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May 29

Irrotational points of a vector field

Is there any general technique, in $\mathbb {R} ^{n}$ space, for finding irrotational points of a vector field? I should add that, in the case of my vector field, not all such points need be isolated.--Leon (talk) 07:02, 29 May 2013 (UTC)

There is a lot of work about detecting singularities of vector fields, a search will yield plenty of papers. You might be able to modify these algorithms for your purpose.--Salix (talk): 15:16, 29 May 2013 (UTC)

Don't have the time to look at it to see if I'm talking nonsense, but a look at Geometrical Methods of Nonlinear Analysis (1984) by Krasnoselʹskiĭ & Zabreĭko, Volume 263 of Springer's Grundlehren might help.John Z (talk) 10:08, 4 June 2013 (UTC)

May 30

Generalized totient function

Is there such a thing as a generalized totient function? Specifically, a function φ_i(n) that is the count of positive integers less than n with exactly i prime divisors. This means that Euler's totient function φ(n) is φ₁(n) in the more general form (i.e., it is the number of positive integers less than n with only one prime divisor). Likewise, φ₂(n) is the number of positive composite integers less than n with only two prime divisors, e.g., numbers from the set {4,6,9,10,14,15,21,22,25,...,n}, which includes all the squares of primes. φ₃(n) includes {8,12,18,27,28,30,...,n}, and so forth. I dimly recall seeing something about Ramanujan studying something similar to this(?). Perhaps such a thing might also be related to the Riemann hypothesis? — Loadmaster (talk) 17:24, 30 May 2013 (UTC)

Do you mean prime counting function rather than totient function? Otherwise I'm very confused. Sławomir Biały (talk) 17:29, 30 May 2013 (UTC)

Yes, that's what I meant. So (replacing φ above with π): Is there such a thing as a generalized prime counting function? Specifically, a function π_i(n) that is the count of positive integers less than n with exactly i prime divisors. — Loadmaster (talk) 18:48, 30 May 2013 (UTC)

Yes, but these can be expressed in terms of the functions pi[n^(1/i)] using Mobius inversion. Count Iblis (talk) 19:54, 30 May 2013 (UTC)

Elliptic integrals and Gaussian integrals

With your help, I would like to find out a mathematical relationship between complete elliptic integrals of the first kind

$K(n)\ =\ \int _{0}^{\pi \over 2}{\frac {d\theta }{\sqrt {1-n^{2}\sin ^{2}\theta }}}\ =\ \int _{0}^{1}{\frac {dt}{\sqrt {(1-t^{2})(1-n^{2}t^{2})}}}$

and gaussian integrals

$G(n)\ =\ \int _{0}^{\infty }{e^{-x^{n}}}dx$

all of which are known to possess the following property

${\frac {G(m)\cdot G(n)}{G({m\ \cdot \ n \over m\ +\ n})}}\ =\ \int _{0}^{1}{\sqrt[{m}]{1-x^{n}}}dx\ =\ \int _{0}^{1}{\sqrt[{n}]{1-x^{m}}}dx$

where

${\tfrac {m\ n}{m+n}}$ is half of the harmonic mean between m and n, and the entire above expression is equal to the product between 1 + ${\tfrac {1}{m}}$ + ${\tfrac {1}{n}}$ and the beta function of arguments 1 + ${\tfrac {1}{m}}$ and 1 + ${\tfrac {1}{n}}$ .

It also goes on without saying that the factorial of every positive number is the gaussian integral of its reciprocal or multiplicative inverse

$n!\ =\ G({\tfrac {1}{n}})\ =\ \int _{0}^{\infty }{e^{-{\sqrt[{n}]{x}}}}dx$

— 79.118.171.165 (talk) 18:33, 30 May 2013 (UTC)

Resolved

 – — See here for more information.

