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Posts posted by John

  1. Your third and fourth equations should be [math]2527 = A \vee (A \wedge B)[/math] and [math]2234 = B \vee (A \wedge B)[/math], respectively, though I don't think Boolean algebra really works here, since it deals with truth values not general numbers. I think you should instead look at this in terms of sets and make use of the principle of inclusion-exclusion.


    We know [math]|A| = 2527[/math], [math]|B| = 2234[/math], and [math] |A \cup B| = 6000 - 1846 = 4154[/math], so we just need to solve for [math]|A \cap B|[/math].

  2. What we have is a set of potential outcomes of an experiment (the "sample space"), which we'll denote with Ω, and individual outcomes are subsets of the sample space. In addition, we have a function P : Ω → [0,1] assigning a probability to each outcome. So when we say P(AB), we're talking about a union of subsets being assigned a probability. The propositional logic symbols don't entirely make sense in this context.


    I suppose you could have a situation where you're randomly generating strings of logical symbols along with the two statements A and B, in which case the string "AB" would be one potential output. But of course, that's not the meaning you're asking about.


    Edit: I should note, as always, that notation isn't sacred. You could use P(AB) to denote the probability of two events A and B both happening. But it'd be a little strange.

  3. Alright, I see what you mean now.


    Well, if we parameterize the parabola as x = t and y = t2, then we have this handy formula for finding parallel curves, which gives us (assuming I've made no mistakes, and letting a = 2)

    [math]x = t - \frac{4t}{\sqrt{1 + 4t^{2}}}[/math]


    [math]y = t^{2} + \frac{2}{\sqrt{1 + 4t^{2}}}[/math]


    which is this funny-looking thing. Of course, a = 2 was arbitrary, just letting each unit along the y-axis represent one inch. We could use something like a = 1/18 (thus letting each unit along the y-axis represent a yard), which yields


    [math]x = t - \frac{\frac{t}{9}}{\sqrt{1 + 4t^{2}}}[/math]


    [math]y = t^{2} + \frac{\frac{1}{18}}{\sqrt{1 + 4t^{2}}}[/math]


    and a slightly nicer-looking curve, which when plotted along with the original parabola, seems to be what we want.


    Converting the parametric equation into a Cartesian equivalent is left as an exercise to the reader. :x

    Edit: A minor point: Also note that, in the last link, the graph is only plotted for t = -2 to t = 2, rather than t = -4 to t = 4. This is because the uneven scaling of the axes visibly screws with the apparent distance over larger intervals, and AspectRatio doesn't seem to work properly.


    Edit 2: Also note that our original result, using a = 2, also works, as we can see in this graph, but I think the second result, using the smaller value for a, will work more nicely for the original problem.

  4. It's not really a pattern. It's just the geometrical interpretation of taking the inverse of a linear function. If we have y = 2x, then the inverse is y = (1/2)x. We can visualize this as reflecting the line y = 2x about the line y = x, as shown here.


    Now, about your definition of γ.


    I'm assuming that xa is some function of x, and ya is the inverse of xa. Given what you've told me about nd (why do you use nd here, and not na?), the first part of the definition seems to be what we've discussed before (i.e. applying the inverse-product-derivative process to a linear function), though I think we still have the restriction that the coefficient of x must be 1 or the coefficient of x0 must be 0. I'm fine with that.


    For the second part of the definition, I guess you're defining each xaya to be the new product at each step of the process, and as you said a few posts ago, this applies to the original process, not the more recent one. Thus the "limit as a goes to infinity" refers to the sequence of functions generated by applying this process over and over starting with our original function. While the exponent on x does (as far as I can tell, at least for positive exponents) approach 1 as the number of iterations increases, that doesn't guarantee the limit is 2x specifically, and even if it did, we'd never actually arrive at 2x (though we would get arbitrarily close).


    Ah, so the author of my book multiplied the top and bottom by (1/dt)/(1/dt). The book didn't say you can "multiply" and the book just did this, which is really uninformative:

    [math]\frac{d}{dx} \frac{dy}{dx} = \frac{\frac{d}{dt} \frac{dy}{dx}}{\frac{dx}{dt}}[/math]


    Which skipped a lot of steps that would have helped me understand the notation better.


