KJW

It's worth pointing out that although for a small black hole, the tidal forces may be so intense as to cause "spaghettification" well outside the event horizon, for an extremely large black hole, the tidal forces may be so weak that one might not even be aware that one has crossed the event horizon. Also, black holes have a peculiar property compared to ordinary matter: the mass of a black hole increases linearly with respect to radius, in contrast to the cube-power for ordinary matter. Thus, extremely large black holes have very low density. One only needs to accumulate enough matter into the large volume to form the large black hole... one doesn't need to compress anything.

No. This is what Wikipedia says about the use of "arbitrary" in mathematics: This is different to "random", which is associated with chance events and probability, neither of which apply to "arbitrary". Are you talking about finite lists of finite sequences, or are you talking about what Cantor is talking about... infinite lists of infinite sequences? It seems to me that you are talking about finite lists of finite sequences, in which case, you can't argue against Cantor, who is talking about infinite lists of infinite sequences. If you are talking about infinite lists of infinite sequences, then you are not taking into account the infinite nature of the sequences and lists.

With the solution to the second problem in place, I decided to provide a solution to the generalised intermediate version of this problem, where the first and second problems are the two extremes. Because much of the solution to the second problem is unchanged in the solution to the generalised problem, I shall only include the parts of the solution that have changed. Let [math]\sigma[/math] be a parameter, [math]0 \leqslant \sigma \leqslant 1[/math], such that [math]\dfrac{\sigma \pi}{n}[/math] is the angle between the line joining the centre of the large circle to the centre of the outermost circle and the line joining the centre of the large circle to the centre of the nearest second outermost circles. Applying the law of cosines and continuing as shown in solution to the second problem: [math](1 + \alpha)^2 \cos^2\dfrac{\pi}{n} - 2 (1 + \alpha) (1 + \cos\dfrac{\sigma\pi}{n}) + 2 (1 + \cos\dfrac{\sigma\pi}{n}) = 0[/math] Note that the angle associated with [math]\cos^2\dfrac{\pi}{n}[/math] has a different origin to the angle associated with [math]1 + \cos\dfrac{\sigma\pi}{n}[/math]. [math]\phi = \dfrac{\pi}{n}[/math] [math](1 + \alpha)^2 \cos^2\phi - 2 (1 + \alpha) (1 + \cos\sigma\phi) + 2 (1 + \cos\sigma\phi) = 0[/math] [math]\phi \approx 0[/math] , [math]\cos\phi \approx 1 - \dfrac{1}{2} \phi^2[/math] : [math](1 + \alpha)^2 (1 - \phi^2) - (1 + \alpha) (4 - \sigma^2 \phi^2) + (4 - \sigma^2 \phi^2) = 0[/math] Solving the quadratic equation for [math](1 + \alpha)[/math] : [math]1 + \alpha = \dfrac{(4 - \sigma^2 \phi^2) - \sqrt{(4 - \sigma^2 \phi^2)^2 - 4 (1 - \phi^2)(4 - \sigma^2 \phi^2)}}{2 (1 - \phi^2)}[/math] [math]1 + \alpha = \dfrac{(4 - \sigma^2 \phi^2) - \sqrt{(16 - 8 \sigma^2 \phi^2 + \sigma^4 \phi^4) - (16 - (16 + 4 \sigma^2) \phi^2 + 4 \sigma^2 \phi^4)}}{2 (1 - \phi^2)}[/math] [math]1 + \alpha = \dfrac{(4 - \sigma^2 \phi^2) - \sqrt{(16 - 4 \sigma^2) \phi^2 - (4 \sigma^2 - \sigma^4) \phi^4}}{2 (1 - \phi^2)}[/math] [math]1 + \alpha = 2 - \sqrt{4 - \sigma^2}\>\phi[/math] [math]\alpha = 1 - \sqrt{4 - \sigma^2}\>\dfrac{\pi}{n}[/math] [math]1 - \alpha^2 = 2 \sqrt{4 - \sigma^2}\>\dfrac{\pi}{n}[/math] [math]\displaystyle \lim_{n\to\infty} P = \dfrac{\pi}{2 \sqrt{4 - \sigma^2}}[/math] For problem #1, [math]\sigma = 0:[/math] [math]\displaystyle \lim_{n\to\infty} P = \dfrac{\pi}{4}[/math] For problem #2, [math]\sigma = 1:[/math] [math]\displaystyle \lim_{n\to\infty} P = \dfrac{\pi}{2 \sqrt{3}}[/math]