May 31

Linear independence of partial derivatives

Consider the general solution y(x,c₁,c₂) of a 2nd-order linear homogeneous ODE, where c₁ and c₂ are two parameters, and y is assumed smooth in x, c₁, and c₂. Consider the partial derivatives

${\frac {\partial y}{\partial c_{1}}}(x,c_{1},c_{2}),\quad {\frac {\partial y}{\partial c_{2}}}(x,c_{1},c_{2})$

Now fix some (c₁,c₂) = (k₁,k₂). Is is true that the functions of x defined by

$u_{1}(x)={\frac {\partial y}{\partial c_{1}}}(x,k_{1},k_{2}),\quad u_{2}(x)={\frac {\partial y}{\partial c_{2}}}(x,k_{1},k_{2})$

are linearly independent? If not, what extra assumptions are necessary? Does it depend on the choice of (k₁,k₂), and if so, in what way? Thanks for any help. —Anonymous Dissident^Talk 00:40, 31 May 2013 (UTC)

A condition is clearly needed to ensure that c1 and c2 are independendent parameters (they could otherwise be the same parameter, for instance). Then the answer would presumably be yes, just by definition of independence of the parameters. Sławomir Biały (talk) 13:22, 31 May 2013 (UTC)

Two questions related to sphere packing...

These are both questions just to start getting my head around sphere packing.

1. Let A and B both be n dimensional balls. Color any point on the surface of A if it is between the center of A and any point of B. For dimension n, What fraction of the surface of A is colored. (The answer for 2 dimensions is exactly 1/6.
2. Let A be an n dimensional ball and let B₁, B₂, ...B_n additional n dimensional balls with the centers making an n-dimensional simplex. Mark the n points (p_n on the surface of A where the lines from the center of A to B_i touch the surface of A. Now connect all p_n the "short way" round to get a n-1 dimensional simplex on the surface of A. Color the surface inside A if it is between the center of A and any point in any of the B_i. For each dimension n, what fraction of the area of the triangle is *uncolored*.

For the three dimensional version of this, the first question is what percentage of the surface area of the sphere is "covered" by a touching sphere. The second question is if you have three spheres all touching each other and the main sphere, how much of a "hole" is there between them.Naraht (talk) 00:41, 31 May 2013 (UTC)

Borsuk's question

In recent edits two anonymous users added an information about new bound for the Borsuk's problem [1]. However it is not clear from the linked abstract, that the claimed bound is true: is says 'We found a two-distance set consisting of 416 points on the unit sphere in the dimension 65 which cannot be partitioned into 83 parts of smaller diameter. This also reduces the smallest dimension in which Borsuk's conjecture is known to be false.', however the Borsuk's question concerns dividing sets in d–space into (d+1) parts. Bondarenko says his set can not be partitioned into 83 subsets, but that does not obviously imply it can't be partitioned into 66 subsets of required size.

Additionally he says 'This reduces the smallest dimension...', but it's not clear in what manner it reduces that dimension (possibly the two dimensions mentioned are related somehow, but not necessarily equal). Anybody has access to the full text, and can verify the 65 is an actual new limit for Borsuk's question, please? --CiaPan (talk) 06:24, 31 May 2013 (UTC)

Is it correct to equate the 2nd equation?

Consider the following two equations-
A) 2^x * 2^y = 2² * 2⁴ From this equation, we can simply write x = 2 and y = 4.
B) 2^x + 2^y = 2² + 2⁴ From this equation, can I write x = 2 and y = 4?
I know, I am right for my 1st equation, here, I just equated LHS to RHS. In the second case also I did the same thing, but I am not sure whether I am right or wrong. So, correct me if I am wrong. Scientist456 (talk) 15:57, 31 May 2013 (UTC)

x=2 and y=4 are *one* solution to the second equation, but they are by no means the only ones. An additional answer to both equations is x=4 and y=2. For the first you can also have x=3 and y=3 (and x=1, y=5, etc.). The second is harder to come up with alternative integer solutions to, but there are plenty of non-integer solutions, e.g. x=3 and y=ln(2.5)/ln(2) y=ln(12)/ln(2) (correct thinko). To be equal, the two sides have to be the same, either before or after algebraic manipulation. If you substitute x=2 and y=4 in both equations, you'll end up with a true statement (both sides are the same), but that can also happen with a number of other values for x and y. -- 71.35.97.37 (talk) 16:13, 31 May 2013 (UTC)

Actually, I was confused in this problem- 3^x + 7^y = 3² + 7⁴
I had to find the value of x and y. Here, can I write x = 2 and y = 4 by equating LHS to RHS. Scientist456 (talk) 17:04, 31 May 2013 (UTC)

