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August 14

Imaginary Fibonacci

What is the Fibonacci root of i? That is, what put into the function ((1+sqrt 5)/2)^(x)-((-1)^(x)/((1+sqrt 5)/2)^(x))))/sqrt 5 = i? Robo37 (talk) 10:57, 14 August 2011 (UTC)

So I think you're asking us to solve $F_{x}=i$, where $F_{x}$ is the expression for the Fibonacci series, which you have slightly wrong, $F_{x}={\frac {({\frac {1+{\sqrt {5}}}{2}})^{x}-({\frac {1-{\sqrt {5}}}{2}})^{x}}{\sqrt {5}}}$.

This is not an easy problem. --COVIZAPIBETEFOKY (talk) 12:56, 14 August 2011 (UTC)

Does Wolfram Alpha's answer make sense? (I was using COVIZAPIBETEFOKY's version of your question – but Wolfram's has several different things about it). Grandiose (me, talk, contribs) 13:43, 14 August 2011 (UTC)

Thanks a lot for that, I really appreciate the help. Yeah I got my equation from the second equation listed at [1] because when I tried the first one with real integars in Google calculator I got incorrect resaults for some reason while the second one got 1 when I entered 1, 1 when I entered 2, 2 when I entered 3, 3 when I entered 4, 5 when entered 5, 8 when I entered 6... ect. Now there's something else that baffles me, howcome when I enter those answers you gave into my function I don't get i? Could you possibly calulate the same thing using my equation, out of curiousity? Thanks again. Robo37 (talk) 14:26, 14 August 2011 (UTC)

Wait, I've got it. Thanks for linking me to that program, it looks like it could come in really handy. I still don't understand why both forumla's work differently to each other in the complex plane but the same in the real plane but that's not a burning issue of mine. Thanks a lot for the link. Robo37 (talk) 14:44, 14 August 2011 (UTC)

The problem with trying to generalise Binet's formula

$F_{x}={\frac {({\frac {1+{\sqrt {5}}}{2}})^{x}-({\frac {1-{\sqrt {5}}}{2}})^{x}}{\sqrt {5}}}$

to non-integer values of x is that ${\frac {1-{\sqrt {5}}}{2}}$ is negative, so its xth power is not well defined for non-integer x (see exponentiation). Even if you extract a factor of -1 to get

$F_{x}={\frac {({\frac {1+{\sqrt {5}}}{2}})^{x}-(-1)^{x}({\frac {-1+{\sqrt {5}}}{2}})^{x}}{\sqrt {5}}}$

(which is what I think you are attempting to do) then you have the same problem because $(-1)^{x}$ is not well defined for non-integer x. You have two possible values for $(-1)^{\frac {1}{2}}$, three possible values for $(-1)^{\frac {1}{3}}$ etc. If you allow x to take irrational or complex values the problem becomes even worse - the set of possible values becomes infinite. Gandalf61 (talk) 14:50, 14 August 2011 (UTC)

The equation is

$e^{ax}+e^{i\pi bx}-i{\sqrt {5}}=0\,$

where

$a=\log(1+{\sqrt {5}})-\log 2\approx 0.481212$

and

$b=\log(-1+{\sqrt {5}})-\log 2=-a.$

Truncate the taylor series of the exponentials, and solve the resulting algebraic equation numerically. Bo Jacoby (talk) 10:38, 15 August 2011 (UTC).

That gives one solution - but my point is that selecting $e^{i\pi x}$ as a value of $(-1)^{x}$ is arbitrary - why not use $e^{-i\pi x}$ or $e^{3i\pi x}$ etc. For almost all non-integer values of x you have an infinite number of possible values for $(-1)^{x}$. Therefore there are (almost certainly) an infinite number of "Fibonacci roots" of i, not just one. Gandalf61 (talk) 16:23, 15 August 2011 (UTC)

The equation

$f(x)=0\,$

where

$f(x)=e^{ax}+e^{-i\pi ax}-i{\sqrt {5}}\,$

has an infinite number of solutions. Not just one solution. $f\,$ is like a polynomial of infinite order. There has been taking care of your point. Bo Jacoby (talk) 17:57, 15 August 2011 (UTC).

The solution −1.206+0.512i was computed in J like this.

  a=.^.-:>:%:5
  f=.^@(a&*)+^@((-j.o.a)&*)-(j.%:5)"_
  {:>{:p.f t.i.14
_1.20636j0.51203

Bo Jacoby (talk) 11:18, 16 August 2011 (UTC).

August 15

"Most"

Can the statement, "Most [partial / nonlinear / etc] differential equations have no analytic solution" be quantified in any way? Obviously it is true, but it seems to come up a lot in textbooks without any further detail. 166.250.65.167 (talk) 00:44, 15 August 2011 (UTC)

No answers for a while, so I'll give one that's only tangentially related. There is the Risch algorithm which tells you which elementary functions have elementary antiderivatives. Once you focus on the class of elementary functions, there are various ways in which you could rigorously define "most", and then (for example) prove that "most" elementary functions do not have elementary antiderivatives (I don't know if that's true actually, or if it's been done). Some similar thing could in theory be done for general DEs, though it would have to be much more complicated. Staecker (talk) 23:29, 16 August 2011 (UTC)

factoring integers

Hi. I have a number 'n' and know its prime factorization. Everything is integers. I seek 'a', 'b' with 'n=ab, a >= b' such that $a/b$ is as close to one as possible. For example, if $n=4896=2^{5}3^{2}17$, I would write $a=2^{3}3^{2}=72$ and $b=2^{2}17=68$, giving a ratio of $72/68=1.0588$. I don't think it's possible to do better. Is this a special case of a more general problem? Are any efficient techniques known for this problem? The numbers I am considering are of the order of a googol Robinh (talk) 01:53, 15 August 2011 (UTC)

Here's one basic method:

              _
S = truncate(√N)
for I = S to 2, step -1
  if N/I is an integer, then
    solution = I and N/I
    stop program
  end
end
N is prime, so no solution found

StuRat (talk) 03:24, 15 August 2011 (UTC)

OK, thanks StuRat. Your method works, but doesn't "use" the prime factorization of 'N'. If 'N' is a googol to within an order of magnitude, and the best answer has a/b=1.1, say, then you'll need to check about $5\times 10^{48}$ suggestions. And, taking a googol minus 10 as a representative example, this has 16 factors counting multiplicity, which would give me just $2^{15}=32768$ options to check. I am just wondering if there is a "nice" way to check these 32768 options, as I can't believe that this problem isn't a commonly encountered one. Thanks again, Robinh (talk) 04:27, 15 August 2011 (UTC)

If you consider the logs of the primes you end up with a problem similar to the partition problem, except that your values aren't integers. According to the article the partition problem is NP-complete so probably there's not a polynomial time algorithm to find the optimum for this problem either. Maybe some of the approximation algorithms would be applicable though. Rckrone (talk) 07:00, 15 August 2011 (UTC)

