Not sure what you mean by that. However one answer can be found it you consider dx as a small parameter and Taylor expand it to first order.
xdx=exp(dx*ln (x))=1+ln x dx
So the integral provided reduces to the integral of ln x, which is x ln x - x
This has some geometrical justification, because if you were to write the integral as a limit, it's only necessary to collect first order terms in dx, since higher order terms do not contribute. Interestingly, if we allow this, we can put dx "anywhere", as long as we can expand the function. Try to calculate the integral of ln(1+dx), for example.
The video makes a similar argument. It notes that the "dx" part is not merely some syntax for the integral, but in fact it represents a value that can be calculated with (more precisely, it represents the infinitesimally small width of the "columns" we are summing in order to get the are under the curve.) There is, in principle, no reason why this "width" value couldn't be somewhere else in the formula than as a multiplicand.
He proceeds to note that: xdx - 1 = ((xdx - 1) / dx) * dx, and proceeds from there to get the same result as you.
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p4wn3r wrote:
Try to calculate the integral of ln(1+dx), for example.
x+C
I've seen the video Warp is referring to. It uses the interperetation of dx as something that approaches zero. Using the same "multiply by dx/dx" approach as in the video, and applying the properties of logs, the term inside becomes (1+dx)^(1/dx). The expression equals e as dx approaches 0.
A hundred years from now, they will gaze upon my work and marvel at my skills but never know my name. And that will be good enough for me.
It turns out (see this article) that if f(x) and g(x) are analytic everywhere and not the zero function, with the limit as x--->0 of both functions being 0, then the limit of f^g must be 1.
By they way, the paper talks about f and g being real functions. Does that mean that the theorem does not hold for complex functions?
It turns out (see this article) that if f(x) and g(x) are analytic everywhere and not the zero function, with the limit as x--->0 of both functions being 0, then the limit of f^g must be 1.
By they way, the paper talks about f and g being real functions. Does that mean that the theorem does not hold for complex functions?
If the complex function is analytic, I don't think there isn't anything in the theorem that prohibits it. So I think it also applies for complex analytic functions.
Note that complex differentiation is more restrictive than real differentiation, in the sense that the following are all equivalent (such a function is termed "holomorphic"):
* complex differentiable (once)
* complex differentiable infinitely many times
* complex analytic
But there are a lot of complex functions; in particular, any function that does not satisfy the Cauchy-Riemann equations is necessarily non-holomorphic. So an example where the limit is not 1 (even when the functions are continuous at z=0) should not be too hard to come up with.
That gave me the idea that the theorem would work for any analytic function, not just ones representable as a closed-form expression. For example, if I understand correctly, the Riemann Zeta function is analytic, so it would follow that
lim(c->a) Zeta(c)g(c) = 1
lim(c->a) g(c)Zeta(c) = 1
and even
lim(c->a) Zeta(c)Zeta(c+someoffset) = 1
for any analytic g(x), and Zeta(a) = 0, g(a) = 0, and Zeta(a+someoffset) = 0.
And I don't even need to know how to raise Zeta to some power, or something to the power of Zeta, or even what Zeta is, as long as Zeta has zeros.
Prove:
If a triangle with integer side lengths has an area which is also an integer, then the area is a multiple of 6.
If a,b,c are the sides of the triangle, then Heron's formula gives the area as sqrt(s(s-a)(s-b)(s-c)), where s=(a+b+c)/2.
First, we show that s(s-a)(s-b)(s-c), and therefore the area, must be even. Suppose that s(s-a)(s-b)(s-c) is odd, which can only happen if all factors are odd. This can only occur when s=1 or 3 mod 4, and of a,b,c, either all of them are 2 mod 4 or one is 2 mod 4 and the other two are 0 mod 4. We make the following table of possibilities (up to reordering of a,b,c):
MOD 4
a b c | s s-a s-b s-c | Product
0 0 2 | 1 1 1 3 | 3
0 0 2 | 3 3 3 1 | 3
2 2 2 | 1 3 3 3 | 3
2 2 2 | 3 1 1 1 | 3
In these cases, s(s-a)(s-b)(s-c) is 3 mod 4, so s(s-a)(s-b)(s-c) cannot be a perfect square.