    Thanks for the help.


    Well, I should say I don't know for sure that the author did it that way, heh, or indeed what his thought process was at all. That's why I was wondering what the context was. This was simply the process that occurred to me to get from the left-hand side to the right-hand side.

  6. What's the context in which this equation comes about?


    Using the Leibniz notation, I guess the way to think about it is [math]\frac{d}{dx} \frac{dy}{dx} = \frac{d \frac{dy}{dx}}{dx} = \frac{d \frac{dy}{dx}}{dx} \frac{\frac{1}{dt}}{\frac{1}{dt}} = \frac{\frac{d}{dt} \frac{dy}{dx}}{\frac{dx}{dt}}[/math]. In general we can just consider "multiplying" the differentials, obtaining [math]\frac{d}{dx} \frac{dy}{dx} = \frac{d}{dx} \frac{d}{dx} y = \frac{d^{2}}{dx^{2}} y = \frac{d^{2} y}{dx^{2}}[/math], which is the standard Leibniz notation for the second derivative of [math]y[/math] with respect to [math]x[/math].


    Leibniz notation is handy for seeing the process of taking derivatives, in certain cases, and in some limited sense we can, as above, talk about "multiplying" and "dividing" and so forth. However, it's not really valid (at least, at the introductory standard calculus level) to treat differentials as quantities that can be manipulated algebraically willy-nilly.

  7. One thing I forgot to mention about this earlier is 2x as a function doesn't follow this pattern, as 2x and 2x would be the same if the function is 2x and inverse is x/2.


    I'm not sure what you mean here. The function y = 2x follows the pattern just fine.


    Here is a way to represent the new type of iterating function:


    [math]\gamma (a)=\begin{cases} \frac{\mathrm{d} }{\mathrm{d} x}\left ( x_{a}y_{a} \right )& \text{ if } n_{d}=1 \\ \lim_{a\rightarrow \infty }\frac{\mathrm{d} }{\mathrm{d} x}\left ( x_{a}y_{a} \right )& \text{ if } n_{d}>1 \end{cases}[/math]


    I think you'll have to clarify a bit. What is nd? What's the purpose of the limit? Keep in mind that the limit involving [math]x^{\frac{a_{n}b_{n} + 1}{a_{n}b_{n}}}[/math] assumed the original algorithm of taking the inverse and differentiating the resulting product, regardless of the starting function, whereas more recently we've been talking about differentiating until we have a linear equation before worrying about the inverse at all.

  8. This concept has been explored in several works of fiction. See, for example, the list in this Wikipedia article (which includes Divided By Infinity): Quantum suicide and immortality.


    While it's a nice idea, and I'm sure it makes for some interesting stories, it's always struck me as an example of quantum woo. While the many worlds interpretation does seem to suggest that a person survives a potentially fatal event in some universes, I wouldn't expect that to mean the person never subjectively experiences death. I guess my thinking is that although there may be other copies of me in many universes, we're all separate entities with separate lives.


    Of course, I'm certainly not a physicist, so perhaps there's better justification for considering the concept than I believe there is.

  9. Oh, one thing I was thinking about earlier but forgot to mention:


    When we get to the point that we have y = rx + s, what we have is a linear equation, and its inverse is also a linear equation. Geometrically, we have a line, and to get the inverse, we're reflecting the line about y = x (example here). To arrive at 2x, we first need to arrive at x2 + k for some real number k, i.e. given our two lines, we somehow want the products of their points over the entire range of x-values to generate an upward vertical parabola with its vertex on the y-axis. While this still isn't a proof, it does seem that such a special result would require special starting conditions.


    Just something to keep in mind, if you're not thinking along those lines already.

  10. What do you mean by this? By naively, do you mean it would not be proper to think it in the light of limits?


    Nah, I just meant that while the reasoning is intuitively correct, it's not a rigorous proof. Don't read too much into it. It's a minor point, and I don't feel like tossing epsilons all around. :)


    EDIT: I think it is actually a sound conjecture. I have tested with many functions and still get results expected. I would have to keep testing in order to prove the conjecture true of false. Either that, find proof of either.