I was thinking something similar, except with subset sum. The approximation algorithm given there should work with multiplication in place of addition.--Antendren (talk) 07:07, 15 August 2011 (UTC)

The LLL algorithm can also be applied here. Count Iblis (talk) 00:00, 16 August 2011 (UTC)

(OP) Thanks for this, Iblis. I think I see that the page is helpful, but am getting tangled up as to how lattices are relevant here. Could you give me a hint as to how to translate my problem to a form where the LLL algorithm helps? best wishes, Robinh (talk) 09:29, 16 August 2011 (UTC)

You run it as an integer relation algorithm as described in the last paragraph of the article. You then search for a relation between 1/2 Log(n) and the logarithms of the prime numbers. You can implement this easily in Mathematica as follows. We define the function latred[ac,x1,x2,x3,....,xn], where "ac" is the accuracy in decimal digits and the x_r are the real numbers amongst which we seek a relation:

latred[ac_, x__] :=

 Module[{w = 10^(ac), y = {x}, ln, ar}, ln = Length[y]; 
   ar = Table[
       If[r == s, 1, If[s < ln + 1, 0, Floor[w yr]]], {r, 1, ln}, {s, 1, ln + 1}]; LatticeReduce[ar]]

Then if the we apply this to n numbers, the output will be a list of vector of length n+1. The list contains n such vectors. Each vector defines a linear relation between the x_r, the rth components of a vector is the coefficients of x_r such that the sum is zero to witin a tolerance defined by the last component of the vector times 10^(-ac).

You need to experiment with chosing the parameter "ac" correctly. In the case of your example, the following input works:

In[2]:= latred[2, 1/2 Log[4896], Log[2], Log[3], Log[17]]

Out[2]= {{-1, 3, 2, 0, 1}, {-1, 2, 0, 1, -3}, {-2, 0, 0, 3, 1}, {-1, -4, 9, -1, -2}}

The first two output vectors in the list correspond to the numbers a and b that you obtained. Count Iblis (talk) 19:08, 16 August 2011 (UTC)

Thank you Iblis, I see it now.

Resolved

Robinh (talk) 19:40, 16 August 2011 (UTC)

I see this as a version of the knapsack problem. The question states that the prime factors are known. So, you don't really care about the original number. You have the prime factors and you want to arrange them into two groups (knapsacks) such that the product of each group is about the same. In other words, you are trying to fill up one knapsack so you get as close as possible to filling it up. The size of the knapsack will be the square root of the original number. Once you fill it up to the maximum amount possible, the other number will be the original number divided by the product of the primes in the knapsack. I know that the real knapsack problem uses the sum of the items, but switching to the product of the items does not (in my mind) change the problem. So, use any of the solutions in the knapsack problem article. Regardless, this is still an NP-complete problem. -- kainaw™ 19:40, 16 August 2011 (UTC)

Actually, can I retract the resolved tag? The LLL algorithm, unlike the knapsack version, does not account for the fact that I have only a limited number of factors. But I am relieved to discover that it is NP-complete (I have been giving myself a hard time for not being able to solve it quickly). Best, Robinh (talk) 20:08, 16 August 2011 (UTC)

Is the probability of guessing the right number out of N choices one in N^2?

If there were a lottery game where one of 10 balls labeled 0-9 were blindly pulled out of a jar, then when I was guessing before the drawing, there would be a one-in-ten chance that I would guess, say, three. Then, when the drawing was actually done, there would also be a one-in-ten chance that the winning ball would also be three. So there are 100 different combinations (e.g., guess four, draw eight, etc...). So isn't it correct that the probability of you guessing a given number out of N and that number in fact being drawn is one in N^2? You can think of your guessing, or pressing the "easy pick" button, as another separate drawing. 20.137.18.50 (talk) 13:43, 15 August 2011 (UTC)

If you did indeed want to model the guessing with probability as well, the probability of correctly guessing the winning ball is still 1/N. The reason is that there are still N ways in which you can correctly guess the winning ball. To continue with your example, while there are 100 possible outcomes in your experiment, there are 10 outcomes which correspond to you picking the winning ball: guess 0/pick 0, guess 1/pick 1, ..., guess 9/pick 9. If each one of those outcomes has a probability of 1/N^2, the probability of the union of those (mutually exclusive) events is N*1/N^2=1/N. Nm420 (talk) 14:57, 15 August 2011 (UTC)

EC: There are N² ways of the game 'playing out', but N of them result in you winning ('pick x, draw x'), so you have an N in N² chance of winning --- or, 1 in N. Icthyos (talk) 15:00, 15 August 2011 (UTC)

This reminds me of a recent case where a woman had three children with the same birthday (not the same year) - some papers claimed that the probability of this was one in 365*365*365 = 48,627,125, whereas it's actually one in 365*365 = 133,225 (or probably much better than that, as she presumably exercised some choice in the likely range of birth dates). Ah, here's an example. AndrewWTaylor (talk) 17:10, 15 August 2011 (UTC)

Note that it being 365×365 assumes she only has 3 children. With more kids the chances are better. Also it assumed they were completely independent events, and, as you noted, they aren't (perhaps her and her hubby only hook up on their anniversary each year :-) ). StuRat (talk) 17:45, 15 August 2011 (UTC)

You laugh at that. I have friends with three children, all born in the same week nine months after the husband's birthday. --Jayron32 20:29, 15 August 2011 (UTC)

And my mother was born exactly 9 months after the repeal of Prohibition. :-) StuRat (talk) 21:06, 15 August 2011 (UTC)

AndrewWTaylor's example is actually mentioned in the biography of the quoted mathematician, Roger Heath-Brown. I doubt he would make that mistake. I suspect that the paper either gave him another question such as "What is the chance of 3 children being born 7 October?", or that they misunderstood his reply. PrimeHunter (talk) 20:59, 16 August 2011 (UTC)

Riemann?

Hi. I'm taking a theory of knowledge course and one of the major topics is different views of things that are usually considered absolute, such as maths. There is an example given (which unfortunately I do not remember to extreme clarity) in geometry, something about parallel lines. I might not have this exactly right so please correct me if you know what I'm talking about: Anyway, in the Euclidean view of geometry, there are an infinite number of lines parallel to a given one, in the Riemannian view (I think) there is only one, and in another view (I forget) there are none. I am familiar with the Euclidean interpretation, but what are the other ones, and why are there fewer (or no) parallel lines? I'd just like an intuitive overview, I'm not really a mathy person but I'd like to have more knowledge on this. Thanks. 203.117.33.23 (talk) 22:14, 15 August 2011 (UTC)

We have an article, parallel postulate, that might be more useful to you than an attempt at an explanation here. Looie496 (talk) 23:08, 15 August 2011 (UTC)

See also Hyperbolic geometry, especially the section on visualisation (as featured in the works of M. C. Escher), and Elliptic geometry. I first learned about stuff this many years ago from W.W. Sawyer's wonderful "Prelude to Mathematics", which has a nice non-technical treatment of the hyperbolic plane. AndrewWTaylor (talk) 07:47, 16 August 2011 (UTC)

I see there's a Dover reprint now. Yes I agree a very good book, it was the first maths book I ever bought with my saved up pocket money when I was twelve, I kept it hidden from everybody and still have it. Dmcq (talk) 08:52, 16 August 2011 (UTC)

August 16

Extracting parameters from a matrix form ellipse

I found an algorithm for fitting the smallest possible ellipse through a set of points, but the solution is provided as the set of points $\mathbf {p}$ for which $(\mathbf {p} -\mathbf {c} )^{\mathrm {T} }\mathbf {A} (\mathbf {p} -\mathbf {c} )=1$ where $\mathbf {A}$ is a symmetric, positive definite matrix.

