Daughters problem.

The sum of an infinite geometric series is 27 and the sum of the first three terms is 19. Find the first term and the common ratio.

and the identity is given:

Sn = a(r^n -1)/(r-1)

So I thought to myself, this is easy and broke it down into two simultaneous equations

27 – 27*r – a = 0

a*r^3 – a -19*r + 19 =0

Threw that into mathcad, because I’m lazy, and got the result

a = 9

r = 2/3

Which seems to be right.

I then tried to work out the simultaneous equations by hand, so I could show the sprog, and I’m stumped.

So stop talking about sex and get to work.

sibeen said:

Daughters problem.

The sum of an infinite geometric series is 27 and the sum of the first three terms is 19. Find the first term and the common ratio.

and the identity is given:

Sn = a(r^n -1)/(r-1)

So I thought to myself, this is easy and broke it down into two simultaneous equations

27 – 27*r – a = 0

a*r^3 – a -19*r + 19 =0

Threw that into mathcad, because I’m lazy, and got the result

a = 9

r = 2/3

Which seems to be right.

I then tried to work out the simultaneous equations by hand, so I could show the sprog, and I’m stumped.

So stop talking about sex and get to work.

Can’t help. I know less about maths than I do about sex (& that’s not much)

Probably a bit long winded, as I am rusty but:

27 – 27*r – a = 0
27 – a = 27*r
a = 27 – 27*r

a*r^3 – a -19*r + 19 = 0
a = (27 – 27*r)*r^3 – 19*r + 19
27 – 27*r = (27 – 27*r)*r^3 – 19*r + 19

8 – 27*r = (27 – 27*r)*r^3 – 19*r
8 – 8*r = (27 – 27*r)*r^3
8 * (1 – r) = 27r^3 * (1 – r)
8 = 27r^3
8/27 = r^3
r = 2/3

a = 27 – 27 * (2/3)
a = 9

furious said:

Probably a bit long winded, as I am rusty but:

27 – 27*r – a = 0
27 – a = 27*r
a = 27 – 27*r

a*r^3 – a -19*r + 19 = 0
a = (27 – 27*r)*r^3 – 19*r + 19
27 – 27*r = (27 – 27*r)*r^3 – 19*r + 19

8 – 27*r = (27 – 27*r)*r^3 – 19*r
8 – 8*r = (27 – 27*r)*r^3
8 * (1 – r) = 27r^3 * (1 – r)
8 = 27r^3
8/27 = r^3
r = 2/3

a = 27 – 27 * (2/3)
a = 9

Egad.

Thanks, furious. I completely missed the substitution a = 27 – 27*r.

Jaysus, quite thick of me.

Been a long time since I’ve done one of these by hand. They do pop up a bit when I’m working, but I always just throw the equations into mathcad and let it do the grunt work.

Thanks again :)

http://journals.aps.org/prd/abstract/10.1103/PhysRevD.72.084013

Information loss in black holes
Phys. Rev. D 72, 084013 – Published 18 October 2005
S. W. Hawking

The question of whether information is lost in black holes is investigated using Euclidean path integrals. The formation and evaporation of black holes is regarded as a scattering problem with all measurements being made at infinity. This seems to be well formulated only in asymptotically AdS spacetimes. The path integral over metrics with trivial topology is unitary and information preserving. On the other hand, the path integral over metrics with nontrivial topologies leads to correlation functions that decay to zero. Thus at late times only the unitary information preserving path integrals over trivial topologies will contribute. Elementary quantum gravity interactions do not lose information or quantum coherence.

——

Fred Wong

For the sum of a GP to converge the absolute value of the common ratio must be less than 1, so I prefer to write the formula for the sum like this:

S_n = a (1 – rⁿ) / (1 – r)
The limit as n → ∞ of rⁿ for |r| < 1 is 0, so
S_∞ = a / (1 – r)

For the problem in the OP,

S₃ = a (1 – r³) / (1 – r) = 19 … (1)
S_∞ = a / (1 – r) = 27 … (2)

Dividing eqn (1) by (2)
1 – r³ = 19 / 27
r³ = 1 – 19 / 27 = 8 / 27
Therefore
r = (8 / 27)^1/3 = 2 / 3

From eqn (2)
a = 27(1 – r) = 27 (1 – 2/3) = 27 / 3 = 9

PM 2Ring said:

For the sum of a GP to converge the absolute value of the common ratio must be less than 1, so I prefer to write the formula for the sum like this:

S_n = a (1 – rⁿ) / (1 – r)
The limit as n → ∞ of rⁿ for |r| < 1 is 0, so
S_∞ = a / (1 – r)

For the problem in the OP,

S₃ = a (1 – r³) / (1 – r) = 19 … (1)
S_∞ = a / (1 – r) = 27 … (2)

Dividing eqn (1) by (2)
1 – r³ = 19 / 27
r³ = 1 – 19 / 27 = 8 / 27
Therefore
r = (8 / 27)^1/3 = 2 / 3

From eqn (2)
a = 27(1 – r) = 27 (1 – 2/3) = 27 / 3 = 9

Elegant, PM :)

I must admit I got caught up in the fact that 0 < r > 1, and thought there must be a winkle in using that, so missed the bloody obvious, ala furious, or your more succinct solution.

That, and I’m an idiot :)