Did you make up this problem or does it come from some source? If it comes from some source, there is a high probability that you have left out important information. Looie496 (talk) 17:14, 31 May 2013 (UTC)

• There are an infinite number of possible solutions to both equations, unless perhaps you are restricted to integers only, but even so, there's likely no single solution. For example, if x=0 in the second equation, then y=log₇2409.--Jasper Deng (talk) 17:18, 31 May 2013 (UTC)

Choose x yourself and then compute y to satisfy your equation.
A) 2^x * 2^y = 2² * 2⁴ From this equation write y = log((2² * 2⁴) / 2^x)/log(2)
B) 2^x + 2^y = 2² + 2⁴ From this equation write y = log((2² + 2⁴) - 2^x)/log(2)
Bo Jacoby (talk) 13:35, 3 June 2013 (UTC).

June 1

Question regarding a 4 x 4 sliding tiles puzzle

When I was young, I had a metal trinket from who knows where. It was a tiles grid, with 16 squares (though one was empty, so you could slide the tiles around.) The squares had the numbers from 1 to 15 on them, and in solved form you would get a grid that looked like this:

1 . 2 . 3 . 4

5 . 6 . 7 . 8

9 .10 11 12

13 14 15 __

Of course, the point was to mix the tiles up and work on getting them back into this solved state.

However I once read something that stated that "if, from the solved state, you switched the 14 and the 15 tiles, the puzzle would become unsolveable."

So, my two questions:

1) Is that true, that such a simple change makes the puzzle unsolveable?

2) Assuming 1 is yes, what if I simplified the puzzle to a 3 x 3 grid, then switched tiles 8 and 9. Would the puzzle then be likewise unsolveable? 169.231.8.137 (talk) 06:28, 1 June 2013 (UTC)

Yes the puzzle is unsolvable from half the starting positions, and the 3 x 3 can always be solved. See 15 puzzle.--Salix (talk): 06:36, 1 June 2013 (UTC)

Ooh, thank you! 169.231.8.137 (talk) 06:41, 1 June 2013 (UTC)

No there's the same rule for 3x3 or any size, if you swap the two tiles it can't be solved. Dmcq (talk) 15:08, 1 June 2013 (UTC)

Wikipedia has the article "15 puzzle".—Wavelength (talk) 16:27, 1 June 2013 (UTC)

Permutations

I am attempting to determine the number of possible combinations of a given set. For the sake of simplicity, the 6 items are A,B,C,D,E, and F. Only 3 of these items can be used in a combination at once. Each letter may be repeated in a combination any number of times up to 3. The order of the letters is irrelevant. For instance, ABB is the same as BAB and BBA. How can I determine the number of unique combinations, taking into consideration the irrelevance of the order of the items mentioned previously?CalamusFortis 07:22, 1 June 2013 (UTC)

First, let's recast this as an equivalent problem: You have 6 buckets, labeled A, B, C, D, E and F. You have 3 identical balls. How many ways can you distribute the three balls among the 6 buckets? Hopefully, you agree that this is equivalent.

Now let's recast this again: How many different sequences are there using 3 stars (*) and 5 bars (|)? Each sequence codes an assignment of balls into buckets, with stars representing balls and bars representing the boundaries between buckets. Buckets are listed in order. So *|*|*||| means 1 ball in bucket A, 1 ball in bucket B and 1 ball in bucket C. ||*|||** means 1 ball in bucket C and 2 in bucket F. Etc.

This is straightforward to calculate: there are 8! sequences of 8 characters, but permuting the 3 stars doesn't generate a new sequence, so divide by 3!. Similarly, permuting the bars doesn't generate a new sequence, so also divide by 5!. 8!/(3! 5!) = 56.--80.109.106.49 (talk) 13:49, 1 June 2013 (UTC)

Your problem is treated here: Multiset#Counting_multisets. Bo Jacoby (talk) 06:53, 3 June 2013 (UTC).