Second, we show that s(s-a)(s-b)(s-c), and therefore the area, must be a multiple of 3. Suppose that s(s-a)(s-b)(s-c) is not a multiple of 3. Then s is 1 or 2 mod 3 and a,b,c must not be the same mod 3 as s, but must be consistent with s=(a+b+c)/2 mod 3. The possibilities (up to reordering of a,b,c) are:
MOD 3
s a+b+c | a b c | s s-a s-b s-c | Product
1 2 | 0 0 2 | 1 1 1 2 | 2
2 1 | 0 0 1 | 2 2 2 1 | 2
In these cases, s(s-a)(s-b)(s-c) is 2 mod 3, so s(s-a)(s-b)(s-c) cannot be a perfect square.
So the area is even and divisible by 3, and so the area is divisible by 6.
Saw this somewhere. Combinatorics and number theory:
A bag contains 100 marbles, and each marble is one of three different colors. If you were to draw three marbles at random, the probability that you would get one of each color is exactly 20 percent.
How many marbles of each color are in the bag?
(If it helps, you may assume that the colors are red/green/blue, with #red>=#green>=#blue. Or not.)
Joined: 12/28/2013
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Location: Rio de Janeiro, Brasil
FractalFusion wrote:
Saw this somewhere. Combinatorics and number theory:
A bag contains 100 marbles, and each marble is one of three different colors. If you were to draw three marbles at random, the probability that you would get one of each color is exactly 20 percent.
How many marbles of each color are in the bag?
(If it helps, you may assume that the colors are red/green/blue, with #red>=#green>=#blue. Or not.)
There are C(100, x, y) [not sure if this notation is confusing - should read as 100!/(x! y! (100-x-y)!)] permutations of 100 marbles where x, y and z are the amounts of marbles of each color. Exactly 20% of such permutations start with one marble of each color, and then are followed by one of the C(97, x-1, y-1) permutations with x-1, y-1 and z-1 marbles of each color. Since there are 6 possible ways to arrange those first 3 marbles, we get:
C(100, x, y) = 6*5*C(97, x-1, y-1) -> xyz = 100*99*98/30 = 2^2 * 3 * 5 * 7^2 * 11
Now we need to split this product into 3 parts that sum up to 100. It requires a bit of trial and error: 7*3 + 7*5 + 2*2*11 = 21+35+44 = 100
Yes, that's right.
I got the same equations: Solve xyz= 22*3*5*72*11 and x+y+z=100.
To see how to get the solution, note that x,y,z are at most 48. This is because if any of them are 49, then the most that xyz can be is 49*25*26= 49*5*2*65 (less than 22*3*5*72*11 = 49*5*2*66), and higher numbers would result in even lower xyz (the numbers would be further apart which decreases the product).
So two of x,y,z are multiples of 7 (let's say x,y). Then the factor of 11 must go with z. Since x+y+z=2 mod 7, z must be a multiple of 11 that is 2 mod 7; the only possibility is z=44=11*22. The factor of 5 goes with one of x,y (let's say y), and so y=7*5=35 (can't be higher), and x=7*3=21. So the distribution of colors must be 44, 35, 21.
This problem was from yesterday's Riddler Express.
To see how to get the solution, note that x,y,z are at most 48. This is because if any of them are 49, then the most that xyz can be is 49*25*26= 49*5*2*65 (less than 22*3*5*72*11 = 49*5*2*66), and higher numbers would result in even lower xyz (the numbers would be further apart which decreases the product).
I'd like to add the information that this can also be used in the same way to calculate a lower bound of numbers being at least 21, since with 20 the maximum product would be 20*40*40 = 32000 < 32340.
Of course, that's very useless in this case, because with the upper bound of 48, you're already only left with three product options, all of which include integers lying in the interval [21,44].
Warning: Might glitch to creditsI will finish this ACE soon as possible
(or will I?)
I don't really get the bolded part, it is not obvious to me why that is what follows out of what came before, or why this specific way of rewriting the equation makes sense. Is that an easy thing to explain?
