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March 10 edit

Error in the fundamental theorem of calculus? edit

I integrate $f(x,y)={\sqrt {y}}$ over

\Omega =\{(x,y)\in R^{2}:0\leq y\leq 1,-{\sqrt {y}}\leq x\leq 0\}=\{-1\leq x\leq 0,x^{2}\leq y\leq 1\}

On one hand,

\int _{0}^{1}\int _{-{\sqrt {y}}}^{0}{\sqrt {y}}\,dx\,dy=\int _{0}^{1}ydy=\left.{\frac {y^{2}}{2}}\right|_{0}^{1}={\frac {1}{2}}

On the other hand, the same integral is equal to

\int _{-1}^{0}\int _{x^{2}}^{1}{\sqrt {y}}\,dy\,dx=\int _{-1}^{0}\left[\left.{\frac {y^{3/2}}{3/2}}\right|_{x^{2}}^{1}\right]dx={\frac {2}{3}}\int _{-1}^{0}(1-x^{3})dx=\left.{\frac {2}{3}}(x-x^{4}/4)\right|_{-1}^{0}={\frac {5}{6}}

Have fun.(Igny (talk) 00:05, 10 March 2008 (UTC))[reply]

Since

x<0

,

{\sqrt {x^{2}}}=-x

and thus

(x^{2})^{3/2}=-x^{3}\;\!

rather than

x^{3}

.

{\tfrac {1}{2}}

is correct. -- Meni Rosenfeld (talk) 00:17, 10 March 2008 (UTC)[reply]

Oh well, that was fast. I didn't hide it well enough. (Igny (talk) 00:22, 10 March 2008 (UTC))[reply]

In general, the value of a double integral may depend on the order you do the integration (though not in this case) 163.1.148.158 (talk) 10:27, 10 March 2008 (UTC)[reply]

Can you give an example? Our Multiple integral doesn't say a lot about this, but I recall that the equality holds under fairly mild conditions. -- Meni Rosenfeld (talk) 10:38, 10 March 2008 (UTC)[reply]

From Hilary Priestley's book: the function

{\frac {x-y}{(x+y)^{3}}}

on the unit square will do it. It's pretty clear what's going to happen I think. Or on the same region,

{\frac {x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}}

. ~~Even $e^{-xy}$ on $[0,1]\times [1,\infty )$ may do it.~~ These are examples when Fubini's theorem tells you the function is not in

L(\mathbb {R} ^{2})

. One thing they tell you is that a function can look fairly innocuous, and still fail to be integrable over a compact region. Edit: a sufficient condition, from Fubini/Tonelli, for the double integrals of

f

to be equal is that one of the repeated integrals of

|f|

exists.163.1.148.158 (talk) 11:01, 10 March 2008 (UTC)[reply]

Another example, from Reed and Simon, is

f(x,y)

defined on

\mathbb {R} ^{2}

as,

f(x,y)={\begin{cases}1,&x>0,\quad y>0,\quad 0\leq x-y\leq 1\\-1,&x>0,\quad y>0,\quad 0<y-x\leq 1\\0,&{\mbox{otherwise}}\end{cases}}

.

This function is pretty easy to visualize. It is only non-zero in the first quadrant, between the lines

y=x-1

and

y=x+1

. It is 1 between

y=x+1

and

y=x

, and -1 between

y=x-1

and

y=x

. It's not hard to check that

|f(x,y)|

is not integrable, and that the double integrals are different. 134.173.93.127 (talk) 06:07, 11 March 2008 (UTC)[reply]

Question edit

What is the correct pronounciation of "kilometre": "ki-loh-mee-tre" or "ki-lo-ma-ta"? 58.168.209.250 (talk) 01:20, 10 March 2008 (UTC)[reply]

You are likely to receive more helpful responses at Wikipedia:Reference_desk/Language. Michael Slone (talk) 01:29, 10 March 2008 (UTC)[reply]

