Wikipedia:Reference desk/Archives/Mathematics/2010 August 28

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August 28

A geometric sequence

Can anyone help me with this question? A geometric sequence is given. If one replaces the signs of the items with even index, the sum of the new sequence will be 3 times smaller then the original one. What is the sum of the original sequence? I get from the equation I write that q, the quotient of the sequence, is 1\2. But what next? —Preceding unsigned comment added by 87.68.249.109 (talk) 07:37, 28 August 2010 (UTC)

Surely if you multiply any such sequence by an arbitrary nonzero constant, you get something else with the same property, and so the only answer is "whatever you like provided it's nonzero"? 95.150.22.219 (talk) 12:05, 28 August 2010 (UTC)

You should double check your calculations for q = 1/2. That's correct if you replace swap the signs on the odd terms. If you replace those of the even terms, you get a different ratio. There is a unique answer. —Preceding unsigned comment added by 203.97.79.114 (talk) 14:24, 28 August 2010 (UTC)

If q is the ratio between term in the original series, then the the original sum is

${\frac {a}{1-q}}$

and the new sum is

${\frac {a}{1+q}}$

so the problem reduces to solving

${\frac {a}{1-q}}={\frac {3a}{1+q}}$

for q.--RDBury (talk) 15:14, 28 August 2010 (UTC)

You've made the same mistake as the OP. That's if you replace the 'odd terms. —Preceding unsigned comment added by 203.97.79.114 (talk) 15:19, 28 August 2010 (UTC)

Surely we can treat this as the sum of two independent geometic sequences. That is if the original sequence is n + a.n + a^2.n + a^3.n + ... then we have two geometric sequences (using b=a^2) which sum as n + b.n + b^2.n + ... and also a.n + b.a.n + b^2.a.n + ... If these sums are Se (sum of even terms) and So (sum of odd terms) and the original total was T then So + Se = T. Then if we change the sign of the even terms we get So - Se = T/3. Solving gives So = 2Se. We can pair the terms of the two sequences: n <-> a.n; a^2.n with b.a.n = a^3.n; a^4.n with a^5.n; ... The tricky bit is do the indices start at 0 or at 1? If they start at 1, then we can pair terms of So is n, a^2.n, a^4.n and Se is a.n, a^3.n, a^5.n so a=2. If the indices start at 0 then think Se = So/2 and a=0.5. If a=2 the the sum of the original sequence is infinite. If a=0.5 then the sum of the sequence = 2.n and the actual value depends upon the chosen 'n'. -- SGBailey (talk) 16:24, 28 August 2010 (UTC)

(ec) The first geometric series is Σqⁱ=1/(1−q). The second geometric series is Σ−(−q)ⁱ=−1/(1+q). The equation −1/(1+q)=(1/3)(1/(1−q)) has the unique solution q=2. The series 1+2+4+8+... is not convergent but by analytic continuation the value is −1. Bo Jacoby (talk) 17:26, 28 August 2010 (UTC).

([[wp:edit conflict|edit conflict]) Assume, for | r | < 1, that you have a geometric series given by S = a + ar + ar² + ar³ … then the standard formula tells us that

$\sum _{k=0}^{\infty }ar^{k}=a+ar+ar^{2}+ar^{3}+\cdots ={\frac {a}{1-r}}.$

I think the OP wants us to reverse the signs of the even-powered terms. If we replace r with −r then we chance the sign of the odd-powered terms. Taking the negative of the whole series will give us what we want:

$-\sum _{k=0}^{\infty }a(-r)^{k}=-a+ar-ar^{2}+ar^{3}-ar^{4}\pm \cdots ={\frac {-a}{1+r}}.$

If the second series is three times smaller than the first then we are left with the following:

$\sum _{k=0}^{\infty }ar^{k}=-{\frac {1}{3}}\sum _{k=0}^{\infty }a(-r)^{k}\ \iff \ {\frac {a}{1-r}}={\frac {-a}{3(1+r)}}\ \iff \ a(2+r)=0.$

Clearly either a = 0 or r = −2. If r = −2 then | r | > 1, and the series cannot converge! So there is no such series. I guess the only answer is the trivial answer where a = 0 and the sum of the series is also zero! — Fly by Night (talk) 19:11, 28 August 2010 (UTC)

Fly by Night, you made the second series three times bigger than the first. If the second series is three times smaller than the first then:

$\sum _{k=0}^{\infty }ar^{k}=-{3}\sum _{k=0}^{\infty }a(-r)^{k}\ \iff \ {\frac {a}{1-r}}={\frac {-3a}{1+r}}\ \iff \ a(2-r)=0.$

Bo Jacoby (talk) 06:25, 29 August 2010 (UTC).

...which gives a=0 or r=2, and thus |r| > 1 still, so Fly by Night's objection that it doesn't converge still applies. --81.158.2.129 (talk) 08:43, 29 August 2010 (UTC)

The function f(x)=(1−x)⁻¹ is an analytic function for x≠1. The series Σxⁱ is convergent for |x|<1 with value f(x). So f(x) is the analytic continuation of the series, and f(2)=−1. Convergence is not the only way to assign a value to a series. Bo Jacoby (talk) 11:32, 29 August 2010 (UTC).