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The following is a solution to the second problem: Let [math]R[/math] be the radius of the large circle. Let [math]r[/math] be the radius of the small circles in the outermost layer. Let [math]\alpha r[/math] be the radius of the small circles in the second outermost layer. Let [math]n[/math] be the number of small circles around the circumference of the large circle. Let [math]P[/math] be the proportion of the area of the large circle covered by small circles for [math]n[/math] small circles around the circumference of the large circle. Partitioning the large circle into [math]n[/math] sectors, the area of each sector is: [math]= \dfrac{\pi R^2}{n}[/math] The total area of small circles covering a sector: [math]= \pi r^2 (1 + \alpha^2 + \alpha^4 + \alpha^6 + \cdots) = \dfrac{\pi r^2}{1 - \alpha^2}[/math] Then the proportion of the area of a sector covered by small circles: [math]P = \dfrac{n r^2}{(1 - \alpha^2) R^2}[/math] [math]\dfrac{r}{R - r} = \sin\dfrac{\pi}{n}[/math] [math]r = (R - r) \sin\dfrac{\pi}{n}[/math] [math]r (1 + \sin\dfrac{\pi}{n}) = R \sin\dfrac{\pi}{n}[/math] [math]\dfrac{r}{R} = \dfrac{\sin\dfrac{\pi}{n}}{1 + \sin\dfrac{\pi}{n}}[/math] [math]P = \dfrac{n\>\sin^2\dfrac{\pi}{n}}{(1 - \alpha^2) (1 + \sin\dfrac{\pi}{n})^2}[/math] Let [math](1 - \alpha^2) = \beta \dfrac{\pi}{n}[/math] [math]\displaystyle \lim_{n\to\infty} P = \dfrac{n\>\left( \dfrac{\pi}{n} \right)^2}{\beta\>\dfrac{\pi}{n}} = \dfrac{\pi}{\beta}[/math] Applying the law of cosines: Note that [math](1 + \alpha) r[/math] is the distance between the centres of the outermost and nearest second outermost circles, and that [math](R - r)[/math] and [math]\alpha (R - r)[/math] are the distances from the centre of the large circle to the centres of the outermost circle and second outermost circle, respectively. [math](1 + \alpha)^2 r^2 = (1 + \alpha^2) (R - r)^2 - 2 \alpha (R - r)^2 \cos\dfrac{\pi}{n}[/math] [math](1 + \alpha)^2 \dfrac{r^2}{(R - r)^2} = 1 + \alpha^2 - 2 \alpha \cos\dfrac{\pi}{n}[/math] [math]\dfrac{r}{R - r} = \sin\dfrac{\pi}{n}[/math] [math](1 + \alpha)^2 \sin^2\dfrac{\pi}{n} = 1 + \alpha^2 - 2 \alpha \cos\dfrac{\pi}{n}[/math] [math](1 + \alpha)^2 \sin^2\dfrac{\pi}{n} = (1 + \alpha)^2 - 2 \alpha (1 + \cos\dfrac{\pi}{n})[/math] [math](1 + \alpha)^2 \cos^2\dfrac{\pi}{n} - 2 \alpha (1 + \cos\dfrac{\pi}{n}) = 0[/math] [math](1 + \alpha)^2 \cos^2\dfrac{\pi}{n} - 2 (1 + \alpha) (1 + \cos\dfrac{\pi}{n}) + 2 (1 + \cos\dfrac{\pi}{n}) = 0[/math] Let [math]\phi = \dfrac{\pi}{n}[/math] [math](1 + \alpha)^2 \cos^2\phi - 2 (1 + \alpha) (1 + \cos\phi) + 2 (1 + \cos\phi) = 0[/math] Also let [math]\phi \approx 0[/math] [math]\cos\phi \approx 1 - \dfrac{1}{2} \phi^2[/math] [math](1 + \alpha)^2 (1 - \phi^2) - (1 + \alpha) (4 - \phi^2) + (4 - \phi^2) = 0[/math] This is a quadratic equation to be solved for [math](1 + \alpha)[/math]. However, only the lesser of the two solutions is of interest. [math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{(4 - \phi^2)^2 - 4 (1 - \phi^2)(4 - \phi^2)}}{2 (1 - \phi^2)}[/math] [math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{(16 - 8 \phi^2 + \phi^4) - (16 - 20 \phi^2 + 4 \phi^4)}}{2 (1 - \phi^2)}[/math] [math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{12 \phi^2 - 3 \phi^4}}{2 (1 - \phi^2)}[/math] [math]1 + \alpha = 2 - \sqrt{3} \phi[/math] [math]\alpha = 1 - \sqrt{3} \phi[/math] [math]\alpha^2 = 1 - 2 \sqrt{3} \phi + 3 \phi^2[/math] [math]1 - \alpha^2 = \beta \dfrac{\pi}{n} = 2 \sqrt{3} \phi = 2 \sqrt{3} \dfrac{\pi}{n}[/math] [math]\beta = 2 \sqrt{3}[/math] Therefore: [math]\displaystyle \lim_{n\to\infty} P = \dfrac{\pi}{2 \sqrt{3}}[/math]

