limiting sums (1 Viewer)

[Yellow]

can somebody please help me with this, it's going to be in my trial exam:

For what range of values of x will the series 1 + x + x^2 + x^3 +..
have a limiting sum? What is this limiting sum when x = 1/2?

 

[ben]

Originally posted by Yellow
can somebody please help me with this, it's going to be in my trial exam:

For what range of values of x will the series 1 + x + x^2 + x^3 +..
have a limiting sum? What is this limiting sum when x = 1/2?

Limiting sums = a/(1 - r)
Limiting sums occur when -1 < r < 1

Therefore it's a limiting sum when -1 < x < 1
(Since r = [(x^2)/x] = x

And use the equation for x = 1/2
1/(1/2) = 2


*I think - No responsibility taken if I'm wrong

 

[Minai]

look for questions that are like "a ball is dropped from a building 20m tall, and bounces 10m on its first bounce...if each consecutive bounce is 0.5 the length of the previous, find the total distance it will bounce"
i got somethin like that in my trial....those types are the more likely applications of limiting sums

 

[ben]

Originally posted by MinAi
look for questions that are like "a ball is dropped from a building 20m tall, and bounces 10m on its first bounce...if each consecutive bounce is 0.5 the length of the previous, find the total distance it will bounce"
i got somethin like that in my trial....those types are the more likely applications of limiting sums

yeah, in our trials we got a question about a tree's growth in height

 

[Yellow]

thank you so much! there's a limiting sum question in my 4u trial and it asks "for which values of x can a limiting sum occur" - or so my teacher says.

 

[ben]

4u?

 

[Nightshade]

4 unit.

aka Extension 2.

 

[ben]

Originally posted by Nightshade
4 unit.

aka Extension 2.

they do basic limiting sum questions in extension 2?

 

[Angelic_Angel88]

Originally posted by ben
they do basic limiting sum questions in extension 2?

I don't think they DO do basic limiting sum questions in 4U. Maybe 'Yellow' just forgot how to do the basic questions.

 

[Yellow]

we don't do that in 4u but we do do harder 3u questions which explains the limiting sum.

 

[Lazarus]

Technically, you're calculating a limiting sum every time you integrate.

 

[spice girl]

Originally posted by Lazarus
Technically, you're calculating a limiting sum every time you integrate.

a limit is different from a limiting sum...

 

[Lazarus]

I know.

 

[amster]

Ok, what about this one then...

Two numbers, p,q (p>q) are such that p, 6, q are in a geometric progression and 1/p, 5/18, 1/q are in arithmetic progression. Find:
a) the common ratio of the G.P. (done)
b) the limiting sum to infinity of the G.P. (not done)
c) the common difference of the A.P. (done)

Ok, so I totally suck and I need some help from you smart people. anyone?

 

[SoulSearcher]

Ok, what about this one then...

Two numbers, p,q (p>q) are such that p, 6, q are in a geometric progression and 1/p, 5/18, 1/q are in arithmetic progression. Find:
a) the common ratio of the G.P. (done)
b) the limiting sum to infinity of the G.P. (not done)
c) the common difference of the A.P. (done)

Ok, so I totally suck and I need some help from you smart people. anyone?

You have the common ratio of the G.P., r, from there you can find the value of the first term, a, and from there you should be able to use the limiting sum formula to calculate the limiting sum of the G.P.

 

[-pari-]

what i dont get is how ppl who haven't studied 2u maths for over a year still remember this stuff?

i studied it LAST TERM and i'm already forgetting.

 

[SoulSearcher]

what i dont get is how ppl who haven't studied 2u maths for over a year still remember this stuff?

i studied it LAST TERM and i'm already forgetting.

I help out a lot at school, plus I'm thinking of tutoring people soon, so it's good to know that you can still remember some of it

 

[Dragonmaster262]

Can somebody refresh my memory and remind me what a limiting sum is?

 

[Aquawhite]

we don't do that in 4u but we do do harder 3u questions which explains the limiting sum.

hehe. Nice English... (sarcasm). Haha, you said a poopey word.

 

[gurmies]

A limiting sum is essentially the sum of a geometric progression, a(1-r^n)/(1-r) where |r|<1 and as n (number of terms) tends to infinity. Since |r|<1 and n tends to infinity , r^n tends to 0 and you end up with the formula as a/(1-r).