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June 2

Integration

$\int {\frac {dx}{\sqrt {1-x^{-1}}}}$

I got ${\frac {1}{2}}\ln |2x{\sqrt {1-x^{-1}}}+2x-1|+x{\sqrt {1-x^{-1}}}$

but Mathematica tells me it's ${\sqrt {x^{2}-x}}+\ln({\sqrt {x-1}}+{\sqrt {x}})$

I found these to be the same function by graphing. I understand how integrals can be in different forms and are still equal. This is true in many trigonometric integrals, where two functions may look entirely different, but are found to be equal or to differ by a constant when some identities are used to simplify. How are these equivalent? I couldn't find a way to simplify either one to make it look like the other expression. And also, how did mathematica arrive at a simpler answer? --Yanwen (talk) 02:19, 2 June 2009 (UTC)[reply]

That two trigonometric expressions look different but are equal is the phenomenon known as trigometric identities.

I'm quite rushed now. More later..... Michael Hardy (talk) 02:35, 2 June 2009 (UTC)[reply]

Oh, I just noticed: this one's easy: the square of

{\sqrt {x-1}}+{\sqrt {x}}\,

is

2x{\sqrt {1-x^{-1}}}+2x-1\,

(just multiply out the square and then collect like terms). Then remember that

{\frac {1}{2}}\log(a)=\log({\sqrt {a}})\,

etc. And

{\sqrt {x^{2}-x}}={\sqrt {x}}{\sqrt {x-1}}={\sqrt {x}}({\sqrt {x}}{\sqrt {1-x^{-1}}}).\,

Michael Hardy (talk) 02:39, 2 June 2009 (UTC)[reply]

babyloian quadratic equations

what is the babylonian quadratic equation, used to solve simultaneous equations? what was their logic to solve quadratic equations in their times? —Preceding unsigned comment added by Thee madhav (talk • contribs) 06:54, 2 June 2009 (UTC)[reply]

There's a description of the Babylonian approach to quadratic equations, as represented by the tablet YBC 6967, in our article on a different tablet, Plimpton 322, which seems to have been a table of quadratic-equation solving exercises and their solutions. —David Eppstein (talk) 07:12, 2 June 2009 (UTC)[reply]

Conic section ellipse

Conic section says one of the conics is an ellipse (not an oval). I know oblique section of a cylinder is an ellipse. Thinking from there, the section of a cone seems to be an oval, because a cone is in a shape of a cylinder augmenting to the bottom (If one end of an ellipse is augmented, it is likely to be an oval. I tried cutting a cone made with paper, and that resulted in subtle, but like a tear drop shape (have I done it wrong?) Like sushi (talk) 10:53, 2 June 2009 (UTC)[reply]

Yep you've got it wrong I'm afraid. You might like the Dandelin spheres article. Dmcq (talk) 11:29, 2 June 2009 (UTC)[reply]

If you want to see the conic sections, project a cone of light on the wall.--194.95.184.202 (talk) 11:32, 2 June 2009 (UTC)[reply]

(EC) Take a simple conic surface

z={\sqrt {x^{2}+y^{2}}}

and cut it with a slope plane

z=s\cdot x+h

. Of course the plane's slope must be less than the cone's slope:

|s|<1\,\!

. Get rid of z and you get

{\sqrt {x^{2}+y^{2}}}=s\cdot x+h\,\!

then

x^{2}+y^{2}=s^{2}x^{2}+2shx+h^{2}\,\!

and

(1-s^{2})x^{2}-2shx+y^{2}=h^{2}\,\!

which is an ellipse equation. Of course this ellipse is a projection of the intersection curve onto the

XY\,\!

plane, but if you plug

x=x^{\prime }/{\sqrt {1+s^{2}}}

then you'll get an equation of the intersection curve in

X^{\prime }Y

coordinates of the intersecting plane, that is a true intersection curve equation. And that will also be an ellipse. --CiaPan (talk) 11:34, 2 June 2009 (UTC)[reply]

If you want to try it experimentally, do not make a conic surface with a paper sheet, as it gets deformed in scissors, so at the moment you cut it, it's no longer a cone! Make a cone solid with putty, plasticine or polymer clay, then cut it with a knife. --CiaPan (talk) 11:42, 2 June 2009 (UTC)[reply]

I tried putting clay into a paper cone. It resulted in an almost perfect ellipse!

But I don't know why. I can not understand the formulas Mr. or Ms CiaPan has shown. I don't even know why the ellipse equation is like that (In Conic section#Parameters, there is "equation", but I don't know if that's it, nor why it is like that). What I know about ellipses is only that they are extended circles.

Like sushi (talk) 13:43, 2 June 2009 (UTC)[reply]

Hey, let the scissors alone, you may hurt yourself... look at a cone of light on the wall instead ;-) --194.95.184.202 (talk) 15:00, 2 June 2009 (UTC)[reply]

I don't have an appropriate light source. I tried with a handy light with a bulb. The light was so ununiform that I could not see the boundary clearly.