I'd go for kill-om-e-tur. -mattbuck (Talk) 09:32, 10 March 2008 (UTC)[reply]

Some think it should be keel-o-meet-ur or keel-om-eat-ur.87.102.94.48 (talk) 15:56, 10 March 2008 (UTC)[reply]

And some pronounce it "stupid" :-) --Carnildo (talk) 19:49, 10 March 2008 (UTC)[reply]

Kill-'em-eat-her? —Keenan Pepper 20:00, 10 March 2008 (UTC)[reply]

Kill-'im, eat her - the cannibals wedding...87.102.94.48 (talk) 22:31, 10 March 2008 (UTC)[reply]

As he said, you're more likely to receive a helpful response at the Language desk. :) Black Carrot (talk) 00:20, 11 March 2008 (UTC)[reply]

Try this and this link (available at Merriam-Webster online dictionary through two red loudspeaker icons at this page). --CiaPan (talk) 15:12, 11 March 2008 (UTC)[reply]

Rings edit

I was wondering why mathematical rings are called "rings"? I can't think of any way in which rings are more "ringlike" than other algebraic systems. What's the history behind the name? Thanks. --Bmk (talk) 04:55, 10 March 2008 (UTC)[reply]

According to Ring theory#History, "The term ring (Zahlring) was coined by David Hilbert in the article Die Theorie der algebraischen Zahlkörper, Jahresbericht der Deutschen Mathematiker Vereinigung, Vol. 4, 1897." —Bkell (talk) 06:17, 10 March 2008 (UTC)[reply]

Ah, thanks - I was looking in the Ring (mathematics) article, and I didn't notice the article on Ring theory. Anyone know why Hilbert called them rings? I don't think I have access to that article. --Bmk (talk) 06:54, 10 March 2008 (UTC)[reply]

A review of the English edition of Hilbert's article[1] contains the phrase: "even though Hilbert uses the word "(Zahl)ring" for orders on algebraic number fields, this must not be taken as evidence that Hilbert employs here parts of our current algebraic terminology the way we would do it; rather than referring to a general algebraic structure, the word "ring" is used for sets of algebraic integers which form a ring in our modern sense of the word." I'm not quite sure what to make of this; it sounds a bit like the statement that the works of Shakespeare were actually not written by Shakespeare but by another person of the same name. It also does not clarify why Hilbert chose to use the word "Zahlring" for these sets of algebraic integers, but it may be a piece in the puzzle. My first speculation on reading the question was that it might have something to do with the cyclic structure of the rings Z/nZ for n > 1, but that is less likely in view of the quotation. --Lambiam 08:50, 10 March 2008 (UTC)[reply]

Dictionary.com doesn't have an etymology on it, but I'm impressed they even have the definition. Black Carrot (talk) 09:25, 10 March 2008 (UTC)[reply]

Searching for -ring group etymology- on Google, however, does turn up this, where he says, "Short for Zahlring (German for number ring). Think of Z[2^(1/3)]. Here the generating element loops around like a ring." Black Carrot (talk) 09:26, 10 March 2008 (UTC)[reply]

Sounds like that's probably the explanation for the name. Thanks folks! --Bmk (talk) 17:35, 10 March 2008 (UTC)[reply]

f(x)=1^x&g(x)=(-1)^x edit

I have to questions here,is the function,f(x)=1^x,afixed point function?what is the value of g(x)if ,x=an irrational number like,2^1/2?thank you.Husseinshimaljasimdini (talk) 13:36, 10 March 2008 (UTC)[reply]

Exponentiation over the complex numbers is inherently a multivalued function. In some cases we can choose a nice branch and it will be single-valued; in other cases we cannot. For

1^{x}

the obvious choice of branch is

f(x)=1

which is a constant function. For

(-1)^{x}

there is no such obvious choice. Its values are

(-1)^{x}=(e^{(2k+1)\pi i})^{x}=e^{(2k+1)x\pi i}=\cos((2k+1)x)+i\sin((2k+1)x)\;\!

