# Wikipedia:Reference desk/Archives/Mathematics/2007 December 17

Mathematics desk
< December 16 << Nov | December | Jan >> December 18 >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

# December 17

## Geometry question

I friend of mine gave me a problem which he could solve, that pretty much amounted to this :

I have a circle of fixed radius r and another circle of variable radius R, with center on the first circle.

I want to find for which value of R is the area of the intersection of the two circles half the area of the first circle.

I did a little diagram : http://img518.imageshack.us/img518/7262/probsw7.png

First I calculated the two points of intersection of the circles :

$x^2 + y^2 - R^2 = x^2+(y-r)^2-r^2 \implies -R^2=-2yr \implies y=\frac{R^2}{2r}$

Then $x^2+y^2=R^2 \implies x^2 + \frac{R^4}{4r^2} = R^2 \implies x = \pm R \sqrt{1-\frac{R^2}{4r^2}}$

Then I said : area of intersection = area of red part + area of blue part

Area of red part = area of sector FEI - area of triangle FEI

Area of sector FEI = $\frac{\angle{FEI}}{2} r^2 =\angle{FEP}\cdot r^2$

But $\cos\left(\angle{FEP}\right) = \frac{EP}{EF}=\frac{r-\frac{R^2}{2r}}{r}$ so $\angle{FEP} = \arccos \left(1-\frac{R^2}{2r^2}\right )$

So area of sector FEI = $r^2\arccos\left ( 1-\frac{R^2}{2r^2}\right )$

Area of triangle FEI = FP * EP = $R \sqrt{1-\frac{R^2}{4r^2}} \left(r-\frac{R^2}{2r} \right)$

Then : area of blue part = area of sector FOI - area of triangle FOI

Area of sector FOI = $\frac{\angle{FOI}}{2} r^2 = \angle{FOP}\cdot r^2$

But $\cos\left(\angle{FOP}\right) = \frac{OP}{OF}=\frac{\frac{R^2}{2r}}{R}$ so $\angle{FOP} = \arccos \left(\frac{R}{2r}\right )$

Then area of sector FOP = $R^2 \arccos \left(\frac{R}{2r}\right )$

Area of triangle FOI = FP * OP = $R \sqrt{1-\frac{R^2}{4r^2}} \frac{R^2}{2r}$

Adding everything together, the $R \sqrt{1-\frac{R^2}{4r^2}} \frac{R^2}{2r}$ cancel, we get :

Area of intersection = $r^2\arccos\left ( 1-\frac{R^2}{2r^2}\right )+R^2 \arccos \left(\frac{R}{2r}\right )-R r \sqrt{1-\frac{R^2}{4r^2}}$

So now I have to solve that = $\frac{\pi r^2}{2}$

But first I think this equation is wrong as I can't seem to find any solutions (numerically that is, let alone analytically).

So where did I make a mistake and how do I solve this in an easier way ? As usual I'm sure there's a much easier and quite obvious way I just didn't think about.

I tried doing it all by giving equations to everything and integrating, it was even worse with really many arcsines and really complicated expressions.

Thanks. --Xedi (talk) 01:32, 17 December 2007 (UTC)

This is backwards: $\frac{\angle{FEI}}{2} r^2 =\frac{\angle{FEP}}{4} r^2$ --tcsetattr (talk / contribs) 03:40, 17 December 2007 (UTC)
Wow thanks that was a silly mistake, corrected. Seems the equation is correct now, though tough to solve. --Xedi (talk) 05:08, 17 December 2007 (UTC)

So, any ideas how to solve this equation or solve the problem in an entirely different way ? --Xedi (talk) 17:07, 17 December 2007 (UTC)