    With your example, however, I am iterating and still haven't found an end to it. I will keep trying though.

    Trigonometric functions and others could be included in some form or fashion. The process might just have to be modified in order to include for these. I would have to work these out to find out.


    EDIT2: It would be interesting if it could be predicted what m would equal in this case. It would provide a way to determine if your conjecture works for all numbers(this smells of Fermat :P )

    I'll leave that to you for now, heh. Between an abstract algebra midterm earlier and thinking about this, I'm somewhat mathed out for the night. But we'll see if that changes later.

  11. I took "regular functions" to include things like exponentials, trigonometric functions, etc. (there's a more technical definition of the term in algebraic geometry, but I don't know much about that), but now I'm assuming we're restricting ourselves to polynomials here. Thus we have the built-in rules that the powers of x must be non-negative integers, the leading coefficient must be non-zero, etc.

    Of course, for a polynomial of degree n, the (n - 1)th derivative will be of the form f(x) = rx + s for some real numbers r ≠ 0 and s. The inverse of this will be f-1(x) = (1/r)x - s/r, giving us a product of x2 - sx + (s/r)x + s2/r = x2 + [(s - sr)/r]x + s2/r. The derivative of this product, then, is 2x + (s - sr)/r.

    Now, if s = sr, i.e. s = 0 or r = 1, then we're done. Otherwise, repeating the process, the inverse of this is (1/2)x - (s - sr)/(2r). Multiplying, we have x2 + [s/2-s/(2r)]x - s2/2 - s2/(2r2) + s2/r. The derivative of this is 2x+ [s/2-s/(2r)].

    Now, if s/2 = s/(2r), i.e. s = 0 or r = 1 again, then we're done. But if s = 0 or r = 1, then we'd have been done earlier.

    So I'm wondering if this process only works in the case that s = 0 or r = 1. It seems like, given the case that s ≠ 0 and r ≠ 1, we end up with 2x + m for some real number m ≠ 0, but never get to just 2x. Of course, completing a few more iterations might prove me wrong, but it's getting tedious keeping track of everything, heh. Unless you already have a counterexample to my conjecture here.


    If I'm correct, then the process works for a polynomial of degree n only if the coefficient of xn-1 is 0 or the coefficient a of xn is such that an(n - 1)(n - 2)...(n - (n - 2)) = 1, i.e. a = 1/(n!). But there's enough arithmetic here (read: opportunities for arithmetic errors) that I'm not entirely confident in my conclusions.

    EDIT3: What I noticed with the pattern you had shown if you were to add a limit to infinity for the steps it would eventually reach x, which would soon be multiplied by its inverse(x) and then the derivative would be taken to get 2x. Correct me if I misinterpret what you stated.


    [math]\lim_{(a_{n},b_{n})\rightarrow \infty }x^{\frac{a_{n}b_{n}+1}{a_{n}b_{n}}}=x[/math]


    This is certainly true. Naïvely, we can just think about the fact that as an and bn approach infinity, the added 1 in the numerator makes a smaller and smaller contribution.

  12. Consider f(x) = x2, so f-1(x) = x1/2. Then f(x)f-1(x) = x5/2. Though the results of applying this sequence of steps quickly become annoyingly tedious to type completely, the general form of the derivative seems to go as follows (letting ri denote some real number):

    1: r1x3/2


    2: r2x7/6


    3: r3x43/42


    4: r4x1807/1806


    and so on. In general, starting with our original function and continuing with each derivative thereafter, we have a term xq where q is some rational number. The inverse involves a term x1/q, thus the product of the derivative and its inverse has a term x(q^2 + 1)/q. Thus the new derivative will have a term x(q^2 + 1)/q - q/q = x(q^2 - q + 1)/q. You may also notice a pattern here, namely that letting qn = an/bn be the exponent of x in the nth derivative, qn + 1 = (anbn + 1)/(anbn).


    Unless I've made some mistake here, then, no matter how many times we repeat the process, starting with x2, we will never end up with 2x.

  13. Alright. Continuing on, we have that a = a + b, so b = 0. In short, to reason that -1 = 2 from what you have above requires that b is nonzero, which we can't assume, and which is ultimately false.