For

k\in \mathbb {Z}

. -- Meni Rosenfeld (talk) 14:00, 10 March 2008 (UTC)[reply]

Learning Calculus fast edit

Anyone have a strategy of getting a decent understanding of Calculus within about a months time? 131.91.80.75 (talk) 15:46, 10 March 2008 (UTC)[reply]

Get a decent tutor.(Igny (talk) 16:20, 10 March 2008 (UTC))[reply]

I grabbed my Additional Mathematics textbook and did every single problem in it but I don't think it's for everyone. x42bn6 Talk Mess 16:27, 10 March 2008 (UTC)[reply]

If you’re just looking for the concept as to what it is, and not the computational ability for doing stuff with it, I’d recommend the book “Calculus for cats.” It’s short, and uses a bit of humor, and I found it to be exceptional at getting across the concepts of calculus. Again, though, it is not a textbook: it’s more for somebody that’s curious about calculus, but that doesn’t actually plan to use it, or a supplement to a textbook to explain concepts that textbooks don’t do well. GromXXVII (talk) 23:18, 10 March 2008 (UTC)[reply]

Special functions edit

Consider:

\,{d \over dx}f(x)=f(x)

$f(x)=e^{x}\,$

The function is its own derivative.

But what about:

${\frac {|f''(x)|}{(1+(f'(x))^{2})^{3/2}}}=f(x).$

The function is its own curvature????? --Goon Noot (talk) 21:43, 10 March 2008 (UTC)[reply]

Well yes. But you don't really have that function as yet.. only a differential equation - I wonder what solution(s) would look like......87.102.94.48 (talk) 22:29, 10 March 2008 (UTC)[reply]

Yes, I think that's the question that's being asked. We all know a function which is equal to its derivative, but is there a function which is equal to its curvature? The answer: I have no idea! --Tango (talk) 22:40, 10 March 2008 (UTC)[reply]

Of course there is a function (infinitely many, actually). There's the trivial zero function; another starts with

1+{\frac {t^{2}}{2}}+{\frac {t^{4}}{6}}+{\frac {79t^{6}}{720}}+\cdots

. Whether it has a closed form is another matter entirely. Does anyone know of a version of Plouffe's inverter which gives a function based on its Taylor coefficients? -- Meni Rosenfeld (talk) 23:05, 10 March 2008 (UTC)[reply]

The zero function would be a dot at (0,0)? There's also a straight line at y=infinity. Is that right?87.102.14.194 (talk) 10:47, 11 March 2008 (UTC)[reply]

Well, yes, I was ignoring the 0 function. Any idea what the radius of convergence is for that power series? A function that's defined over the whole real line/complex plane would be nice. --Tango (talk) 00:34, 11 March 2008 (UTC)[reply]

I have no idea about this question other than if it has a full Taylor series (i.e., a positive radius of convergence), I would try to express the problem of finding it as one of recurrence relations. But otherwise no idea. Neat question though. Baccyak4H (Yak!) 01:56, 11 March 2008 (UTC)[reply]

I tried entering a power series expansion, and amazingly enough it turned ugly within the first couple of terms. If a_0 and a_1 are the constant and linear coefficients respectively (and are used to take care of the two degrees of freedom the DE allows) the expansion starts with (assuming I got the algebra right):

$a_{0}+a_{1}x+{\tfrac {1}{2}}a_{0}\left(1+a_{1}{}^{2}\right)^{\tfrac {3}{2}}x^{2}+{\tfrac {1}{3}}a_{1}\left(1+a_{1}{}^{2}\right)^{\tfrac {3}{2}}\left(1+3a_{0}\left(1+a_{1}{}^{2}\right)^{\tfrac {1}{2}}\right)x^{3}+{\mathcal {O}}\left(x^{4}\right)$

Don't ask me what the radius of convergence is like, though. I'm going to go out on a limb and suggest that the function doesn't have a nice closed form. However, like Baccyak4H says, you could always get the recurrence relation going so you at least know something about the coefficients. Confusing Manifestation(Say hi!) 03:53, 11 March 2008 (UTC)[reply]

It occurred to me that it might be more useful to work with the parametric form

y^{2}(x'y''-y'x'')^{2}=(x'^{2}+y'^{2})^{3}\,\!

together with some extra constraint to fix the parameterization. With x'=1 you get the f(x) form above. With y'=1 you get

y^{2}x''(y)^{2}=(x'(y)^{2}+1)^{3}\,\!