Like sushi (talk) 02:01, 3 June 2009 (UTC)[reply]

If I understand your doubt, you are arguing that an oblique section of a cone should look like a section of a cylinder with small radius near the origin of the cone, while it should look like a section of a cylinder with a larger radius in the opposite extremity: therefore you would expect an oval shape, that is, one narrow extremity near the origin, the other with larger radius of curvature. BUT the point is that the angle of incidence is different: it is true that you can compare with oblique section of a cylinder, but not only the radius varies, the angle too. The effect is that you get a perfect ellipse, as shown above (or with no computations: the section is a bounded quadratic curve, hence it is an ellipse, qed) --194.95.184.202 (talk) 15:16, 2 June 2009 (UTC)[reply]

You understand my doubt perfectly. What angle is the "angle of incident"? "A bounded quadratic curve" is out of my reach.

Like sushi (talk) 02:01, 3 June 2009 (UTC)[reply]

I just mean the angle that the cutting plane does with the surface of the cone, like in figure 2 here. You may well compare a small portion of the surface of the cone around a point on the section curve with a similar small portion of a cylinder, but you have to take into account both radius and angle. If you look at the picture, at the two extrema of the ellipse you have : on the right, large radius, shaving cutting; on the left, small radius, almost normal cutting. I told this just as a qualitative argument to say that you don't need to expect an asymmetry. To prove that the curve is actually an ellipse of course require a proof. The other thing is a triviality: imagine a cone in a 3 dimensional Cartesian coordinate system. The key point is that its equation is P(x,y,z)=0 with P a polynomial of degree 2 in x,y,z. The intersection with the plane z=0 has of course equation P(x,y,0)=0, that is, it is a polynomial of degree 2, or less, in x and y. If you already know that these curves are ellipses/parabolas/hyperboles plus degenerate cases, you also know what you will get as intersection.--194.95.184.202 (talk) 07:42, 3 June 2009 (UTC)[reply]

If a waterproof open conical vessel is available (it doesn't need to have a sharp point, a tapering vase would do), study the shape of the water surface as the vessel is tilted. When it is upright a circle is seen, the more the tilt the more eccentric the indubitable ellipse is. Incidentally, here's a nice exercise - consider a fixed volume of water in an inverted true cone; as the cone is tilted, up to the point of spillage all resulting ellipses have equal minor axis (obviously the same value as the diameter of the "upright" circle).86.146.175.44 (talk) 15:25, 2 June 2009 (UTC)[reply]

Dandelin spheres!

If you can't understand the algebraic equations given in response to your question about ellipses, then do look at the article titled Dandelin spheres. Michael Hardy (talk) 15:28, 2 June 2009 (UTC)[reply]

I have looked at Dandelin spheres. I see that, in "Proof that ellipse has constant sum of distances to foci", the article is provong that the section is an ellipse, by showing the sum of distance from two points to the point P on the shape is constant. I understand the approach, but I don't understand the proof. One of the obstacles is that I don't see what

F_{i}

and ,

P_{i}

mean there. Another is although I know pins-and-string method produces an ellipse from experience, it is a surprise to me (I don't see why).

Like sushi (talk) 02:01, 3 June 2009 (UTC)[reply]

The concept of ellipse is often defined by saying it's the set of points for which the sum of the distances from two fixed points is a fixed constant. You're saying you don't know why that's an ellipse? You must be using some other definition. What is it? Michael Hardy (talk) 06:36, 3 June 2009 (UTC)[reply]

The only definition that seems clear to me of an ellipse is that it is an extended circle.

Like sushi (talk) 07:44, 4 June 2009 (UTC)[reply]

BTW, as to the name "Dandelin spheres", it seems to me another small historical distortion by Moderns against the Ancients. Proofs by Pappus and Apollonius are perfectly clear and rigorous, and if something is less clear, that's just because after more that 2000 years notations style and language have changed. I'd like to see Mr Dandelin brought back in 300bc to explain the "originality" of his spheres to Pappus or Apollonius. --pma (talk) 08:03, 3 June 2009 (UTC)[reply]

F_{i}

and ,

P_{i}

mean F1 P1 or F2 P2 in the picture. F1 and F2 are where the plane touches the small sphere and the big sphere. The distance from P to P1 is the same as the distance from P to F1 because both lines are tangent to the small sphere. Anyway answering your original question you can see that an ellipse defined by the pins and strings method cannot be an oval and fatter at one end because the pins are treated exactly the same as each other. Dmcq (talk) 09:35, 3 June 2009 (UTC)[reply]

Thank you. I understand now that F1 P + F2 P is constant and that it is not an oval. I also understand that it is reflectionally symmetrical with two perpendicular axes

Like sushi (talk) 07:44, 4 June 2009 (UTC)[reply]

Looking for a textbook on solid geometry

Hi, I'm looking for a good textbook on solid geometry, preferably with problems and answers. I'd like it to cover stuff like proving that the shortest surface distance between any two points on a sphere lies along a great circle between them. Also conic sections and intersections between, say, planes and various solids. Any suggestions welcome, thanks, It's been emotional (talk) 13:51, 2 June 2009 (UTC)[reply]