$\mathbf {c}$ is already useful as the center of the ellipse, but in order to plot it I need the whole thing in parametric form. This means I need to decompose $\mathbf {A}$ into some parameters for the parametric form:

$X(t)=X_{c}+a\,\cos t\,\cos \varphi -b\,\sin t\,\sin \varphi$

$Y(t)=Y_{c}+a\,\cos t\,\sin \varphi +b\,\sin t\,\cos \varphi$

Is there a simple formula for extracing $a$, $b$ and $\varphi$ from $\mathbf {A}$?

Eonzo (talk) 10:08, 16 August 2011 (UTC)

If you write $\mathbf {p} -\mathbf {c} ={\begin{bmatrix}x\\y\end{bmatrix}}$ and $\mathbf {A} ={\begin{bmatrix}a&b\\b&c\end{bmatrix}}$, then $(\mathbf {p} -\mathbf {c} )^{\mathrm {T} }\mathbf {A} (\mathbf {p} -\mathbf {c} )=ax^{2}+2bxy+cy^{2}=1$. To finish up, see Rotation of axes#Elimination of the xy term by the rotation formula. 130.76.64.117 (talk) 00:45, 17 August 2011 (UTC)

Thank you for pointing me there, that's exactly what I needed :) Eonzo (talk) 07:51, 17 August 2011 (UTC)

August 17

Air Resistance Terminal Velocity

I have an assignment which involves dropping an object of a certain mass and recording the time that it takes to hit the ground, then using this to determine the coefficient of air resistance and terminal velocity of the mass.

I am told to assume that the resistance is given by $kv$ where $v$ is the velocity in meters per second. I am told to calculate the air resistance $k$ and the terminal velocity.

I don't really understand the question. For instance it doesn't mention where I should take mass into account. The wiki articles on air resistance have complicated formulae and this is a high school math question that is not supposed to require any physics knowledge.

If the mass falls according to an acceleration constant of $-g$, and that means that it has a velocity of $-gt$ at time $t$, does that mean that its actual velocity then becomes $-gt+kgt$ when incorporating the effect of air resistance?

And if the terminal velocity is the velocity when the acceleration is equal to zero should I calculate this finding $v$ when $kv=gt$? — Preceding unsigned comment added by 118.208.93.112 (talk) 05:57, 17 August 2011 (UTC)

In order for this to be a high school math problem, you should have been given a formula to use. Working this out from the information given requires solving a differential equation, and high school students generally aren't expected to be able to do that, even if they have taken calculus. (The result ends up being that the velocity decays exponentially to the terminal velocity.) Looie496 (talk) 06:30, 17 August 2011 (UTC)

$kv$ is the force. To find the acceleration caused by it you need to divide by the mass (alternatively, you equate this force with the force of gravity $mg$).

The terminal velocity satisfies $kv_{0}=mg$.

The problem can be solved with high-school math if you assume the time to reach the terminal velocity is negligible. Then you have $v_{0}=h/t$ where h is the height from which the object is dropped, and from this you can solve for k.

For reference, the kv formula is only correct for low enough speeds. For higher speeds, resistance is proportional to the square of the velocity. -- Meni Rosenfeld (talk) 09:13, 17 August 2011 (UTC)

Drop the object from a range of different heights and measure the time it takes to reach the ground as accurately as possible. Make the range of heights as wide as possible, and do several runs at each height, averaging the times to reduce experimental error. Plot your times t (vertical axis) against your heights h (horizontal axis). For small heights the graph of t against h will be a curve, but for large enough heights, once the object reaches its terminal velocity, the graph will become a straight line. The reciprocal of the slope of this line is the object's terminal velocity (you have to take the reciprocal because you want units of m/s, not s/m). As Meni says, once you have the terminal velocity you can calculate k from

$k={\frac {mg}{v_{0}}}$

k is specific to this one object, so you can regard the object's mass m as fixed. If you wanted to shoot for extra credit, you could repeat the experiment for different objects with different shapes and densities and see what value k takes for each object.Gandalf61 (talk) 10:26, 17 August 2011 (UTC)

In addition to gravity and drag the terminal velocity depend on buoyancy. A pebble moves differently from a balloon. (See also added mass and lift.) Bo Jacoby (talk) 09:56, 18 August 2011 (UTC).

DeeperQA's nearly the last theorem

Whereas: (3^3) + (4^3) + (5^3) - (6^3) = 0

"I have discovered a truly marvelous proof of this theorem, which this moment is too short to contain."

Okay, all joking aside does anyone have graphical proof that Laplace's Fermat's Last Theorem and this theorem are valid and correct? --DeeperQA (talk) 08:06, 17 August 2011 (UTC)

(Do you mean Fermat's last theorem?) According to the article Euler's sum of powers conjecture, 2682440⁴ + 15365639⁴ + 18796760⁴ = 20615673⁴, so I'm afraid your "theorem" isn't true. AndrewWTaylor (talk) 08:42, 17 August 2011 (UTC)

Google calculator verifies so it must not be true. --DeeperQA (talk) 15:49, 17 August 2011 (UTC)

Are W, X, Y, and Z always to be consecutive integers ? Even so, I can't imagine how you could show that graphically, for all possible values of W, X, Y, Z, and N. You could make a 2D graph showing one value of N, with the horizontal axis being small values of W, X, Y, and Z, and the vertical axis being the values of W^N, X^N, Y^N, W^N+X^N+Y^N, and Z^N. I suppose you could extend that into a 3D graph, with the third dimension being different values of N. However, you could still only show a small portion of all the possible values, which doesn't constitute a "general proof". StuRat (talk) 08:50, 17 August 2011 (UTC)

Real Number

Can we define the real numbers by defining them as a complex number with imaginary part equal to zero and then defining the complex numbers as the smallest set of numbers ( or smallest field) that we must work in to ensure that every nth order polynomial has precisely n roots? Or does this leave us open to ambiguities, such as the cardinality of sets? Thanks. asyndeton talk 11:49, 17 August 2011 (UTC)

You'll have to be specific about which polynomials you are considering. If you only at first consider polynomials with rational coefficients, then the "smallest field" obtained will be the algebraic closure of Q, which is not C. It doesn't even contain all of R. For instance, you can never get a real transcendental as a root of a rational polynomial. The way to get all of C by this method is to start with polynomials with real (or complex) coefficients, which would defeat the purpose of your construction. Staecker (talk) 12:00, 17 August 2011 (UTC)

I should have realised that. Thanks for your very helpful answer. asyndeton talk 12:04, 17 August 2011 (UTC)

There's a few problems with saying reals are the same as complex numbers with the imaginary part zero. Simple arithmetic is okay but then you get to comparisons which give real trouble with complex numbers. And after comparisons you get to things like e^z always has a single well defined value but (e+0i)^z does not where I've added a zero imaginary to show we're now talking about a power of a complex number. Dmcq (talk) 12:52, 17 August 2011 (UTC)

I respectfully disagree with Dmcq. The positive numbers are all real numbers, and the function a^z is tricky unless a is positive, but that is no reason why reals should not be identified with complex numbers having zero imaginary part. You are really in trouble if you insist that e ≠ e+0i. Bo Jacoby (talk) 15:23, 17 August 2011 (UTC).