, which has a closed-form solution for x' (not x):

x'(y)=\pm {\frac {\ln(y/y_{0})}{\sqrt {1-[\ln(y/y_{0})]^{2}}}}=\pm \tan \sin ^{-1}\ln(y/y_{0})

. Numerically integrated and plotted sideways (x(y),y) it looks like this:

for the case y₀ = 1, x(1) = 0. Other values of y₀ and x(y₀) just scale and translate it. I assume the general solution can be assembled from pieces of this. -- BenRG (talk) 16:10, 11 March 2008 (UTC)[reply]

Oops, I think I got this backwards—f was supposed to be the reciprocal of its radius of curvature. In that case the parametric form is

(x'y''-y'x'')^{2}=y^{2}(x'^{2}+y'^{2})^{3}\,\!

and I get

x'(y)=\pm {\frac {y^{2}+C}{\sqrt {4-(y^{2}+C)^{2}}}}=\pm \tan \sin ^{-1}{\tfrac {1}{2}}(y^{2}+C)

. This does seem to have an antiderivative in terms of elliptic integrals, but it's fairly nasty and I'm not sure if it works for all C. I think Meni Rosenfeld's series is the case C = −3. For C > 0 you get a sinusoidal curve that looks like it could repeat across all of

\mathbb {R}

. -- BenRG (talk) 18:09, 11 March 2008 (UTC)[reply]

I'm guessing that the sinusoidal curve is very roughly like the cycloid curve, except more curvy at y=zero .. anyway how about using f(x)=a0+a1 sin x + a2 sin 2x +etc and attempting to solve in a similar fashion to the methods above that gave the first few terms easily.. 87.102.74.53 (talk) 19:07, 11 March 2008 (UTC)[reply]

Might just be me but I'm getting x'³(x'y''-y'x'')=y²(x'²+y'²)³ (where x' = dx/dt) - can someone point out where and if I'm,going wrong.. kindly please.87.102.74.53 (talk) 18:03, 11 March 2008 (UTC)[reply]

I don't think this can be right because it's not symmetric in x' and y'. -- BenRG (talk) 18:09, 11 March 2008 (UTC)[reply]

I just used the equation for curvature and put in the parametric derivatives.. maybe I made an obvious mistake ~~- could you give a first step of what you did so I know I'm not barking up the wrong tree?87.102.74.53 (talk) 18:13, 11 March 2008 (UTC)~~[reply]

~~OOPS sorry I get x'³(x'y''-y'x'')²=y²(x'²+y'²)³.. hang on a minute —Preceding unsigned comment added by 87.102.74.53 (talk) 18:15,~~ ~~11 March 2008 (UTC)~~

Ignore (boloks) I get the same, msut be drunk or getting old ignore previous87.102.74.53 (talk) 18:18, 11 March 2008 (UTC)[reply]

Mathematica churns for a few minutes, spits out several warnings exhorting us (humans) to check the answer by hand and various other terrible diagnostics, then says:

       {{y[x] ->    InverseFunction[(\[ImaginaryI] Sqrt[
        2] ((1 + 
             C[1]) EllipticE[\[ImaginaryI] ArcSinh[
              Sqrt[1/(2 - 2 C[1])] #1], (-1 + C[1])/(1 + C[1])] - 
          EllipticF[\[ImaginaryI] ArcSinh[
             Sqrt[1/(2 - 2 C[1])] #1], (-1 + C[1])/(
           1 + C[1])]) Sqrt[(2 + 2 C[1] - #1^2)/(1 + C[1])] Sqrt[(
        2 - 2 C[1] + #1^2)/(1 - C[1])])/(Sqrt[1/(1 - C[1])] Sqrt[
        2 - 2 C[1] + #1^2]
         Sqrt[-2 (1 + C[1]) + #1^2]) &][-\[ImaginaryI] x + 
    C[2]]}, {y[x] -> 
  InverseFunction[(\[ImaginaryI] Sqrt[
        2] ((1 + 
             C[1]) EllipticE[\[ImaginaryI] ArcSinh[
              Sqrt[1/(2 - 2 C[1])] #1], (-1 + C[1])/(1 + C[1])] - 
          EllipticF[\[ImaginaryI] ArcSinh[
             Sqrt[1/(2 - 2 C[1])] #1], (-1 + C[1])/(
           1 + C[1])]) Sqrt[(2 + 2 C[1] - #1^2)/(1 + C[1])] Sqrt[(
        2 - 2 C[1] + #1^2)/(1 - C[1])])/(Sqrt[1/(1 - C[1])] Sqrt[
        2 - 2 C[1] + #1^2]
         Sqrt[-2 (1 + C[1]) + #1^2]) &][\[ImaginaryI] x + C[2]]}}

Enjoy, Robinh (talk) 08:48, 11 March 2008 (UTC)[reply]

Could someone convert that back into maths? Or maybe not - it really looks like mathematica(TM) has 'gone insane' over this question - specifically 'square root insanity' would be my diagnosis... Poor old computer.87.102.14.194 (talk) 08:57, 11 March 2008 (UTC)[reply]

What does the graph look like????--Goon Noot (talk) 10:06, 11 March 2008 (UTC)[reply]

If you mean meni rosenfeld's answer .. it looks like a 'steep' parabola, or a cosh function.. that sort of shape (assuming the curvature is always considered positive ie magnitude.87.102.14.194 (talk) 10:35, 11 March 2008 (UTC)[reply]

Someone asked about a "A function that's defined over the whole real line/complex plane" technically menirosenfeld's function (as I'm now calling it) or in general functions that satisfy the equations given by Confusing Manifestation will work for imaginary numbers. But what if the complex function used to describe a scalar value ie f(x)=a+ib g(x)=sqrt(aa+bb) and the curvature of that scalar at x is equal to the the scalar at x.. That's impossible to analyse right? or am I missing some more maths to learn?87.102.14.194 (talk) 11:02, 11 March 2008 (UTC)[reply]

What does complex curvature mean??--Goon Noot (talk) 15:48, 11 March 2008 (UTC)[reply]

difficult to give a 'real world' equvivalent - as curvature is as you know the radius of a circle corresponding to the rate of change of slope at a point, then a complex curvature means the circle has a complex radius.. If the slope is changing complexely (ie the complex part is changing) then the radius of curvature will have a complex coefficient. Obviously this won't happen if x and y are always real.. Did that explain at all, oe help?87.102.74.53 (talk) 18:09, 11 March 2008 (UTC)[reply]

Is it just me or Jakob Bernoulli's "Spira Mirabilis" is its own evolute? So the equation

r=e^{a\theta }

is this curve that he is asking about, right?A Real Kaiser (talk) 05:02, 12 March 2008 (UTC)[reply]

Mmh I assumed they meant radius of curvature = y (rectangular), but if radius or curvature = r (polar) you are probably right.(or very close) luckily Spira_mirabilis#Properties saves me the bother of having to work it out, you are right it is indeed its own evolute.87.102.17.32 (talk) 13:40, 12 March 2008 (UTC)[reply]

As far as I can tell im certain that the radius of curvature of (sin(x)e^x,cos(x)e^x) is not e^x and can't find an exact solution87.102.17.32 (talk) 17:21, 12 March 2008 (UTC)[reply]

