limitng (1 Viewer)

[Kutay]

cant do this one

Find x if the series 1, 2x -1, x^2 +3x +1 ....... has a limiting sum???

how you do that >>?? my teacher couldn't get it

 

[Trev]

Ah.... I don't have a clue! I'll keep tryin.

 

[Trev]

Find x if the series 1, 2x -1, x^2 +3x +1 ....... has a limiting sum???

Where did you get this question from? I have never come across a question like this!

 

[Xayma]

If a series a+ar¹+ar²+ar³+...... has a limiting sum than |r|<1.

Also note that yours is a sequence it doesnt really have a limiting sum but anyway.

Now property of a geometric series is that T₃/T₂=T₂/T₁
(x²+3x+1)/(2x-1)=(2x-1)/1
x²+3x+1=4x²-8x+1
3x²-11x=0
x(3x-11)=0
x=0 OR 3x=11
x=0 OR x=11/3


As you can see we get different answers. So the question is wrong in any case it isnt a geometric progression.

Only one of which fits, however, the limiting sum is just alternating not approaching.

 

[Raginsheep]

However, r is given by T₂/T₁=2x-1
x=1/2

As you can see we get different answers. So the question is wrong in any case it isnt a geometric progression.

Huh? Why is T₂/T₁=0?

 

[Trev]

Also note that yours is a sequence it doesnt really have a limiting sum but anyway.


As you can see we get different answers. So the question is wrong in any case it isnt a geometric progression.

Hard to answer a question that cannot be answered
Where did you get this question from Kutay?

 

[Slidey]

cant do this one

Find x if the series 1, 2x -1, x^2 +3x +1 ....... has a limiting sum???

how you do that >>?? my teacher couldn't get it

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

Maybe.

On that note, examine the sequence:

1+x+x^2+...+x^(n-1)

Sum equals: (x^n-1)/(x-1)
From this we note that:
xⁿ-1=(x-1)(1+x+x²...+x^n-1)

 

[Xayma]

Huh? Why is T₂/T₁=0?

Just realised I put that.

x=11/3 but that doesn't fit in with the assumption that |r|<1.
So it must be x=0.

But then if we take the assumption that r=2x-1
if x=0, |r|=1.

Which wont allow us a limiting sum.

If we take x=11/3
We get |r|>1

 

[Raginsheep]

Yeah...that bit I realised. Maybe there a typo in the oringial Q or something?

 

[Xayma]

Isn't the Mathematics Extension 1 course limited to Arithmetic and Geometric progressions?

If it isn't a GP we are most likely looking for a range of values of x (and it is a tad hard with only three terms, one of which contains no x variable).

 

[Raginsheep]

Keep in mind that the series can have a limiting sum without being a GP!!

Err....explain please?

 

[Slidey]

I'm covering convergence currently. Neato.

 

[Raginsheep]

Hmm......yeah, I doubt thats in the 3u course.

 

[Xayma]

Convergence of series is a vast topic with GP's occupying a tiny fraction of the space.

Most converging series are not GP's...on occasion the 4U paper will wander off in this direction but not too far as the general idea of convergence has not been dealt with. I think the question is stuffed

Yes I know that but you have to remember what forum this was placed in, and what techniques you should be looking at to sovle it.

 

[Xayma]

Keep in mind that the series can have a limiting sum without being a GP!! Also just because a series alternates does not mean it can't have a limiting sum!
1-1/2+1/3-1/4+1/5....converges to ln(2) for eg

Except for those three it was alternating between 1 and 0. Not limiting towards either.

Thats why it is expressed |r|<1.

 

[Slidey]

Uh, Derek?

 

[velox]

No wasn't this some other *guy* who Derek was pissing off?

 

[Slidey]

Looks like a typo. Have a crack at this one

For which values of x will the geometric progression

x^2-8x+16, x-4, 1 have a limiting sum. NIce sting in the tail.

a=(x-4)^2, r=1/(x-4)

|r|<1, 1<|x-4|

1 < x-4, x>5
1<-x+4, x<3

For x<3, x>5, the geometric series will converge.

 

[Trev]

I'm covering convergence currently. Neato.

Haha, great Slide, how can you be bothered to do 'extracurricular' maths? I find it hard to be bothered to do the stuff we learn in school.

 

[Kutay]

i got this question from a past exam from my school