Why are you "in trouble"? You seem to feel very strongly about this kind of point, and I think you are very incorrect. The reals may be considered a subset of the complex numbers, or not, according to convenience. In their usual set-theoretic coding, they are definitely not literally a subset of the complex numbers. Ordinarily that's a fact of little importance, but occasionally a context where it's important does come up. --Trovatore (talk) 07:02, 18 August 2011 (UTC)

Whilst I appreciate this is now a moot point, would the cardinality of sets also have been an issue? asyndeton talk 19:02, 17 August 2011 (UTC)

As said above you would have problems with the cardinalities. If you haven't defined the reals and only have rationals then the only numbers you can define your way are the algebraic numbers. There's only $\aleph _{0}$ of those. You can enumerate all the polynomials with rational coefficients, in fact even the ones with algebraic coefficients only give rise to algebraic numbers so they are factors of polynomials with rational coefficients.

As to e the real being different from e+0i there is no difficulty in normal working as when you compare a real and a complex number you automtically convert the real to a complex number. It is the same when comparing integers with reals. 0 the integer is however a different entity from 0 the real or 0 the complex number or 0 the quaternion or 0 (mod 7) in modular arithmetic. Whe exponentiating one should not promote e the real to a complex number or all sorts of things go wrong, see exponentiation#Failure of power and logarithm identities, the last example - the paradox by Clausen - shows what can happen. Dmcq (talk) 22:51, 17 August 2011 (UTC)

Programmers distinguish between integers and reals, because the names refer to different data types. Zero is represented differently as integer and as real, but still the two representations refer to the same mathematical object, zero, and expressions like 0==0.0 evaluate to true. Mathematically the set of integer numbers $\mathbb {Z}$ is considered a subset of the set of real numbers $\mathbb {R}$ , and the set of real numbers is considered a subset of the set of complex numbers $\mathbb {C}$ . $\mathbb {Z} \subset \mathbb {R} \subset \mathbb {C}$. You are not 'promoting e the real to a complex number', because any real number is also a complex number. Functions that are defined for real numbers may not generally be defined for complex numbers, but of course they are defined for some complex numbers, namely the real numbers. Bo Jacoby (talk) 06:23, 18 August 2011 (UTC).

Integers are completely different conceptually from reals. As a ring, they are isomorphic to a subring of the reals, so without harm you can consider them to be a subset of the reals if you like, at least as long as you limit yourself to what can be expressed in the language of rings (addition and multiplication). But there is no reason to expect such a strong identity between integers and reals as to exclude differences in any language whatsoever. They're a completely different breed of cat. --Trovatore (talk) 07:12, 18 August 2011 (UTC)

Consider the first degree equation 2x=x. Mathematicians think that this equation has the unique solution x=0, but according to Trovatore and Dmcq it has the integer solution x=0, the real solution x=0.0, the complex solution x=0+i0, a quaternion solution, a vector solution for each vector space, and many many more solutions. This is very far from main stream mathematics. Bo Jacoby (talk) 14:21, 18 August 2011 (UTC).

I'm with Trovatore and Dmcq on this one. Once we have established that there is a subset of $\mathbb {R}$ that behaves like $\mathbb {Z}$ and a subset of $\mathbb {C}$ that behaves like $\mathbb {R}$ then it is common and convenient to blur the difference, but strictly speaking $0\in \mathbb {Z}$, $0\in \mathbb {R}$ and $0\in \mathbb {C}$ are three distinct mathematical objects. Gandalf61 (talk) 14:39, 18 August 2011 (UTC)

It's completely mainstream mathematics. Normally people don't make a big deal about it, because it usually doesn't matter, and of course no one wants to bother with distinctions that don't make a difference. But absolutely no mainstream mathematician is going to identify the zero of SO(3) with the zero of ℓ²; that makes no sense at all.

When mathematicians say that 2x=x has a unique solution, it's understood that that's after specifying a structure in which the equation is to be interpreted. Usually the structure is obvious from context.

Bo, are you trying to eliminate context as a factor in how mathematics is interpreted? You can't do that, sorry. It is important and it's always going to be. --Trovatore (talk) 16:44, 18 August 2011 (UTC)

Gandalf, what is the difference between 'a subset of $\mathbb {R}$ that behaves like $\mathbb {Z}$' and $\mathbb {Z}$ itself? Trovatore, what is the zero of SO(3)? Bo Jacoby (talk) 21:35, 18 August 2011 (UTC).

Actually, I was too glib in talking about SO(3). Of course SO(3) doesn't have a zero. I just noticed that now. Replace it with the zero of the ring whose elements are linear transformations from R^3 to R^3 (multiplication being composition). The zero of that ring is the linear transformation that sends every vector to the zero vector. --Trovatore (talk) 21:47, 18 August 2011 (UTC)

Bo - equivalence is not identity. For example, the set of matrices

${\begin{pmatrix}a&b\\-b&a\end{pmatrix}}\quad a,b\in \mathbb {R}$

behaves like (i.e. is isomorphic to) $\mathbb {C}$, but it is not $\mathbb {C}$ itself because $\mathbb {C}$ consist of numbers, not matrices. Gandalf61 (talk) 13:52, 19 August 2011 (UTC)

Gandalf, you may like to study the article When is one thing equal to some other thing?. The point is that $\mathbb {C}$ is only defined modulo isomorphism. The representation by pairs of numbers is not better than the representation by matrices. Both representations define $\mathbb {C}$ 'itself'. Bo Jacoby (talk) 22:41, 19 August 2011 (UTC).

Trovatore, This ring also contains $\mathbb {Z}$ as a subring. Answering your question: algebra is abstract and independent on context or interpretation. Zero apples = zero bananas = zero. Bo Jacoby (talk) 22:22, 18 August 2011 (UTC).