Question regarding a congruence and the multiplicative order

I am investigating the congruence 2ⁿ ≡ 1 (mod (n + 1)²). I try to prove that if $n\in \mathbb {N}$ is a solution to this congruence, then n is a multiple of ord_{(n + 1)²}(2). How could this be done (at least, what strategy could one pursue to do this)? I appreciate any hints. -- Toshio Yamaguchi 11:32, 1 June 2013 (UTC)

Unless I'm misunderstanding your question, this follows immediately from the definition of order; in general, for any element g of a group if m is the smallest integer so g^m = 1 and gⁿ = 1, then m | n; you can take a look at out article on the order of group elements. So if it has a solution, then the order divides it.Phoenixia1177 (talk) 00:10, 2 June 2013 (UTC)

Thanks for the reply. So if I understand correctly, for a specific m = (n + 1)², the powers of two modulo m form a cyclic group, because the residues of 2^k (with k = 1, 2, 3, 4, ...) modulo m can only take on a finite number (say x) of values for that m. And a specific residue will "reoccur" with period x for increasing k. -- Toshio Yamaguchi 09:04, 2 June 2013 (UTC)

If 2 is coprime to m, then yes; in your case, if there is a solution, then (n + 1)² | 2ⁿ - 1, if n + 1 where even this couldn't occur. Hence, n + 1 and 2 are coprime. If you look at our article Bézout's identity, you'll see that if d is the greatest divisor of a and b, then there are x and y so ax + by = d, so ax = d mod b. If d is 1, meaning a and b are coprime, then there is x so ax = 1 mod b, so a is in the group of units mod b. On the other hand, if a is in the group of units mod b, then there is c so ac = 1 mod b and we can write ac = by + 1, for some y. Then, ac - by = 1, from the same article you can see this means a and b are coprime. So, elaborating on what you wrote above, a^x = 1 mod b has a solution x iff a is a unit mod b iff a and b are coprime.Phoenixia1177 (talk) 20:44, 2 June 2013 (UTC)

On the subject of your problem, n can never be the order unless n + 1 is prime, it must be strictly larger. Write phi(k) for the totient function, by a quick generalization of Fermat's Little Theorem a^phi(k) = 1 mod k for all k and a coprime k. Then, if n is the order of 2 mod (n + 1)², n | phi((n + 1)²). Let n + 1 have prime factorization
p₁^a₁...p_s^a_s; phi((n + 1)²) = ((n + 1)² /p₁...p_s) * (-1 + p₁)...(-1 + p_s). Since n and n + 1 are coprime, n | t = (-1 + p₁)...(-1 + p_s), but n = p₁^a₁...p_s^a_s - 1; however, unless s = 1 and a₁ = 1, then t < n implying not n | t.Phoenixia1177 (talk) 23:21, 2 June 2013 (UTC)

Have you found any such n? I didn't see any for n < 18 (then I stopped checking), so I'm guessing this isn't an accidental pattern you noticed given that 2 ** n get's pretty big at that point. Have you found any solutions? For some reason, I get the feeling they don't exist or are rare, but haven't really looked into it.Phoenixia1177 (talk) 09:56, 3 June 2013 (UTC)

Indeed. If 2ⁿ ≡ 1 mod (n + 1)² then n+1 is a Wieferich prime. Only two Wieferich primes are known: 1093 and 3511. Gandalf61 (talk) 10:51, 3 June 2013 (UTC)

There are no solutions for n < 1000, as Gandalf says. The first solution is 1092, which is due to the fact that 1093 is a Wieferich prime. It is easy to see that if an odd prime p is a Wieferich prime (i.e. satisfies 2^p-1 ≡ 1 (mod p²)), then the exponent p-1 (1092 in this case) is a solution to the congruence 2ⁿ ≡ 1 (mod (n + 1)²). These are exactly the cases where n is a solution such that n + 1 is prime, and according to the latest result from the ongoing Wieferich prime search here, there are no others up to approximately 10¹⁷. However, I suspect that there might be cases where n + 1 is composite. If such cases exist, they should be EXTREMELY rare, because in that case, m = n + 1 must be a Fermat pseudoprime to base 2 having two additional properties:

• m must satisfy 2^m-1 ≡ 1 (mod m²) and
• all prime factors of m must be Wieferich primes