The functions e^z, e^kz have (eg the function used as radius when z is angle in polar coords) have radius of curvature sqrt(2)e^z, sqrt(1+k²)e^kz ie the radius of curvature is always proportional to the function, but not the same.. 87.102.32.239 (talk) 23:15, 12 March 2008 (UTC)[reply]

To GOON NOOT - I realised the answers I've been giving were based on radius of curvature not curvature - ie the reciprocal, apologies for any confusion.87.102.17.32 (talk) 16:15, 12 March 2008 (UTC)[reply]

One way of looking at the problem is to consider a parametric representation of the curve, in which the point (x(s), y(s)) is a function of the arclength s. If we also introduce φ(s), giving the direction of the curve expressed as the angle of the tangent with the x-axis, then, writing x =x(s) etc. as usual,

{\frac {dx}{ds}}=\cos \varphi ,\,\,\,{\frac {dy}{ds}}=\sin \varphi ,\,\,\,{\frac {d\varphi }{ds}}=\kappa ,\,

where κ is the curvature, which may itself be a function of s, x, y, and φ. Here it is given that |κ| = y, which is a bit indeterminate: whenever y reaches 0, in general two different continuations are possible.

Assuming κ = y, we have

{\frac {dy}{d\varphi }}={\frac {\sin \varphi }{y}},

which is separable and can be solved to give

{\tfrac {1}{2}}y^{2}+\cos \varphi =C.\quad (\ast )

This is the generic form of the integral curves of the vector field assigning to the point (φ,y) the directional vector (y,sin φ). Each value of C > −1 in (*) gives one curve. Assume we start in an area where −1 < C < 1. By inspection of the vector field diagram, we see anti-clockwise cycles around (π,0), which correspond to a sinus-like curve in the x-y-plane wiggling around the x-axis in the negative direction. As we move away from φ = π and pass φ = π/2 (or φ = 3π/2), the wiggling gets more pronounced and becomes like meandering.

If we start in a position with C > 1, y cannot vanish, so the sign of y is invariant. According to the sign, φ will monotonically increase or decrease, and in the x-y-plane we get to see cycles. I haven't analyzed the critical case C = 1.

Another approach I have not further looked into is to put v = dy/ds, so that on the one hand

{\frac {d^{2}y}{ds^{2}}}={\frac {dv}{ds}}={\frac {dy}{ds}}\,{\frac {dv}{dy}}=v{\frac {dv}{dy}},

while also

{\frac {d^{2}y}{ds^{2}}}={\frac {d\,\sin \varphi }{ds}}=\cos \varphi \,{\frac {d\varphi }{ds}}=\pm y{\sqrt {1-v^{2}}},

.

which combine to give another separable equation:

v{\frac {dv}{dy}}=\pm y{\sqrt {1-v^{2}}}.

I hope I did not make many mistakes, since this was scribbled on a (too small) napkin. --Lambiam 18:12, 13 March 2008 (UTC)[reply]

division by zero edit

We all learn in elementary school that any number divided by itself is 1. Later, we learn that division by zero is undefined. I wondered why it is undefined, because if any number divided by itself is one, shouldn't $0/0$ be 1? I thought about this...

Consider $f(x)=r/x$ where r is any real number.
As lim_x→0⁺, x_→+∞
Also, as lim_x→0^–, x_→-∞

So does $0/0$ =+∞ and -∞?
J.delanoy ^gabs _adds 23:28, 10 March 2008 (UTC)[reply]

We have an article on Division_by_zero. Black Carrot (talk) 23:29, 10 March 2008 (UTC)[reply]

Sorry, I didn't realize that... J.delanoy ^gabs _adds 23:31, 10 March 2008 (UTC)[reply]

No problem at all. Feel free to expand on your question if the article isn't detailed enough. Black Carrot (talk) 00:06, 11 March 2008 (UTC)[reply]

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