Interpretation of mathematics is very much context-dependent, and always will be. You're certainly free to try to get rid of the dependency. You are guaranteed to fail. --Trovatore (talk) 22:29, 18 August 2011 (UTC)

The fathers of abstract algebra did not fail in getting rid of context dependency. Bo Jacoby (talk) 10:31, 19 August 2011 (UTC).

This argument seems to boil down to what it means for objects to be isomorphic. I have to agree with gandalf and Trovatore that isomorphism has to be distinct from considering two objects to be "the same object". The main example above is the ring of integers as a subset of the ring of reals. It so happens that there's only one injective homomorphism from the integers to the reals, which might make it easy to falsely believe that there are no problems saying that 3 in Z is the same thing as 3 in R. The relationship is un-ambiguous. Even your example equation 2x = x, seems conveniently chosen to avoid problems of context because it just so happens that the unique solution in every Z-module is the zero element. As soon as you move away from these special cases, this interpretation of isomorphism is going to have problems. What if I'm talking about the ring of complex numbers C and the quaternions H? Clearly C is isomorphic to a subring of H and one such homomorphism send the imaginary unit to i in H. Would you claim that i in H is the same object as the imaginary unit in C? What if I wanted to map the imaginary unit to j instead? Is the imaginary unit the same as more than element of H? Similarly, what's the solution to 2x = 0 without context? Rckrone (talk) 03:25, 20 August 2011 (UTC)

Trovatore asked: why are you in trouble if you insist that e≠e+0i? Because it follows from the rules of arithmetic that 0=0 → 0=0i → e+0=e+0i → e=e+0i. I am amazed that you guys seriously accept contradictions like e≠e+0i. Rckrone's quaternion example can be simplified to complex conjugation: "Is i is the same object as −i? Is the imaginary unit more than one element of $\mathbb {C}$?" The answers are that i≠−i but they are indistinguishable mirror images, and that there are two imaginary units in $\mathbb {C}$. But $\mathbb {Z}$ only occurs once as a subring of $\mathbb {R}$, and $\mathbb {R}$ only occurs once as a subring of $\mathbb {C}$, so there is no ambiguity in $\mathbb {Z} \subset \mathbb {R} \subset \mathbb {C}$, and there is no doubt that −0.0+0.0i is an integer, even if it is written in the form of a noninteger, just like $e^{\pi i}$ is an integer even if it written in terms of transcendental and imaginary numbers. Bo Jacoby (talk) 09:59, 20 August 2011 (UTC).

3x=x (mod 4) has the solution 2 as well as 0 in modulo 4 arithmetic. Exponentiation is defined differently for a real to a complex power from a complex number to a complex power, it is not a ring operation. The complex logarithm is used when the base is complex and its value is not unique or you have to use branch cuts. Comparison does not carry over either. Is 0 greater than or less than 1-i? Saying 0i is 0 is a convenience and done according to the circumstances, it is normally pretty obvious when it is sensible and when it is not. Using the complex number definition of exponentiation $(e+i0)^{\pi i}=e^{\pi i\operatorname {Log} (e+i0)}=e^{\pi i(\ln {|e+i0|}+2\pi in)}=e^{\pi i(1+2\pi in)}=e^{\pi i-2n\pi ^{2}}=-e^{-2n\pi ^{2}}$. So the value can be any old value like this depending on the particular integer n chosen here by using any particular branch cut for the complex logarithm. Dmcq (talk) 11:30, 20 August 2011 (UTC)

Bo - it seems clear to me that ${\text{e}}\in \mathbb {R}$ cannot be the same object as ${\text{e}}\in \mathbb {C}$ because there are operations you can do to ${\text{e}}\in \mathbb {C}$ that you cannot do to ${\text{e}}\in \mathbb {R}$. An example is taking the imaginary part; the imaginary part of ${\text{e}}\in \mathbb {C}$ is 0, whereas the imaginary part of ${\text{e}}\in \mathbb {R}$ is undefined. Of course you can forget the additional structure of $\mathbb {C}$ and say that ${\text{e}}\in \mathbb {C}$ is equivalent to ${\text{e}}\in \mathbb {R}$ for any operations that are defined on both $\mathbb {C}$ and $\mathbb {R}$ - but they are still distinct and separate mathematical objects. Gandalf61 (talk) 12:10, 20 August 2011 (UTC)

Dmcq - The power a^b is defined for nonzero a and integer b, and for zero a and positive b, and for complex a and zero b, and for positive a and complex b. Here 'complex' includes the reals, and 'positive' means positive and real. The definitions match where several cases apply simultaneously (such as 3² where 3 is positive and 2 is integer). Comparison: a>b when a−b is positive, and a<b when b>a. This makes sense for real a and b, and also for some nonreal a and b, but general complex numbers are not comparable. The number 1−i is neither >0 nor <0 nor =0. Logarithms of nonpositive numbers are ill defined and their use leads to contradictions, as you show. All this has nothing to do with the case. It is no reason for letting e≠e+0i.

Gandalf - The imaginary part of a real number is zero. "distinct mathematical objects"? e and e+0i are not mathematically different. The number e=e+0i is both real and complex. Bo Jacoby (talk) 13:31, 20 August 2011 (UTC).

I see. And presumably, if e is both real and complex, then it is also a quaternion because e=e+0i=e+0i+0j+0k ? But why stop there - is it an octonion and a sedonion too ? And, since it shares some properties with the matrix

${\begin{pmatrix}e&0\\0&e\end{pmatrix}}$

is e also a 2x2 matrix, and a 3x3 matrix etc. etc. ? Is it also a vector ... and a tensor ... and a constant polynomial ? Are you saying that one and the same mathematical object is simultaneously a member of all of these structures ? Or is there a line somewhere where the object stops being the same e as the real number e ? Gandalf61 (talk) 16:13, 20 August 2011 (UTC)

Yes. If the structure contains a unique subset satisfying the axioms of $\mathbb {R}$, then this subset is $\mathbb {R}$, and the element e is the real number e. Bo Jacoby (talk) 22:52, 20 August 2011 (UTC).

That's called isomorphism and embedding is where one has something isomorphic to a bit of something larger. Dmcq (talk) 23:13, 20 August 2011 (UTC)

Self study mathematics

Hello all. I am looking to self study mathematics. However, I am still in high school and have yet to go through calculus. I was wondering if I could get any recommendations on branches of mathematics to study that don't require large amounts of difficult calculus (I understand how to differentiate pretty well and I can integrate, but not much else.) I've already taken Algebra and Euclidean geometry and I understand set theory. I'm more interested in pure math, but I'm open to anything. I'm apprehensive to self-study calculus and then test out of it, because a friend said that especially with calculus, classroom experience is a good thing, but if you think it's possible, I'm open to suggestions. Thanks Aacehm (talk) 16:07, 17 August 2011 (UTC)

If you are interested in pure maths a good place to start would be Linear Algebra. MIT has some great video lectures on the internet. You could view those.-Shahab (talk) 16:17, 17 August 2011 (UTC)