One could download the file annotated-psps-below-2-to-64.txt.bz2 from here and check, whether there are entries whose only prime factors are 1093 or 3511. The only such numbers I am aware of are 1093² and 3511², but it seems extremely unlikely to me that either (1093² - 1) or (3511² - 1) satisfies 2ⁿ ≡ 1 (mod (n + 1)²). Suppose that were the case: It means we have (for (1093² - 1)) 2^10932-1 ≡ 1 (mod (1093²)²), ie. 2^10932-1 ≡ 1 (mod (1093⁴)). According to theorem 5 of [2], 2^p2-1 ≡ 1 (mod p²) if and only if 2^p-1 ≡ 1 (mod p²), so if say (1093² - 1) satisfied 2ⁿ ≡ 1 (mod (n + 1)²), 1093 would have to satisfy 2^p-1 ≡ 1 (mod p⁴). So I guess the only numbers that could reasonably be expected to be solutions such that n + 1 is composite would be base 2 pseudprimes numbers n such that n + 1 is a base 2 pseudoprime all of whose prime factors are distinct Wieferich primes. No such number seems to exist below 10¹⁵. -- Toshio Yamaguchi 11:07, 3 June 2013 (UTC)

Indeed, in PARI

Mod(2,1194648)^(1194649^2%eulerphi(1194648))

gives Mod(298664, 1194648) and

Mod(2,12327120)^(12327121^2%eulerphi(12327120))

gives Mod(3106352, 12327120)

confirming that neither (1093² - 1), nor (3511² - 1) satisfy 2ⁿ ≡ 1 (mod (n + 1)²). -- Toshio Yamaguchi 12:03, 3 June 2013 (UTC)

Circulation

I have been looking into hyper-operators and indeterminate forms and have found several references in PDF's and such to something known as circulation. Circulation, by several articles I have come across is defined as;

$a\uparrow ^{\infty }b$

But I cannot seem to find any real information on the subject. Does anyone know anything about circulation? Who came up with the concept? Has any work been done on the convergence of circulated functions? — Preceding unsigned comment added by 109.153.168.240 (talk) 17:18, 1 June 2013 (UTC)

Note: I have attempted to fix your broken syntax -- Wikipedia's "math" system does not allow special characters such as "∞" to be mixed in with math code. Hopefully the thing I changed it to is what you intended. Looie496 (talk) 22:42, 2 June 2013 (UTC)

Are you asking about what that notation means, or about what this circulation is? I don't know about the latter, but you might get a sense of the operation at Knuth's up-arrow notation. SemanticMantis (talk) 17:13, 3 June 2013 (UTC)

I can't find any definitions like that at a glance. If you give us the reference to at least one paper that has this usage, we could probably help you track down the origin and other work. SemanticMantis (talk) 17:17, 3 June 2013 (UTC)

Well thats the problem. I found several inconsistent references but I couldn't find any of them elsewhere. I understand the notation but I can't find any actual information on it. Also cheers for fixing it! I tried several times but just gave up in the end!

June 2

Optimization problem

Imagine you have a rectangular piece of fabric of dimension 3x4. I want to cut a circular piece out, and obviously the largest diameter possible is 3. But if I want a larger circle I can make a straight cut through the rectangle and then sew the pieces back together in any orientation and position (the pieces cannot be joined at points and cannot overlap). Now with this 1 cut and sewing operation what is the largest circle I can make? What about with n (finite) operations? 24.255.30.187 (talk) 05:44, 2 June 2013 (UTC)

Intuitively, I'd expect the best config for one cut to be like so (forgive the crude pic):

+        +------+
|\      /___    |
| \____//   \   |
|      /     \  |
+-----+       +-+

That is, the two semi-circles should be cut out from opposite sides, and be tangent to each other. StuRat (talk) 06:05, 2 June 2013 (UTC)

Each cut has to be a single straight line. 24.255.30.187 (talk) 06:08, 2 June 2013 (UTC)

The trivial upper bound for the diameter is ${\sqrt {2\times 3\times {4 \over \pi }}}$ which is 3.90882 . Bo Jacoby (talk) 06:16, 2 June 2013 (UTC).