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The flaw is his definition of a diagonal sequence/string/s. It depends on all s listed, and differs from those. He defines the initial list as consisting of horizontal s. That's why the random properties of the s and the list which is itself a random s are included. If the s are parallel there can't be any influence of one on any other. I.e. they are independent. Where did you get the idea that the sequences are random? The more correct term is "arbitrary", which is not the same as "random". In fact, Cantor attempts to construct an infinite list of all possible infinite sequences. But regardless of how that is attempted (represented by a list of arbitrary sequences), the list will always be missing the sequence constructed from the negation of the diagonal. In the case of finite sequences, which Cantor does not consider, one can form a complete list of all possible sequences. But then the diagonal argument becomes establishing that the list of sequences is longer than the sequences themselves. It seems to me that you are placing too much importance on the structure of the list of sequences while not placing enough importance on what the diagonal argument is saying. False. I use the infinite set N. False. Where do you use the infinite set [math]\mathbb {N}[/math]? In an earlier post you said: "Per the constructivist view, no one can nor have formed an infinite list. No one can even form an infinite sequence!" If that's not denying the existence of infinite lists and infinite sequences, then what is? I see nothing wrong with the notion of going from the finite to the infinite. It's a qualitative transition that does not imply a "magic [math]n[/math]". Those are the only ones we can form. No one has or can produce an infinite list. It's not an abstraction but a fantasy. No one will ever see a list for the set N. I realised after had I posted that what I said was ambiguous. And you chose the unintended interpretation. I didn't mean that it was incorrect to apply Cantor's diagonal argument to finite lists of finite sequences. I meant that your application of Cantor's diagonal argument was incorrect. Specifically, you drew the wrong conclusion from the length of the list being longer than the length of the sequences. Also, because Cantor was dealing with infinite lists of infinite sequences, your dealing with finite lists of finite sequences is largely irrelevant. I should remark that although Cantor's diagonal argument applied to infinite lists of infinite sequences only produces one missing sequence (the negation of the diagonal), in fact the cardinality of missing sequences is the same as the cardinality of all possible sequences. Any list of sequences is a proverbial drop in the ocean compared to the set of all possible sequences.

I should point out that I am considering what time is to be distinct from what time does, although it is understandable if people conflate these two notions.

But you have not shown that Cantor's diagonal argument is false. All you have done is denied the existence of infinite lists and infinite sequences. And you are incorrectly applying Cantor's diagonal argument to finite lists of finite sequences.