If you enjoy doing proofs, I think self-study of calculus could be a good thing. Most high school calculus classes focus on teaching formulas and tend to gloss over the mathematical foundations -- the concept of a limit is pretty challenging to many students, and usually doesn't get taught that well. Linear algebra is also good, and the other thing I might recommend would be probability theory, which is extraordinarily useful. Looie496 (talk) 16:25, 17 August 2011 (UTC)

I started reading Spivak's Calculus A month ago but the rigor proved to be a little too much. I might try to go through it again. — Preceding unsigned comment added by Aacehm (talk • contribs) 16:45, 17 August 2011 (UTC)

Dont worry about mathematics being "too rigorous". I was like you when I was in high school, not understanding a thing they're saying, and I just kept bashing through until my mind matured enough to accept the level of rigor. Now I think too rigorously! Many proofs I used to accept now I reject as too handwavy and non-rigorous. Just keep trying, you'll get there. Money is tight (talk) 13:22, 18 August 2011 (UTC)

Quadratic equation with a mod operator

Usually, when I get a mod tossed into an equation, attempting to solve it is a waste of time. So, I thought I'd ask about this one before attempting to reduce it. All letters except X are constants:

$\left(bx+ax^{2}\right)\%\left(e+n-x\right)=n-x$

If it wasn't for the mod, I could easily reduce this to the form $ax^{2}+bx+c=0$ and solve it. I figure that if the mod operator didn't include the x term, I could do the same. But, with the x term in the mod operator, I assume this cannot be reduced. -- kainaw™ 16:37, 17 August 2011 (UTC)

How about rewriting it as

$bx+ax^{2}=k(e+n-x)+(n-x)$?

Looie496 (talk) 17:08, 17 August 2011 (UTC)

Yep. Doing that will give me a bunch of values for x instead of two. So, it introduces a whole new problem of limiting k, which turns the whole thing into a waste of time. I'll approach the original problem with a matrix solution instead of attempting a quadratic one. Thanks. -- kainaw™ 17:20, 17 August 2011 (UTC)

Could you please give us more details? As it stands, your "equation" is (bx + ax²); which isn't actually an equation, because there is no equals sign. Besides that, you use ordinary parentheses; while the modulus is usually denoted with vertical lines, e.g. | y |. Also, you seem to introduce the constant c later on without any explanation. Are you trying to solve | ax² + bx + c | = 0, or | ax² + bx | = | c |, or what? — Fly by Night (talk) 22:55, 17 August 2011 (UTC)

"Mod" here stands for modular arithmetic, not the modulus function. x%y is computer science notation for "the residue of x mod y".--Antendren (talk) 23:51, 17 August 2011 (UTC)

knapsack problem problem

Hi. In the classical knapsack problem, the aim is to optimize the value in the knapsack. I want to list *all* ways of achieving the optimum (and I know that there are many ways of doing it in the cases I'm interested in). Is there a simple way to to this using classical knapsack algorithms, or if not, what is the best way to cast my problem?

Example: knapsack of weight 6. I have 4 objects with (weight,value) pairs (1,1), (2,2), (3,3) and (4,4). knapsack algorithms give (4,2) [ie object 4 and object 2] but I need ((4,2),(3,2,1)). Cheers, Robinh (talk) 22:10, 17 August 2011 (UTC)

Somehow your edit broke the page, so I undid it and re-posted you question. Rckrone (talk) 22:35, 17 August 2011 (UTC)

(OP) yes, some weird error at this end too. Thanks for the fix Robinh (talk) 01:25, 18 August 2011 (UTC)

If you have a hefty supply of memory and time, there is a very simple method of listing all optimal solutions to the knapsack problem. Your function takes the weight capacity of a sack and a collection of items. What the function does is very simple. I'm writing this in a bit of pseudo code since this isn't the computer desk:

function max_capacity(volume_left, items)

Max capacity = 0

For each item in items

If item weight <= volume_left

New items = items with item removed

New volume = volume_left - item weight

This capacity = max_capacity(New volume, New items)

If this capacity > max capacity, set max capacity=this capacity

Return the max capacity

Now, you know what the maximum capacity is. Repeat this function again, but change it so it prints each combination that meets the maximum capacity. Because of the way it is written, we want to print the combination when the volume_left reaches the minimum level. So, min_volume = sack capacity - max capacity.

function print_combos(volume_left, items, min_volume, items used)

If volume_left = min_volume, print the items used

Otherwise...

For each item in items

If item weight <= volume_left

New items = items with item removed

New used = items used with item added

New volume = volume_left - item weight

Call print_combos(New volume, new items, min_volume, new used)

That will print every permutation of the solutions. If you want to avoid redundant permutations, you can have it sort the output when it prints the items used and then remove repeated lines. -- kainaw™ 12:34, 18 August 2011 (UTC)

(OP) Thanks Kainaw, but I fear that removing repeated lines will kill me in terms of computer time (I have maybe 40 elements). I guess this shows that I can't somehow alter the classical knapsack problem to do what I want. I've also realized that my problem is actually a subset sum problem but this isn't helpful because again, I need *all* solutions. Maybe optimization problems are the wrong paradigm for this and I should look instead at Diophantine equations. Best wishes, Robinh (talk) 22:23, 18 August 2011 (UTC)

I did this recently by treating my problem as a multi-dimensional knapsack and - after every successful solution was returned - adding a new constraint outlawing that solution, before re-calling and repeating. I was using CPLEX which is very fast and this was fine for my highly dominated 500 item dataset, but apparently CPLEX actually has specific features to return multiple items efficiently ("solution pooling" apparently, not that I can work out how to use it). --Iae (talk) 12:48, 19 August 2011 (UTC)

How I would do it, if I was allowed infinite time and resources to run the program... First, place all your items in an array. I'd also sort the array to make my output have a nice pattern. Then, when you do the "try every item" thing, change it to "try every item that has an array index higher than what is already in the sack". So, your items are 0:A, 1:B, 2:C, 3:D such that the number is the index and the letter is the item. Once you place item 0 in the sack, you can try item 1. But, once you place item 1 in the sack, it is impossible to put item 0 in the sack. So, if AB comes up as one combination, BA won't come up later. Note that I use array indexes, not item sizes. If you have two items of the same size, you will get them showing up twice. This does have the issue of absolute uniqueness for the same reason. If you have 0:A, 1:B, 2:B, 3:C, you can get ABBC with item 1 followed by item 2. But, technically, you can have ABBC with item 2 before item 1, which is unique from the first ABBC. -- kainaw™ 13:28, 19 August 2011 (UTC)

(OP) Thanks guys. Two perfectly reasonable solution methods here, but I am most interested in the fact that my problem does not seem to be a special case of a standard problem (for if so, someone would have spotted it). I am very intrigued by Iae's suggestion of finding a solution, then calling the routine again with that solution forbidden. Is there a name for this technique? It is certainly clever, but is it substantially faster than Kainaw's approach, which seems to traverse through all $2^{n}$ elements of the power set of my items? Robinh (talk) 09:48, 20 August 2011 (UTC)