Two ways to cut the cloth

You can do StuRat solution with a single cut by cutting along the tangent of the two circles this gives a radius of approximately 1.56. Cutting along the diameter of the circle gives a circle with a radius of about 1.75.--Salix (talk): 08:42, 2 June 2013 (UTC)

How do you get 1.56? The diagonal of the rectangle (length 5) should pass through the point of tangency, so the radius should be 1.25.--80.109.106.49 (talk) 09:10, 2 June 2013 (UTC)

I've got my numbers experimentally. For the first case if the radius is r and the angle of the tanget is θ then for the semi circles to fit we require $2r+2r\sin \theta =4$ and $2r\cos \theta =3$. Getting expression for cos and sin and using the pythagorean identity gives an equation in r with exact solution 25/16 ≈ 1.5625. The second is harder to give an explicit solution for, an aproximate solution has an angle of 45.4334° and r ≈ 1.75189.--Salix (talk): 10:31, 2 June 2013 (UTC)

I just realized my mistake. The diagonal isn't a diameter.--80.109.106.49 (talk) 13:06, 2 June 2013 (UTC)

R=1.75

Here is r=1.75
However, if the grey stripe was a trapezoid or even a triangle instead of a rectangle, the circle might get a bit bigger. --CiaPan (talk) 09:09, 3 June 2013 (UTC)

Should Bo Jacoby's upper bound not be ${\sqrt {4\times 3\times {4 \over \pi }}}$? A quick back of the envelope estimate (by assuming that you can make the whole area of the sheet into the circle) puts that at $4\times {\sqrt {3 \over \pi }}$, which seems to match the numerical value given MChesterMC (talk) 15:13, 3 June 2013 (UTC)

You are right. Thanks. Bo Jacoby (talk) 19:45, 3 June 2013 (UTC).

June 4

What topics in mathematics should I read to understand the article Einstein field equations?

I read the article Einstein field equations and understood everything except the mathematical equations. So, my question is - What topics in mathematics should I read to understand these equations? Once I was suggested a nice link by someone. Suggest me some useful links (like the previous one) related to those mathematical topics. Thank you! Concepts of Physics (talk) 12:29, 4 June 2013 (UTC)

There is mathematics of general relativity for a start. Sławomir Biały (talk) 12:35, 4 June 2013 (UTC)

In addition to Sławomir's recommendation, there is a wikibook on general relativity that while incomplete, has some math sections that may be worth checking out. Outside of the wikiverse, Bernard F. Schutz' book A First Course in General Relativity is an approachable book for self study and this site has a number of links to online resources for learning GR. Good luck, it is a beautiful subject. --Mark viking (talk) 22:17, 4 June 2013 (UTC)

To solve a simple mathematical equation, can we take derivative or integral of both sides and then solve the equation?

I have an equation like this one: 2x²*31 = 5x + 50 . I can add, subtract, multiply and divide any number on both sides of the equation. I can also square (or take log of) both sides and then solve the problem. My question is - Is it correct to take derivative or integral (as in the process of squaring) of both sides and then solve the problem? Concepts of Physics (talk) 13:23, 4 June 2013 (UTC)

No, because the equality is only valid of specific values of x. If you integrate for differentiate, you would use that the equation is an identity that is valid for all values of x (or at least for x in some interval). Count Iblis (talk) 13:30, 4 June 2013 (UTC)

The original question is equivalent to asking for the value of x where the graph of y = 2x²*31 meets the graph of y = 5x + 50. Taking the derivatives and comparing is equivalent to asking for the value of x where the two graphs have the same slope, which is unlikely to give the same answer (in fact only at points where the graphs are tangents to each other). AndrewWTaylor (talk) 13:37, 4 June 2013 (UTC)

And sometimes it may not happen at all. For a simple equation x = 1, which is solved by the only real number, namely by x = 1, differentiating with respect to x gives 1 = 0, which is not satisfied by any real number. On the other hand x = x+1 has no solution, while (x)' = (x+1)' is equivalent to 1 = 1, which is solved by every real x. --CiaPan (talk) 05:22, 5 June 2013 (UTC)

There are lots of things you can't do to both sides of an equation and preserve equality. For instance, you can't add one to the x variable on both sides. It's not something special about calculus. Sławomir Biały (talk) 15:42, 4 June 2013 (UTC)