I unfortunately can't find the specific book I used in my first undergraduate course, but you might want to broaden your search and take a look in the CAD (Computer Aided Design) or Computer Graphics section of your library (or internet querying). These subjects often treat geometry with a very practical combination of analytic and numerical methods, while a pure mathematics book will probably be much more analytical. I'm also holding my copy of Curves and Surfaces for CAGD (Computer Aided Geometric Design), by Gerald Farin (~$75 @ Amazon), which treats affine maps, splines, and surfaces in a variety of contexts. It does do a few analytic proofs, but since you specifically are looking for solid geometry, it may be less relevant (it deals largely with surface representations). I had another really great introductory book on Constructive Solid Geometry CAD, but I'm not able to locate it right now... Nimur (talk) 14:39, 2 June 2009 (UTC)[reply]

Functional Analysis

Engaged in a seminar about functional anlysis, we came across a problem. Let us consider the space $C^{1}[a,b]$ which is the set of all real continuously differentiable functions on the interval [a,b] given. And define the functional

$f(x(t))=x'(c)=x'\left({\frac {a+b}{2}}\right)$

on this space. So the functional maps a function x to its derivative evaluated at the midpoint of the interval. On this set, define the norm to be

$||x(t)||=\max _{t\in [a,b]}|x(t)|+\max _{t\in [a,b]}|x'(t)|$

and define the norm of a functional on this space to be

$||g||=\sup _{x\neq 0}{\frac {||g(x(t))||}{||x(t)||}}=\sup _{||x||=1}||g(x(t))||.$

Now basically I am trying to determine the norm of my above defined functional f. What I have done is that I have shown that this norm is bounded, meaning

$||f(x(t))||=|x'(c)|\leq ||x(t)||$

so the constant is one in this upper bound. Now I know that the norm of f is less than or equal to one. My guess is that the norm of f is exactly one but how can I show that the norm of f is exactly equal to one? If it is not exactly one, then what is it and how can I find it? Is there a lower bound I can put on the norm at least? Can I say anything about the norm of f? Can the norm even be exactly determined? Thanks!-Looking for Wisdom and Insight! (talk) 23:26, 2 June 2009 (UTC)[reply]

The norm is indeed 1. Let's construct a function x with

||x||=1

and

|x'(c)|=1-\epsilon

. We'll want x' to assume its max of

1-\epsilon

at c, and we'll need the max of |x| to be very small. Thinking of x as the integral of x' (starting at a), can you think what x' would have to look like to make

|x|\leq \epsilon

? Let me know if you want more hints. J Elliot (talk) 23:52, 2 June 2009 (UTC)[reply]

That was amazing. That is exactly what I needed. I got it. Thanks!-Looking for Wisdom and Insight! (talk) 04:10, 3 June 2009 (UTC)[reply]

What inequality should I use here

As part of some math I am trying to figure out, I need to show that $(1+x)^{n}\geq 1+nx$ for some small $x$ ( $|x|<1$ ) for $n\geq 1$ ; and I am having trouble figuring it out. I feel like there is some famous inequality I should use here. I know that the right side is the first two terms of the binomial expansion of the left side, but that is only a convincing argument if $x$ is positive; when $x$ is negative, I know that the inequality is still true (from graphing it), but the binomial expansion argument isn't that simple anymore because the binomial expansion contains negative terms also, and so it is harder to say that the contribution from those terms will add up to a positive number. Other simple things I can do, like $1+x\leq e^{x}$ , don't work because the inequality is going the wrong way. I know from graphing it that I could argue that the left side is concave up (as a function of $x$ ), and that the right side is the first order Taylor polynomial at 0, and so has to be a lower bound, but that seems way too complicated. Thanks, --131.179.33.215 (talk) 23:38, 2 June 2009 (UTC)[reply]

This is Bernoulli's inequality. —David Eppstein (talk) 01:19, 3 June 2009 (UTC)[reply]

OK, let's try mathematical induction:

{\begin{aligned}&{}\quad (1+x)^{n+1}=(1+x)^{n}(1+x)>(1+nx)(1+x)\\&{}=(1+nx)+x(1+x)>(1+nx)+x=1+(n+1)x\end{aligned}}

There you have it. Michael Hardy (talk) 06:31, 3 June 2009 (UTC)[reply]

Don't forget the (trivial) proof for the n=1 case. --Tango (talk) 16:42, 4 June 2009 (UTC)[reply]

Alternate proof: Define $f(x)=(1+x)^{n}-(1+nx)$ and show that:

$f(0)=0$
$f'(x)=n\left((1+x)^{n-1}-1\right)\geq 0$ for $x\geq 0$ , which means that $f(x)$ is a monotonically increasing function for non-negative $x$ .
Similarly show that $f(x)$ is a decreasing function for $-1\leq x\leq 0$ .

Combine (1)-(3) to get the result. Note that this proof does not rely on $n$ being an integer, and the proof technique works for many other inequalities, including $1+x\leq e^{x}$ . Abecedare (talk) 07:18, 3 June 2009 (UTC)[reply]