Thanks for pointing out my error. I don't think the OP was set the question to test their knowledge of analytic continuation. It seems quite a simple question put in simple language. The OP didn't mention complex numbers, let alone analytic continuation. My guess is that the question was set to see how s/he could manipulate simple series. And I think the correct answer would have been "No such non-zero, convergent series exists." A series does not have a well defined value outside of its radius of convergence. Finding an analytic function that agrees with the series in the region of convergence and then defining the value of the series outside of that region as equal to the value of the function is subtly different. It gives a value to the series where the series does not have a value of its own. (Do you not feel slightly dishonest assigning the value of −1 to the sum of positive numbers given by 1 + 2 + 2² + 2³ + 2³ + 2⁴ + …?) — Fly by Night (talk) 20:00, 30 August 2010 (UTC)

The series s=1+2+4+8+... satisfies the equation s−1=2s, (because s−1=2+4+8+.. and 2s=2+4+8+... too) and s=−1 is the unique solution to this equation. The series t=−1+2−4+8... satisfies the equation t+1=−2t, and t=−1/3 is the unique solution. That the sum of positive numbers is positive is not necessarily the case for infinite series. That the sum of integers is integer is not necessarily the case for infinite series either. The OP did not require the series to be convergent. No, I don't feel dishonest. Bo Jacoby (talk) 22:06, 30 August 2010 (UTC).

But the infinite sums you mention in you last post; the ones you then perform arithmetic upon, aren't defined unless they converge. It doesn't make sense to say that ∞ − 1 = 2×∞, and then conclude that ∞ = −1 is the unique solution to the "equation". — Fly by Night (talk) 23:10, 30 August 2010 (UTC)

I didn't say that ∞ = −1. If the value of the series s is not defined, then it is free to be defined. You define s = ∞ and I define s = −1. These are different definitions, and we can investigate which definition makes better sense. Note that the binary number 1111111111111111 represents −1 in two's complement, so I am not the only one to define s = −1. The Riemann zeta function is also defined as the analytic continuation of a convergent series. Bo Jacoby (talk) 07:17, 31 August 2010 (UTC).

There are two different ideas at play here. The sum

$g(z)=\sum _{k=0}^{\infty }(2z)^{k}=1+2z+4z^{2}+8z^{3}+16z^{4}+\cdots$

Is a function germ where g : (C,0) → (C,1). It is a local object, and is defined in a neighbourhood of the origin. On the other hand, there is a function

$f(z)={\frac {1}{1-2z}}$

where ƒ : C − {½} → C. This is a global object. It just so happens that g(z) = ƒ(z) for all z ∈ U, where U is an open neighbourhood containing 0. The key is that g is only defined and considered locally. Trying to use ƒ to give global information about g is a flawed strategy. By definition g is a germ of ƒ and it gives the behaviour of ƒ in a certain neighbourhood. If you want to know how ƒ behaves near another point then consider a different function germ; in this case a power-series expanded about a different point.

A germ is a local concept and trying to get global information about g doesn't make sense, and as we've seen: it gives absurd consequences. Your attempts bring you to the conclusion that because the globally defined ƒ is such that ƒ(1) = −1 then the locally defined g is such that g(1) = −1. Let us define

$s_{n}=\sum _{k=0}^{n}=1+2+4+8+\cdots +2^{n}=2^{n+1}-1.$

In your last post you defined s = 1 + 2 + 4 + 8 + …, i.e. to be the limit of s_n as n → ∞, and you declared that to be s = −1. Even though

• s_n > 0 for all n ∈ N.
• s_m < s_n for all m < n.
• s_n → ∞ as n → ∞.

So defining s = −1 is clearly absurd. But that what happens when you try to treat a local object as though it were a global object. Of course, some function germs are globally well-defined; but not all! — Fly by Night (talk) 21:35, 3 September 2010 (UTC)

Proper Form for Factoring Polynomial

I'm doing this summer work for high school and the instructions are "factor completely". The problem is -15x²-5x and I used [1] to factor and it came out as -15(x+1⁄3)*x. But, if I were to factor it such that it has only integers it comes out as -5(3x+1)x. I want to know if factoring it properly would be with only integers (-5(3x+1)x) or such that the monomial portions of the result are in the form of x+a and not nx+a (eg. -15(x+1⁄3)*x). --Melab±1 ☎ 00:50, 28 August 2010 (UTC)

That really depends on what your teacher means by the phrase. It doesn't have a standard mathematical meaning, at least at this level of detail. --Trovatore (talk) 01:04, 28 August 2010 (UTC)

For some purposes you definitely want −15x(x + 1/3). In particular, if you want to know that the whole thing is equal to 0 when x is either −1/3 or 0, but not otherwise, then that's the form you need. But for some other purposes using only integers is approrpriate. But as "Trovatore" notes, the meaning of the phrase "factor completely" is context-dependent. Michael Hardy (talk) 01:24, 28 August 2010 (UTC)

In the high school context, though, 99 times out of 100, factorisation is all about integers (lowest common factor, prime factorisation, etc. etc.). I'll bet you dollars to donuts Melab's teacher is expecting the answer $-5x(3x+1)$ (and depending on how competent they are, would or wouldn't accept the arguments being made here for the context dependence of the instructions). If I'm right, the teacher would argue that you obtain the x=-1/3 solution by setting $3x+1=0$ and doing trivial algebra (that's certainly the way I was taught it). --81.158.2.129 (talk) 08:40, 29 August 2010 (UTC)

... and that's the way I taught it, though I wouldn't necessarily mark other complete factorisations as wrong, I'd just comment that this way is usually considered the simplest and neatest. The "standard" order (given in all text-books) is integer; powers of x; other linear factors without fractions. Dbfirs 14:28, 29 August 2010 (UTC)