x^2+ (2x)^2/(1+x^2)^2+...

I understand that -1< r< 1, however, when I substitute r, I keep getting the wrong answer. The answer to
a) is -(2^1/2) < x < (2^1/2) if that helps.

Thank you :kiss::kiss:, much appreciated!!
 
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ssukarieh1

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Yes I know for the second part of the question, but that doesn't apply to the first part, which asks for the condition for the GP to have a limiting sum
 

enigma_1

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no worries! :)

open this View attachment 31298

bos is not letting me write the inequalities properly so I had to attach it lel

then limitins sum is = a/(1-r)

= 1/(1-(x^2 -1))
= 1/(2-x^2)
 
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enigma_1

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Thanks Enigma!!! But what about question g) ??
no worries :))) for 17g first you find the limiting sum which is (x^2 +1)/(x-1)^2 and then the denominator cannot be equivalent to zero so the condition x=/= 1 and x=/= -1 should be stated for the GP to have a limiting sum
 
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enigma_1

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When you solve 0 < x^2 <2, you have to exclude x=0 from your solution.
this

sorry didn't realise it, it's coz if x=0 then r would be equal to r=-1 which cannot occur for a limiting sum to occur
 

ssukarieh1

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Would you consider this somewhat of a difficult question for a 2 Unit student?
 

braintic

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1+2x/(1+x^2)+ (2x)^2/(1+x^2)^2+... Like that?
-1 < 2x/(1+x^2) < 1

-1-x^2 < 2x < 1+x^2

LHS:
x^2 + 2x + 1 >0
(x+1)^2 > 0
x not equal to -1

RHS:
x^2 -2x +1 > 0
(x-1)^2 >0
x is not equal to 1


S = 1 / [1 - 2x/(1+x^2) ]

= (1+x^2) / (1+x^2 - 2x)

= (1+x^2) / (1-x)^2
 

braintic

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no worries :))) for 17g first you find the limiting sum which is (x^2 +1)/(x-1)^2 and then the denominator cannot be equivalent to zero so the condition x=/= 1 and x=/= -1 should be stated for the GP to have a limiting sum
The restriction x not equal to -1 doesn't come from the denominator.
 

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