It looks like the various quantities are integers and one would with 500 of them expect a lot of overlap. So I'd probably do a divide an conquer of splitting the items in two and enumerating the best value for each weight and then joining those solutions when going the next level up. Then you can find the best solution at the top and work back down the tree again. This would depend on not having too large a maximum weight and it being an integer, turning this into a general solution would be a bit gruesome. Dmcq (talk) 14:48, 20 August 2011 (UTC)

August 18

Binomial theorem and convergence

Using the fact that $(1+x)^{n}=\sum _{k=0}^{n}{\frac {n!}{(n-k)!k!}}x^{k}$ show that $\lim _{n\rightarrow \infty }n^{\frac {1}{n}}=1$ How does one get from A to B here? I tried the substitution $x=-1+n^{\frac {1}{n^{2}}}$ but that didn't make it any easier. Widener (talk) 08:55, 19 August 2011 (UTC)

I assume you're meant to do this without logarithms or L'Hôpital's rule. Let

$x=n^{\frac {1}{n}}-1,\,y={\sqrt {\frac {2}{n}}}$

Then the binomial theorem implies

$(1+y)^{n}>1+ny+{\frac {n(n-1)}{2}}y^{2}=1+{\sqrt {2n}}+n-1>n=(1+x)^{n}$

Take nth roots of both ends to get 1+y>1+x or y>x. The squeeze theorem, using x>0 and y→0 as n→∞, implies x→0 as n→∞.--RDBury (talk) 12:08, 19 August 2011 (UTC)

Thanks. Widener (talk) 23:28, 19 August 2011 (UTC)

August 19

Recursive sequence

Define a sequence $(x_{n})$ such that $x_{1}=1$ and $x_{n}={\sqrt {1+x_{n-1}}}$ for $n\geq 2$. Prove that $1\leq x_{n}<x_{n+1}$ for all $n\geq 1$.
The most obvious approach is to use induction. The statement trivially holds for $n=1$.
Assuming $1\leq x_{n}<x_{n+1}$
$\Rightarrow 1\leq x_{n+1}$
All that's left is to show that $x_{n+1}<x_{n+2}$ but this is where I come unstuck. It is not obviously true that $x_{n+1}<{\sqrt {1+x_{n+1}}}$. Could someone show me what to do next? Widener (talk) 01:38, 20 August 2011 (UTC)

You pretty much have it. The last inequality is obviously not true for all reals, but it is true in a certain range, so it suffices to prove that x never leaves that range. A quick calculation will tell you what that range is (mainly the maximum value). Induction suffices to show that x never exceeds this maximum value. Invrnc (talk) 02:17, 20 August 2011 (UTC)

Yes, the upper bound is the golden ratio interestingly. But how does the induction show that x never exceeds this value? Widener (talk) 02:24, 20 August 2011 (UTC)

Prove that inductively. Suppose $x_{n}<\Phi$ and prove that $x_{n+1}<\Phi$. —Bkell (talk) 02:49, 20 August 2011 (UTC)

Thank you. Widener (talk) 03:51, 20 August 2011 (UTC)

Subsets

Let $I_{n}=[a_{n},b_{n}]$ be a closed interval in $\mathbb {R}$ such that $I_{n}\subseteq I_{n-1}$ for $n\geq 2$. Show that $\bigcap _{k=1}^{\infty }I_{k}$ is nonempty.
I would like to know if this is correct reasoning:
$\bigcap _{k=1}^{\infty }I_{k}$
$=\lim _{n\rightarrow \infty }\bigcap _{k=1}^{n}I_{k}$

$=\lim _{n\rightarrow \infty }I_{n}$ which contains $\lim _{n\rightarrow \infty }{a_{n}}$.
I'm not sure if taking the limit of the intersection of sets is really allowed because it's not really compatible with the formal definition of a limit. I feel that I need to be more precise and formal in my proof. Widener (talk) 04:34, 20 August 2011 (UTC)

You're right that taking the limit of the intersections is rather sketchy. Remember each intersection is a set; what's the limit of a sequence of sets? (There are ways to define it, but they would basically just come back to intersection.) However, you've also correctly identified the point you want: $a=\lim _{n\rightarrow \infty }{a_{n}}$. Prove that $a$ is an element of every $I_{n}$ (prove that it's greater than every $a_{n}$ and less than every $b_{n}$), and conclude that it's in the intersection.--Antendren (talk) 07:51, 20 August 2011 (UTC)

Statistical hypothesis testing of Poisson processes

I'm performing an experiment in which there are two Poisson processes that have been producing events for the same amount of time. So the data gathered for the experiment just consists of two numbers, n₁ and n₂, the number of events that have occurred so far for each of the two processes. The null hypothesis of the experiment is that the rate parameters of the two Poison processes are the same. How can I tell if I can reject that null hypothesis at, say, a 95% confidence level? I'd know how to test the hypothesis if I knew one of the two rate parameters exactly, but I don't have any other information available other than n₁ and n₂. n₁ and n₂ are both at least 179, in case that helps determine whether some approximation would be valid. Red Act (talk) 04:48, 20 August 2011 (UTC)

I think a chi-square test should work for you. Looie496 (talk) 06:25, 20 August 2011 (UTC)

A Bayesian approach is this. Knowing only that the observed $n_{1}$ and $n_{2}$ had poisson distributions with unknown parameters $\lambda _{1}$ and $\lambda _{2}$, the parameters have gamma distributions with mean values $n_{1}+1$ and $n_{2}+1$ and standard deviations ${\sqrt {n_{1}+1}}$ and ${\sqrt {n_{2}+1}}$. The difference $\lambda _{2}-\lambda _{1}$ has mean value $(n_{2}+1)-(n_{1}+1)$ and standard deviation ${\sqrt {(n_{2}+1)+(n_{1}+1)}}$. Now the problem is reduced to decide if you believe that zero is a sample from a distribution having mean value ${\frac {n_{2}-n_{1}}{\sqrt {n_{2}+n_{1}+2}}}$ and standard deviation one. This is the same as the chi-square test except for the '+2' in the denominator. Bo Jacoby (talk) 07:34, 20 August 2011 (UTC).

Thanks! Red Act (talk) 10:26, 20 August 2011 (UTC)

(1+x)^n ≥ 1+nx

Show that $(1+x)^{n}\geq 1+nx$ for natural n and $x\geq -1$.

$(1+x)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}x^{k}=1+nx+\sum _{k=2}^{n}{\binom {n}{k}}x^{k}$ (for n≥2. The cases n=0 and n=1 are trivial)

So now we need to show that $\sum _{k=2}^{n}{\binom {n}{k}}x^{k}\geq 0$ for $x\geq -1$. How? Widener (talk) 10:12, 20 August 2011 (UTC)

Of course it's obvious for $x\geq 0$. Widener (talk) 10:16, 20 August 2011 (UTC)

Minimize $(1+x)^{n}-(1+nx)$ on [−1,0]. Sławomir Biały (talk) 12:40, 20 August 2011 (UTC)

August 20

Ratio from percent

Five oil containing trucks loaded fully with oil travel from City P to City Q. However, because of bad driving conditions, seepage occurs and so the actual quantity of oil delivered to the destination stations reduces.