    Edited because I originally misread your post.


    Edit 2: The array syntax only works for me when I have the entire thing on one line.

  14. The second equation assumes a = 40, which means all questions were answered correctly. Setting your two expressions equal to each other, we have


    [math]\begin{array}{rcl}a - \frac{b}{3} & = & 40 - \frac{b}{3} - (40 - (a + b)) \\ a &= & 40 - 40 + (a + b) \\ a & = & a + b\end{array}[/math]


    thus b = 0 as expected. I'm not sure what reasoning leads you to conclude from this that 2 = -1.

  15. [math]\Leftrightarrow \binom{n}{n-m}(n-m)!(n+1)=\binom{n+1}{n-m+1}(n-m+1)(n-m)![/math]


    [math]\Leftrightarrow \binom{n}{n-m}(n-m)!(n+1)=(n+1)\binom{n+1}{n-m+1}-m\binom{n+1}{n-m+1}[/math]

    You seem to have forgotten to factor out (n-m)! on the left side here, but I guess that's just a typo since you seem to continue as if you had factored it out.


    [math]\Leftrightarrow \left(\binom{n}{n-m}-\binom{n+1}{n-m+1}\right)(n+1)=-m\binom{n+1}{n-m+1}[/math]


    [math]\Leftrightarrow -\binom{n}{n-m}(n+1)=-m\binom{n+1}{n-m+1}[/math]

    I don't think these two equations are equivalent. Consider n = 2 and m = 1. Then [math] {2\choose1} - {3\choose2} = 2 - 3 = -1[/math] while [math]-{2\choose1} = -2[/math].


    Edit: And just as a general piece of advice, you might want to learn how to align equations in LaTeX (if you don't know already). Separating everything out made this OP somewhat hard to follow, especially on the mobile site where the different formula lengths meant the font size varied from line to line. It's not a huge deal, I guess, but for strings of equations this long, the added clarity would be nice.

  16. The notation is sometimes used interchangeably, but f is the function itself, while f(x) is the value of f when evaluated at x.


    So if we consider f(x) = x4 - 6, where x is a real number, then what we have is a function f : ℝ → ℝ defined by f(x) = x4 - 6.


    The question is asking us what formula defines f if f has these various properties. I guess the answer to both parts of your question is "yes," i.e. we're asked to find the equation defining f, and f will be defined by f(x).

  17. The added braces might be nice for clarity, I suppose.


    If we take f'(x) = f(x)2 to indicate that f' is a quadratic function, then we can no longer determine f uniquely from the information given. Also, polynomials are closed under integration, so the choice of the text authors to mention the interval (-∞, 1) seems odd, since if f is a polynomial, then it's differentiable on the entirety of (-∞, ∞).


    That is, of course, unless we take f and f' to be piecewise-defined, but that seems a bit convoluted for a problem I'm assuming is intended to help solidify the reader's knowledge of common derivatives and understanding of function transformations and translations.

  18. If the derivative of f(x) equals f(x)^2, does this mean that to find the original equation for the function, I will have to reverse differatiate (I think this is called integration in Calc 2 but I am nowhere near utilizing this yet)?


    The process is called antidifferentiation, and solving the differential equation would involve doing just that. However, in an introductory calculus class, it's probably better to just consider the derivatives you've already learned, and think about what sort of function has a derivative equal to its square (or at least, close to its square, since you'll have some decisions to make about certain terms).


    As an example, consider f(x) = 0. Then f'(x) = 0 = 02 = f(x)2. However, this doesn't work, since the exercise requires that f(0) = 1.


    Although the fact that f is differentiable on (-∞, 1) doesn't necessarily mean it's not differentiable on [1, ∞), the fact that they specify that interval is suggestive as well.

  19. What class is this?

    As for a hint, [math]f'(x) = f(x)^{2}[/math] is a separable differential equation.


    If the class doesn't assume knowledge of differential equations, then we may want to try another avenue.

  20. Ok, so I read that and it makes sense. This graph that is on the Multiplicative inverse kind of says that you can never get to zero.300px-Hyperbola_one_over_x.svg.pngWikipedia: http://en.wikipedia.org/wiki/Multiplicative_inverse

    And due to this graph, you can never get to 0, because you can't divide by zero. Instead, the number can get closer to zero, to a point were it is 1/infinity.

