2unit Question 10; Limiting sums. (1 Viewer)

[Graceofgod]

From Baulkham hills trial 2008:

10c) It is known that the geometric series 1 + x + x^2 + x^3 + ... has a limiting sum.

i) What are the possible values of x?

ii) Observe that;

1 = 1
2x = x + x
3x^2 = x^2 + x^2 + x^2
4x^3 = x^3 + x^3 + x^3 + x^3

By studying the above arrangement, or otherwise, find in simplest algebraic form, an expression for the limiting sum of the series;

1 + 2x + 3x^2 + 4x^3 + ... + nx^(n-1) + ...

Please use the spoiler code to hide your answers [ spoiler ] [ /spoiler ]
http://community.boredofstudies.org/misc.php?do=bbcode#spoiler

Spoiler

like thishttp://community.boredofstudies.org/misc.php?do=bbcode#spoiler

http://community.boredofstudies.org/misc.php?do=bbcode#spoiler

 

[lyounamu]

From Baulkham hills trial 2008:

10c) It is known that the geometric series 1 + x + x^2 + x^3 + ... has a limiting sum.

i) What are the possible values of x?

ii) Observe that;

1 = 1
2x = x + x
3x^2 = x^2 + x^2 + x^2
4x^3 = x^3 + x^3 + x^3 + x^3....

By studying the above arrangement, or otherwise, find in simplest algebraic form, an expression for the limiting sum of the series;

1 + 2x + 3x^2 + 4x^3 + ... + nx^(n-1) + ...

Please use the spoiler code to hide your answers

Spoiler

like this

Spoiler

i) -1< x <1

ii) RED: 1+x+x^2+... = a/(1-r) = 1/(1-x)
BLUE: x+x^2+.... = a/(1-r) = x/(1-x)
LIGHT GREEN: x^2 +x^3 +... = a/(1-r) = x^2/(1-x)
BROWN: x^3 + x^4 + ... = a/(1-r) = x^3/(1-x)
....continued infinitely

Spoiler

TOTAL:
1/(1-x) + x/(1-x) + x^2/(1-x) +...
= 1/(1-x)(1+x+x^2+...)
= 1/1-x (1/1-x)
= 1/(1-x)(1-x)

By the way, how do you spoiler thingy?

 

[Graceofgod]

Check my original post, I added it.

But yes, you have the correct answer Pretty simple once you see that.

 

[crammy90]

i) -1< x <1

ii) RED: 1+x+x^2+... = a/(1-r) = 1/(1-x)
BLUE: x+x^2+.... = a/(1-r) = x/(1-x)
LIGHT GREEN: x^2 +x^3 +... = a/(1-r) = x^2/(1-x)
BROWN: x^3 + x^4 + ... = a/(1-r) = x^3/(1-x)
....continued infinitely

TOTAL:
1/(1-x) + x/(1-x) + x^2/(1-x) +...
= 1/(1-x)(1+x+x^2+...)
= 1/1-x (1/1-x)
= 1/(1-x)(1-x)

By the way, how do you spoiler thingy?

i have no idea why we had to do this...
so basically we had to see there were many series in the one thing?
i didnt even get i lol
ive dont so much maths and i still suck lol it might have to be my one that doesnt count

 

[Graceofgod]

i have no idea why we had to do this...
so basically we had to see there were many series in the one thing?
i didnt even get i lol
ive dont so much maths and i still suck lol it might have to be my one that doesnt count

Yes, the main part is seeing that it can actually be divided into infinite geometric series.

Finding the limiting sum of each of these gives a geometric series of limiting sums.

Finding the limiting sum of the new series gives you your answer.

 

[Timothy.Siu]

good question...but now we know it wont be in our exam

 

[trailblazer]

Spoiler

i) -1< x <1

ii) RED: 1+x+x^2+... = a/(1-r) = 1/(1-x)
BLUE: x+x^2+.... = a/(1-r) = x/(1-x)
LIGHT GREEN: x^2 +x^3 +... = a/(1-r) = x^2/(1-x)
BROWN: x^3 + x^4 + ... = a/(1-r) = x^3/(1-x)
....continued infinitely

Spoiler

TOTAL:
1/(1-x) + x/(1-x) + x^2/(1-x) +...
= 1/(1-x)(1+x+x^2+...)
= 1/1-x (1/1-x)
= 1/(1-x)(1-x)

By the way, how do you spoiler thingy?

Wow nicely set out.

 

[trailblazer]

good question...but now we know it wont be in our exam

it can be

 

[Timothy.Siu]

Wow nicely set out.

lol wud u do that in a test?

 

[trailblazer]

lol wud u do that in a test?

lol. You can only use black and blue pen, so i would use columns. Why arent you on ba or msn?

 

[Graceofgod]

lol. You can only use black and blue pen, so i would use columns. Why arent you on ba or msn?

Hmm =S Back to dota..

 

[axlenatore]

Spoiler

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

By the way, how do you spoiler thingy?

How come you factorised, couldnt you just say its a gp with a= 1/(1-x) and a common ratio of x/(1-x) / 1/(1-x) = x

Therefore limiting sum is
1/(1-x)/ 1-x
= 1

Yer.. please explain haha

 

[H4rdc0r3]

I prob wouldn't have got this in an exam. You can probably see them easy with practice.

 

[trailblazer]

Hmm =S Back to dota..

What's wrong with DotA?

 

[Timothy.Siu]

lol. You can only use black and blue pen, so i would use columns. Why arent you on ba or msn?

lol i'm on ba now,msn lags me lol

I prob wouldn't have got this in an exam. You can probably see them easy with practice.

lol noob

 

[lyounamu]

How come you factorised, couldnt you just say its a gp with a= 1/(1-x) and a common ratio of x/(1-x) / 1/(1-x) = x

Therefore limiting sum is
1/(1-x)/ 1-x
= 1

Yer.. please explain haha

It's actually not 1/(1-x)/(1-x),

it's 1/(1-x)(1-x) = 1/(1-x)^2

 

[Graceofgod]

What's wrong with DotA?

Nothing is wrong with dota. Just when people with pudge in their avatar play it... =S

Whats your ba name? I will play you after hsc.

 

[3unitz]

From Baulkham hills trial 2008:

10c) It is known that the geometric series 1 + x + x^2 + x^3 + ... has a limiting sum.

i) What are the possible values of x?

ii) Observe that;

1 = 1
2x = x + x
3x^2 = x^2 + x^2 + x^2
4x^3 = x^3 + x^3 + x^3 + x^3

By studying the above arrangement, or otherwise, find in simplest algebraic form, an expression for the limiting sum of the series;

1 + 2x + 3x^2 + 4x^3 + ... + nx^(n-1) + ...

Please use the spoiler code to hide your answers [ spoiler ] [ /spoiler ]

Spoiler

like this

one thing you could do is just differentiate both sides 1 + x + x^2 + x^3 + ... = 1/(1 - x)

1 + 2x + 3x^2 + 4x^3 + ... + nx^(n-1) + ... = 1/(1 - x)^2

 

[lyounamu]

one thing you could do is just differentiate both sides 1 + x + x^2 + x^3 + ... = 1/(1 - x)

1 + 2x + 3x^2 + 4x^3 + ... + nx^(n-1) + ... = 1/(1 - x)^2

How does that work out? Tell us.

 

[Graceofgod]

one thing you could do is just differentiate both sides 1 + x + x^2 + x^3 + ... = 1/(1 - x)

1 + 2x + 3x^2 + 4x^3 + ... + nx^(n-1) + ... = 1/(1 - x)^2

I too missed what you are saying.