I don't really get the bolded part, it is not obvious to me why that is what follows out of what came before, or why this specific way of rewriting the equation makes sense. Is that an easy thing to explain?
Because 100-x-y = z, what the first equation actually tells us is:
So after cross multiplying and cancelling the factorials you're left with 100*99*98/30 at the LHS and xyz at the RHS
Saw this one on the internet and it was fun.
What are the last two digits of 7^7^7^7^7^7^7?
Note: of course if you do lots of computations the problem is pretty easy. Try to solve it in the style of math contests, finding some property that makes the solution come out easily.
Saw this one on the internet and it was fun.
What are the last two digits of 7^7^7^7^7^7^7?
Note: of course if you do lots of computations the problem is pretty easy. Try to solve it in the style of math contests, finding some property that makes the solution come out easily.
Do you mean 7^(7^(7^(7^(7^(7^7)))))?
(It would be easier to see this if it was typeset in proper math notation.)
I found by experimenting that the final two digits of powers of 7 cycle through four different values: 01, 07, 49 and 43. Therefore we need to reduce the exponent of the bottom 7 modulo 4. This can be done using an application of Euler's theorem. Basically in reducing numbers mod 4 you can reduce powers mod phi(4) = 2.
The exponent of the bottom 7 is itself a tower of 7s. Writing it as 7^(7^...^7), clearly the thing in brackets is odd, so 7^...^7 = 1 mod 2. Therefore, modulo 4 we have 7^(7^...^7) = 7^1 = 3. This was the power of the bottom 7 of the original number and 3 corresponds to an ending of 43.
I was thinking about egg-like shapes and their volumes, when I created this.
The graph below is the part of the elliptic curve y^3 - y = -3x^2 above the x-axis:
Rotate this graph around the y-axis to get a 3D shape that looks like an egg standing on end. What is the volume of this egg?
Ooo it's been awhile since I got to use my calculus muscles but I'll give this a shot.
My brain wants to use the x axis instead, so by swapping x and y the curve becomes: -3y^2 = x^3 - x
It was nice that the volume equation canceled out the square root.
More abstract nonsense. While studying some basic homological algebra, I ended up proving a splitting lemma. I haven't read it anywhere, but it definitely seems something that someone thought of before.
Let R be a ring with unit element distinct from 0. A (left) R-module is an abelian group A with an operation rA ∈ A, with r ∈ R and a ∈ A, satisfying the following properties with the abelian group operation +:
r(a1+a2) = ra1+ra2
(r1r2)a = r1(r2a)
(r1+r2)a = r1a+r2a
1a = a
Given two R-modules A and B, a module homomorphism is a function f: A -> B such that f(x+y) = f(x)+f(y) and f(rx)=rf(x) for x,y ∈ A and r ∈ R.
If we denote by 0 the module containing only the zero element, we can define the kernel of f, denoted by Ker(f), the set of all x ∈ A such that f(x)=0. It turns out that Ker(f) is a submodule of A, it might be a good idea to prove this if it's the first time you're seeing this.
We can also define the image of f as the set of all y ∈ B such that there is an x ∈ A satisfying f(x)=y. Im(f) is also a submodule of B, and it might be a good idea to prove this fact.
A (possibly infinite) sequence of homomorphisms, denoted by
... -> A0 -> A1 -> ... -> An -> ...,
is said to be exact if Im(An-1 -> An) = Im(An -> An+1), that is, the image of a homomorphism equals the kernel of the next. In particular, we are interested in short exact sequences:
0 -> A -> B -> C -> 0 (*)
The sequence (*) is said to split if B is isomorphic to A + C, where + denotes the direct sum of modules, defined in a similar way as the direct sum of groups.
Finally, we can state the exercise! Prove that, given the exact sequence (*), if one can "reverse the arrows", that is, find an exact sequence of the form
0 -> C -> B -> A -> 0,
then the sequence (*) splits.
Rotate this graph around the y-axis to get a 3D shape that looks like an egg standing on end. What is the volume of this egg?