Another important consideration is that there are three space dimensions and one time dimension. If instead there were two space dimensions and two time dimensions, as indicated by the signs of the metric coefficients, then although space and time would still be distinct, they would be indistinguishable because the sign of a metric coefficient does not itself indicate whether the component is a space or time component. This would manifest itself in the solution of the wave equation. If my quest not was meaningful - it would be metaphysics. If my quest is meaningful - it is physics. Because physics is meaningful. In that case, time is what is measured by clocks. Can any other statement of what time is be truly meaningful?

Thank you. The flaw in your argument is that Cantor was dealing with infinite lists of infinite sequences, whereas you chose to reject the notion of infinite lists of infinite sequences and instead deal with finite lists of finite sequences, and in doing so, misapplied the diagonal argument. In fact, your argument against the diagonal argument actually proved Cantor's theorem in the case of finite lists of finite sequences. In the case of finite sequences, one can always create a finite list of all of them. Therefore, if one applies the diagonal argument to a finite list of all the possible finite sequences, then naturally the negation of any diagonal will form a sequence that is part of the list not covered by the diagonal. That the total number of sequences is greater than the length of the sequences is actually what Cantor's theorem is saying in the case of finite sequences. But when one considers infinite sequences, the total number of possible sequences cannot be listed, as proven by the diagonal argument.

Space and time are distinct in that the metric of spacetime has oppositely signed coefficients for the space and time components. The existence of an invariant speed of light (though not necessarily light itself) is a manifestation of this distinction.

Time and space are different components of the one notion that we call "spacetime". If two observers are in motion relative to each other, then what is purely time to one observer is a combination of space and time to the other observer. Relativity provides geometry to spacetime.

But saying that time is what is measured by clocks is not simply saying that time is measured by clocks. It is saying that is what time is. Time is a physical quantity and so must have a physical definition if it is to be meaningful. The problem with what you are asking is whether or not the answer you seek is truly meaningful.

First I have to "declare" that my quest here in this thread is the issue of what time is. -Not how we count it or how it works. This quest is on the agenda of modern physics (as well as the issue of what space is). Therefore I'm continuing in this modern approach of physics. I don't think you grasp the significance of what swansont said. Although it may not be philosophically satisfying, the statement that time is what is measured by clocks is in fact a genuine statement of what time is. Note that one can define a clock by instructions on how to build it. What you seek is ultimately not an explanation of what time actually is. The same can be said about other physical quantities... they are what is measured by their respective measuring instruments, given that those measuring instruments are defined by the instructions to build them. Unless the explanation actually connects to the physical realm, then you don't have an actual explanation of the physical quantity.

This asked me to sign in to Google, so please present it on the forum.

Ok, so go on, explain those things.

But one-time pad keys are not public keys.

No, according to Birkhoff's theorem, the spacetime inside a non-rotating spherical shell of matter is flat.