The filling is done at the starting station and the containers are emptied at the destination station.

The table below gives the details about amount of oil filled at the starting station, seepage that occurred in each truck and the actual quantity of oil that was delivered to the destination (all in terms of total percentage).

Truck	Original Qty (Gallons, %)	Seepage (Gallons, %)	Delivered Qty (Gallons, %)
A	13.33	8	14.4
B	26.66	32	25.6
C	33.33	34	33.2
D	10	6	10.8
E	16.67	20	16
Total	100%	100%	100%

What is the ratio of seepage in truck B to the actual quantity of oil delivered by truck D in City Q?

I have been trying hard for the solution, but in vain. please help out - Samantha 12:49, 20 August 2011 (UTC)

Suppose there are originally X gallons of oil altogether, Y gallons lost due to seepage and Z gallons delivered. Then truck B loses 32Y/100 gallons due to seepage and truck D delivers 10.8Z/100 gallons, so the ratio of these quantities is 32Y/10.8Z. If you knew tha ratio Y/Z then you would be home and dry.

But we know that truck A starts out with 13.33X/100 gallons, loses 8Y/100 gallons and delivers 14.4Z/100 gallons. So

13.33X - 8Y = 14.4Z

and so

13.33(X/Z) - 8(Y/Z) = 14.4 (Equation 1)

Similarly, by considering the quantities for truck B, we have

26.66(X/Z) - 32(Y/Z) = 25.6 (Equation 2)

Eliminate X/Z from Equations 1 and 2 and you can find the value of Y/Z. Multiply this by 32/10.8 and you have the ratio of the seepage in truck B to the quantity delivered by truck D. Gandalf61 (talk) 13:27, 20 August 2011 (UTC)

WoW. Thanks. y/z = 1/5 and therefore Ans = 16:27. Samantha 13:36, 20 August 2011 (UTC) — Preceding unsigned comment added by 117.201.252.65 (talk)

One part of oil seeps for every five delivered? I think I'd have rephrased the question in terms of five fat men who were asked to deliver bowls of chips to some tables! ;-) Dmcq (talk) 14:33, 20 August 2011 (UTC)

derivative

I was trying to prove that the derivative of e^x is itself.I used the fact that f(a+b)=f(a)f(b) and since I don't know how to write math here, :D, I summarize it this way: I concluded that f'(x)=f'(0)f(x). now all I need to do is to prove that f'(0)=1... help!--Irrational number (talk) 17:11, 20 August 2011 (UTC)

What definition of $e^{x}$ are you using? A common definition of $e^{x}$ includes the fact that the derivative at $x=0$ is 1, as part of the definition. In other words, the number $e$ is defined to be the unique real number such that the graph of the exponential function $e^{x}$ has a tangent line of slope 1 at $x=0$. If you aren't using this definition, what is the definition of $e$ that you are using? —Bkell (talk) 17:20, 20 August 2011 (UTC)

do you mean that the f'(0)=1 is part of the function's definition? if that's the case my proof is complete...--Irrational number (talk) 17:35, 20 August 2011 (UTC)

Yes, sometimes. But you might not be using that definition. There are other ways to define the number $e$. For example, $e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}$ is another common one. If you're using that definition, then the definition does not include the fact that the derivative of $e^{x}$ at $x=0$ is 1. So you need to go back and find out exactly what definition of $e$, and what definition of $e^{x}$, you are working with. —Bkell (talk) 17:38, 20 August 2011 (UTC)

actually, I knew from Wikipedia that the derivative for e^x is itself, using the fact that it's an exponential function (and therefore f(a+b)=f(a)f(b)) I proved that f'(x)=f'(0)f(x) which is apparently true for all exponential functions.then I wondered that if I can prove that for e^x f'(0)=1, then my proof is complete.is it like "e is the number for which f'(0) (in my equation is) is 1" and therefore my proof is somehow wrong?(sorry for my stupidity :D!)--Irrational number (talk) 17:49, 20 August 2011 (UTC)

It depends on what you have to start with. Every proof in mathematics starts with definitions, postulates, and previously proved theorems, and uses these to prove something new. So it is important to know what definitions you are using. What do you already know about $e^{x}$? Here are two common ways to define the number $e$:

1. $e$ is the unique real number such that the graph of the exponential function $e^{x}$ has a tangent line of slope 1 at $x=0$.
2. $e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}$.

Some textbooks use the first definition, and some textbooks use the second definition. (There are other definitions too—these are not the only ones.) As it turns out, both of these definitions define the same real number, 2.718281828…, so either definition is fine; but this fact requires proof. If you use definition 1, you know that the derivative of $e^{x}$ at $x=0$ is 1, because that's part of the definition—but you do not know immediately from that definition that $e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}$; that requires proof. Likewise, if you use definition 2, you know from the definition that $e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}$, but you do not know immediately that the derivative of $e^{x}$ at $x=0$ is 1; that requires proof.

In your proof, you are starting with some definition of $e$, along with some previously proved theorems about the function $e^{x}$ (for example, you are using the previously proved theorem that $e^{a+b}=e^{a}e^{b}$). If the definition of $e$ that you are starting with is definition 1 above, then you do not need to prove that the derivative of $e^{x}$ at $x=0$ is 1, because that fact is part of the definition. However, if the definition of $e$ that you are starting with is definition 2 above, or some other definition, then you do not know from that definition that the derivative of $e^{x}$ at $x=0$ is 1, so you will need to prove that fact from the definition you have. —Bkell (talk) 18:02, 20 August 2011 (UTC)

so what I understand is whether I need to prove that f'(0)=1 or not depends on which definition I'm using...well I prefer the easy one :D.but what about others like 2^x? how is f'(0) found for that?(which in this case is ln2)?--Irrational number (talk) 18:10, 20 August 2011 (UTC)

You say that you prefer the easy one. Right—so does everyone. :-) But if you are proving this statement for a class, you probably do not get to choose what definition to use—your instructor or your textbook has probably given a definition for $e$, and that is the definition you must use.

For $2^{x}$, use the properties of exponents and logarithms to write $2^{x}=(e^{\ln 2})^{x}=e^{(\ln 2)x}$, and then use the things you know about derivatives and about the derivative of $e^{x}$. —Bkell (talk) 18:14, 20 August 2011 (UTC)

cool! I get it now... thanks!--Irrational number (talk) 18:17, 20 August 2011 (UTC)

?

Hey, I've been trying to figure this out: what's infinity times zero? 64.166.145.51 (talk) 23:51, 20 August 2011 (UTC)

• See Indeterminate form. --Kinu ^t/_c 00:00, 21 August 2011 (UTC)

Uhh... 64.166.145.51 (talk) 00:11, 21 August 2011 (UTC)