    While this particular figure is drawn well, be careful not to place too much trust in graphs and diagrams. They can be misleading.


    But, isn't there a way to prove that .9999...(repeating) is equal to 1.


    I think it goes something like this:






    It is true that 0.999... = 1 (in fact, every nonzero terminating decimal has a second representation ending in infinitely many 9's). I've seen some people object to this particular line of reasoning, but that's a minor point. There are many ways to show that the equality holds.


    Couldn't you do the same thing for zero, because it is on the other side. What I mean it to prove that 1 over infinity is 0.


    Well, again we must remember we're talking about the real numbers here, and since infinity is not a real number, an expression like 1/∞ isn't really valid. We can append ±∞ to the real numbers to obtain the extended real number line, but this is a new structure not equivalent to the field of real numbers. At the level we're talking about here, appending ±∞ is mainly useful to give us a shorthand notation for describing limits involving values that increase or decrease without bound. So while it's true that [math]\lim_{x \to \infty} \frac{1}{x} = 0[/math], it's incorrect to say that [math]\frac{1}{\infty} = 0[/math].

  21. In Letters to a Young Mathematician, Ian Stewart claims "...the working philosophy of most mathematicians is a mostly unexamined Platonist-Formalist hybrid."


    While I'm still a student and I would hesitate to really call myself a "mathematician" yet, I think this is where I fall as well, perhaps with a bias towards the platonist side of that coin. As my education continues, depending on how far down the rabbit hole my ultimate area of focus takes me, perhaps I'll become more concerned with the philosophical foundations of my work. Of course, I may just end up with a thoroughly examined platonist-formalist hybrid at the end. :P

    Besides not really being a mathematician, I also spend my time not being a philosopher of mathematics, but with my current understanding, I don't find the schools besides platonism and formalism very appealing or convincing (and even formalism loses some luster in light of the incompleteness theorems).

    All this is to say: I guess my answer is "Yes."


    Then how is zero a real number. If you can't get it, then what is it. Is zero even a real number. I have always been taught that it is on the number line, so it must be a real number. But because it can not be divided in any way, what is it? Why is it on the number line if it isn't a real number? Is it being used as a metaphor as something more? If zero is nothing, then what is it?


    Zero is indeed a real number. In particular, zero is the unique element that serves as the additive identity and multiplicative annihilator for the field of real numbers, i.e. for any real number x, we have that x + 0 = x and x * 0 = 0.


    I'm not sure what you mean by "if you can't get it." One can arrive at zero by subtracting a number from itself.


    Also, zero can be divided by any non-zero real number, and this will again yield zero.


    Division by zero is undefined because given a/0 = b for some a ≠ 0, there is no real value of b such that 0b = a.


    As discussed above, conceptually, we might think of a/0 as meaning either "how many times must we subtract 0 from a to arrive at 0" or "starting at 0, how many times must we add 0 to arrive at a." In either case, the answer cannot be a real number, as adding 0 to itself any real number of times still yields 0 and subtracting 0 from a any real number of times still yields a.


    Each of these also lends itself to conceptualization in terms of grouping or partitioning. How many groups of 0 objects must we collect to have a objects? Alternatively, given a objects, how many groups of 0 can we form if we must completely exhaust our collection? For any non-zero a, we run into the same issues as before.


    If we consider 0/0, then it turns out that every real number b satisfies 0b = 0. Therefore, instead of having no possible real values, 0/0 has infinitely many possible real values. This is why, as some posters mentioned before, we might call 0/0 "indeterminate" rather than (or perhaps in addition to) "undefined."


    Looking at the number line, including zero is part of achieving a very nice property of the real numbers, which is that the reals have no "gaps." This property is called "completeness," and without it, many important results (including, notably, much of calculus) are invalid.


    Alternatively, if we take zero to be a natural number (or even if we leave it until we've defined the integers), then by definition zero is a real number, as [math]\mathbb{N} \subset \mathbb{Z} \subset \mathbb{R}[/math].

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