What I do understand is that Cantor's theorem is about sets, their elements, and their subsets, as well as things that are represented by sets, their elements, and their subsets, and also things that represent sets, their elements, and their subsets. My vision is not limited to an 1891 paper. Infinite sequences of binary symbols can be placed into one-to-one correspondence with subsets of the natural numbers. As such, it doesn't really matter if one considers them as sequences or subsets. I actually avoided considering the sequences as real numbers due to a complication that I did not wish to discuss. Cantor was a mathematician, not a snake oil salesman. What Cantor demonstrated was a mathematical truth concerning infinite sets. He wasn't "selling" anything. It is disappointing that you didn't see any value in the mathematics I presented about non-convergent infinite series. You started a topic that was ultimately about the infinite, but you failed to accept that the infinite is more complicated than the common simplistic understanding of it. I do wish to conclude with the following: Consider the negative of the derivative: [math]-\dfrac{d}{dx} \displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{k} = -\displaystyle \sum_{k=0}^{\infty} (-1)^{k} \dfrac{dx^{k}}{dx} = \displaystyle \sum_{k=1}^{\infty} (-1)^{k-1} k\,x^{k-1}[/math] [math]= -\dfrac{d}{dx} \dfrac{1}{1 + x} = \dfrac{1}{(1 + x)^2}[/math] For [math]x = 1[/math]: [math]\displaystyle \sum_{k=1}^{\infty} (-1)^{k-1} k = \eta(-1) = \dfrac{1}{4}[/math] [math]\zeta(z) = \dfrac{1}{1 - 2^{1-z}}\>\eta(z)[/math] [math]\zeta(-1) = -\dfrac{1}{3}\>\eta(-1) = -\dfrac{1}{12}[/math] Note that I have evaluated [math]\zeta(-1)[/math], not simply obtained its value from elsewhere. [math]\dfrac{d}{dx} \left(x\displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{k}\right) = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} \dfrac{dx^{k+1}}{dx} = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)x^{k}[/math] [math]= \dfrac{d}{dx} \dfrac{x}{1 + x} = \dfrac{1}{1 + x} - \dfrac{x}{(1 + x)^2} = \dfrac{1 + x}{(1 + x)^2} - \dfrac{x}{(1 + x)^2} = \dfrac{1}{(1 + x)^2}[/math] This is the same as above. However, by multiplying by [math]x[/math] before differentiation, the exponents of [math]x[/math] in the series remains unchanged, leading to coefficients that are powers of [math]k + 1[/math] (equivalent to powers of [math]k[/math]) rather than factorials when the operation is iterated. [math]\dfrac{d}{dx} \left(x\displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1) x^{k}\right) = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)^2 x^{k}[/math] [math]= \dfrac{d}{dx} \dfrac{x}{(1 + x)^2} = \dfrac{1}{(1 + x)^2} - \dfrac{2x}{(1 + x)^3} = \dfrac{1 + x}{(1 + x)^3} - \dfrac{2x}{(1 + x)^3} = \dfrac{1 - x}{(1 + x)^3}[/math] [math]\dfrac{1}{1 - 2^{3}} \dfrac{1 - x}{(1 + x)^3} = \zeta(-2) = 0[/math] for [math]x = 1[/math] [math]\dfrac{d}{dx} \left(x\displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)^2 x^{k}\right) = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)^3 x^{k}[/math] [math]= \dfrac{d}{dx} \dfrac{x - x^2}{(1 + x)^3} = \dfrac{1 - 2x}{(1 + x)^3} - \dfrac{3x - 3x^2}{(1 + x)^4} = \dfrac{1 - x - 2x^2}{(1 + x)^4} - \dfrac{3x - 3x^2}{(1 + x)^4} = \dfrac{1 - 4x + x^2}{(1 + x)^4}[/math] [math]\dfrac{1}{1 - 2^{4}} \dfrac{1 - 4x + x^2}{(1 + x)^4} = \zeta(-3) = \dfrac{1}{120}[/math] for [math]x = 1[/math] [math]\dfrac{1}{1 - 2^{n+1}} \overbrace{\dfrac{d}{dx} \Biggl( \Biggr. x \cdots \dfrac{d}{dx} \Biggl( \Biggr. x}^{n}\displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{k} \Biggl. \Biggr) \cdots \Biggl. \Biggr) = \dfrac{1}{1 - 2^{n+1}} \displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)^n x^{k}[/math] [math]= \dfrac{1}{1 - 2^{n+1}} \overbrace{\dfrac{d}{dx} \Biggl( \Biggr. x \cdots \dfrac{d}{dx} \Biggl( \Biggr. x}^{n} \>\>\dfrac{1}{1 + x} \Biggl. \Biggr) \cdots \Biggl. \Biggr) = \zeta(-n)[/math] for [math]x = 1[/math]

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[math]\overbrace{\dfrac{d}{dx} \Biggl( \Biggr. x \cdots \dfrac{d}{dx} \Biggl( \Biggr. x}^{n}\displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{k} \Biggl. \Biggr) \cdots \Biggl. \Biggr) = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)^n x^{k} = \eta(-n)[/math] for [math]x = 1[/math]