When you "do an operation" to both sides of an equation, you are applying a function. One of the defining properties of a function, say f, is that if a and b are equal, then f(a) and f(b) are equal. The reason you can't differentiate the equation you have above is an issue of domains, things like "multiplying by 2" are functions of the real numbers, things like "take the derivative" are functions of differentiable functions. Your equation works with numbers, not differentiable functions. Best case scenario, you could argue that you are working with constant functions, but in that case both sides have a derivative of 0, which while true, is not helpful.Phoenixia1177 (talk) 07:26, 5 June 2013 (UTC)

Indeed. You can, however, differentiate or integrate both sides of a functional equation, in which the two sides of the equation are equal as functions throughout the whole of a domain, not just at a single point. For example, if you are told that

$f'(x)=f(x)$

for all real x and that f is infinitely differentiable, you can conclude that

$f''(x)=f'(x)$

and by induction, that

$f^{(n)}(x)=f(x)\quad \forall \,x\in \mathbb {N}$

and thus

$f(x)=f(0)(1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+\cdots )=f(0)e^{x}$

Gandalf61 (talk) 08:08, 5 June 2013 (UTC)

Deriving the vector equation of the intersection of two planes

I'm taking a calculus class and I didn't catch the entire example of how to find the line at the intersection of two planes in three-dimensional space. Say I'm given these planes:

$\pi _{1}\equiv A_{1}\cdot x+B_{1}\cdot y+C_{1}\cdot z+D_{1}=0$

$\pi _{2}\equiv A_{2}\cdot x+B_{2}\cdot y+C_{2}\cdot z+D_{2}=0$

How do I determine the equation of the line in the form of ${\vec {r}}={\vec {r_{0}}}+t\cdot {\vec {d}}$? --Melab±1 ☎ 00:58, 5 June 2013 (UTC)

I have fixed your math code -- \\ does not work in math mode. Looie496 (talk) 01:32, 5 June 2013 (UTC)

Plane (geometry)#Line of intersection between two planes gives a fairly good explanation.--Salix (talk): 01:49, 5 June 2013 (UTC)

Convergence of an Infinite Series

I'm having trouble with a past exam question which is whether $\sum _{n=1}^{\infty }{\frac {cos(nx)}{\sqrt {n!}}}$ converges? — Preceding unsigned comment added by 211.31.22.140 (talk) 10:01, 5 June 2013 (UTC)

If a sum is absolutely convergent, then it is convergent. Since $\left|cos(nx)\right|\leq 1$ for all x and n, then by the limit comparison test your series converges if $\sum _{n=1}^{\infty }{\frac {1}{\sqrt {n!}}}$ does. Clearly, ${\frac {1}{\sqrt {n!}}}\leq {\frac {1}{2^{n}}}$ for large n, so by the same test, since $\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}$ converges, your series does. Phoenixia1177 (talk) 10:44, 5 June 2013 (UTC)

Continuity and Limits

Thankyou! I was also having trouble with proving that if f and f' are continuous on the reals, that if they both have finite limits at infinity, this limits the limit of f' is zero and what the limit of f would be? — Preceding unsigned comment added by 211.31.22.140 (talk) 11:15, 5 June 2013 (UTC)

Consider the function $f(x)=a+{\frac {b}{x}}$. The limit of f is a, which can be chosen to anything. Bo Jacoby (talk) 11:38, 5 June 2013 (UTC).

Thankyou Bo, that answers half the question nicely :) But what about proving that the limit of f' is always 0? — Preceding unsigned comment added by 211.31.22.140 (talk) 12:09, 5 June 2013 (UTC)

Apply the mean value theorem to $\lim _{x\to \infty }(f(x+1)-f(x))$. On the one hand this limit is zero (since f has a limit at infinity) and on the other hand it's equal to the limit of $f'$ at infinity. Sławomir Biały (talk) 13:35, 5 June 2013 (UTC)

Can anybody help me ?

I posted a question last week about finding a relationship between elliptic integrals and the factorial/gamma function, to which I got no reply whatsoever. Did you guys miss by any chance my question, or is there truly nobody out there able to help me ? :-( — 79.113.220.191 (talk) 13:46, 5 June 2013 (UTC)