[math]\zeta(z) = \displaystyle \sum_{k=1}^{\infty} k^{-z}[/math] [math]\eta(z) = \displaystyle \sum_{k=1}^{\infty} (-1)^{k-1} k^{-z} = \displaystyle \sum_{k=1}^{\infty} (2k-1)^{-z} - \displaystyle \sum_{k=1}^{\infty} (2k)^{-z} = \displaystyle \sum_{k=1}^{\infty} k^{-z} - 2 \displaystyle \sum_{k=1}^{\infty} (2k)^{-z}[/math] [math]= \displaystyle \sum_{k=1}^{\infty} k^{-z} - 2^{1-z} \displaystyle \sum_{k=1}^{\infty} k^{-z} = (1 - 2^{1-z}) \displaystyle \sum_{k=1}^{\infty} k^{-z}[/math] [math]= (1 - 2^{1-z})\>\zeta(z)[/math] For [math]z = -1[/math], [math]\eta(-1) = \displaystyle \sum_{k=1}^{\infty} (-1)^{k-1} k = 1 - 2 + 3 - 4 + 5 - 6\>+\>...\>= -3\>\zeta(-1) = \dfrac{1}{4}[/math] Although the series [math]1 + 1 + 1 + 1 + 1 + 1\>+\>...\>[/math] can't be determined as the geometric series, it can be determined from the Riemann zeta function: [math]\zeta(0) = \displaystyle \sum_{k=1}^{\infty} k^0 = 1 + 1 + 1 + 1 + 1 + 1\>+\>...\>= -\dfrac{1}{2}[/math] Note that this could also have been obtained from: [math]\zeta(z) = \dfrac{1}{1 - 2^{1-z}}\>\eta(z)[/math] For [math]z = 0[/math], [math]\zeta(0) = -\eta(0) = -\dfrac{1}{2}[/math], where [math]\eta(0) = 1 - 1 + 1 - 1 + 1 - 1\>+\>...\>= \dfrac{1}{2}[/math] had already been obtained in a previous post.

Given that the above infinite series are alternating, one might assume that their values are some sort of average of the terms. But the non-alternating series: [math]\displaystyle \sum_{k=0}^{\infty} x^k[/math] where [math]x > 1[/math] dispels that notion: [math]\displaystyle \sum_{k=0}^{\infty} x^k = \displaystyle \sum_{k=0}^{m-1} x^k + \displaystyle \sum_{k=m}^{\infty} x^k =\dfrac{x^m - 1}{x - 1} + x^m \displaystyle \sum_{k=0}^{\infty} x^k[/math] [math]\displaystyle \sum_{k=0}^{\infty} x^k - \dfrac{x^m - 1}{x - 1} = x^m \displaystyle \sum_{k=0}^{\infty} x^k[/math] Note that removing initial terms from the infinite series yields the same infinite series multiplied by some value, and that removing more initial terms yields an even larger multiple of the infinite series. This behaviour of the infinite series is clearly different to the behaviour of any corresponding finite partial series. Also note that if [math]x = 10[/math], then the removed partial sum has a decimal value of the form [math]11111 \cdots 11111[/math] which is [math]\dfrac{1}{9}[/math] of [math]99999 \cdots 99999[/math], hence the general form for base-[math]x[/math] numbers. Continuing: [math](x^m - 1) \displaystyle \sum_{k=0}^{\infty} x^k = -\dfrac{x^m - 1}{x - 1}[/math] [math]\displaystyle \sum_{k=0}^{\infty} x^k = \dfrac{-1}{x - 1} < 0[/math] for [math]x > 1[/math] All the partial sums of the infinite series are positive-valued, yet the infinite series itself is negative-valued. In the case of [math]x = 2[/math], the value of the infinite series is [math]-1[/math]. But treated as an infinite sum, the value of the infinite series can be expressed in binary as: [math]\cdots 1111111111[/math] This is the same value that was excluded by Cantor's diagonal argument in an earlier post for the mapping of the set of all finite subsets of the natural numbers to the natural numbers. In other words, the sum of the infinite series goes beyond [math]\aleph_0[/math]. However, the binary expression for the infinite sum also corresponds to the two's complement representation of [math]-1[/math] used in computing, albeit extended leftward to infinite bits.

[math]\displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{nk} = \dfrac{1}{1 + x^{n}} \\ y = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{nk}[/math]