I don't think this is entirely different, but it is a sketch of how I would approach this. First, since it is clearly only the ratio between r and R that matters, I would fix r at 1, and have used E as the origin of the unit circle in the Cartesian plane. Out of habit, perhaps, I usually use the x-axis as the axis of symmetry in symmetric diagrams involving circles around the origin, so that your point O would have been at (R, 0). Fixing instead R = 1, leaving O at the origin and putting E at (r, 0) is equally feasible.
We can confine our attention to one half of the diagram, the other being symmetric. So we have two semicircles, both lying on the x-axis.
Now there are two "pie slices" contained in the region of overlap between the semicircles. One is a slice of the larger semicircle, that with radius R (assumed to be > 1), with apex O and whose boundary is formed by the edges OI and OG and the circular arc IG. The other is a slice of the unit semicircle, with apex E, edges EO and EI, and arc OI.
These two pie slices cover the region of overlap between the semicircles completely, but they themselves overlap, the region of overlap being formed by the equilateral triangle EOI. The area of the region of overlap between the two semicircles is therefore the sum of the areas of the two pie slices, minus the area of the triangle. We want this to come out at half the area of the unit semicircle, which equals π/4.
Instead of taking R as the variable to solve for, I would go with the angle ∠OEI. R is easily expressed in terms of this angle; calling it α, we have R = 2 sin(α/2). The first pie slice has area 1/2R2(π−α)/2, the second 1/2α, and the triangle 1/2sin α.
This comes out numerically to approximately α = 1.23590 radians, and R = 1.15873. Essentially the same solution can be found at the end here (my R is their L).  --Lambiam 18:59, 17 December 2007 (UTC)
All right that works out fine (though I think you meant EOI is isosceles not equilateral).
Is there an expression for this constant in terms of other constants ? If not, has anyone given a name to this constant (or related, like the square of this constant or something) ?
Thanks. -- Xedi (talk) 05:38, 18 December 2007 (UTC)
Comparing your diagram with your written description of the problem, I find the two disagree, your decription says the center of the new circle is on the old circle, where you diagram has the new circle crossing the center of the old one. With this in mind I would check the special case that r=R first before trying a more generalized approach, your general approach will still work, but your answer will be different. A math-wiki (talk) 10:47, 18 December 2007 (UTC)
Sorry, I don't see that. E is the center of the given circle with radius r, and then O is the point on that circle that is the center of the other circle (of radius R).
But you are right that I made some assumptions when I drew the diagram and followed them during the resolution :
$R > r$ for the fact that G is above E, and that D and H actually exist
$R < \sqrt{2} r$ for the fact that D and H are inside the circle of radius r (maybe the resolution still works without this assumption but it's true anyway so why not)
Or were you thinking of something else ? -- Xedi (talk) 15:48, 18 December 2007 (UTC)

## Negative logarithms

A while ago, I was playing around with my calculator, and I discovered this series of equations regarding negative logarithms:

• The one I found most interesting: ln(−$x$) = ln $x$ + π$i$
• log10(−$x$) = log10$x$ + A$i$, where A ≈ 1.36437635384
• log9(−$x$) = log9$x$ + B$i$, where B ≈ 1.42980043369
• log8(−$x$) = log8$x$ + C$i$, where C ≈ 1.51078671394
• log7(−$x$) = log7$x$ + D$i$, where D ≈ 1.61445925708
• log6(−$x$) = log6$x$ + E$i$, where E ≈ 1.75335624426
• log5(−$x$) = log5$x$ + F$i$, where F ≈ 1.95198126583
• log4(−$x$) = log4$x$ + G$i$, where G ≈ 2.26618007091
• log3(−$x$) = log3$x$ + H$i$, where H ≈ 2.85960086738
• log2(−$x$) = log2$x$ + I$i$, where I ≈ 4.53236014182

Is there a theorem for this on Wikipedia? Quin 07:41, 17 December 2007 (UTC)

Yes. This is briefly discussed in logarithm#Generalizations for the natural logarithm, which includes the formula
ln(z) = ln(|z|) + i arg(z)
for any complex number z. Applying the change-of-base formula yields
logb(z) = logb(|z|) + i arg(z) / ln(b),
which agrees with your results above. You may also be interested in complex logarithm, which goes into more detail. Michael Slone (talk) 08:33, 17 December 2007 (UTC)
It's also worth noting that the complex log, as an analytic function, cannot be defined on the entire complex plane, but must exclude a ray with origin at 0. Most commonly, the ray excluded is precisely the nonpositive real numbers. Remember that arg(z) is a multivalued function: You can just as easily get -x by taking eln xi or any other odd integer times πi. 76.90.44.176 (talk) 08:47, 17 December 2007 (UTC)
Since the logarithm is just the inverse of the exponential function, you might find that article interesting. The formula you have found most interesting, $\ln(-x)=\ln x +\pi i$, is a direct consequence of Euler's identity. -- Meni Rosenfeld (talk) 09:25, 17 December 2007 (UTC)
Not that it's useful to know, but Poisson's formula for the logarithm is really nasty to derive. But yes, that ln|x| + argx is a standard analytic form of the logarithm. mattbuck (talk) 12:13, 17 December 2007 (UTC)

## Equation rearrangement

A couple of questions on this subject:

Page 2 says that we can rearrange $a^l + b^l = c^l \,$ to $a^l + b^l - c^l = 0 \,$ which is clear, but then it goes on to say we can replace c with -c for a more "symmetric form". Why? Doesn't this change the equation's original form? Or do we simply at some point need to remember to reverse the transformation in the future?

Page 3 says the aforementioned equation can be graphed, but it graphs the equation $y^2 = x^3 - x \,$ instead. Is this because through rearrangement, $a^l + b^l + c^l = 0 \,$ is the same as $y^2 = x^3 - x \,$, and if so, can someone show me the process? —Preceding unsigned comment added by 91.84.143.82 (talk) 14:06, 17 December 2007 (UTC)

1. $a^l+b^l-c^l=0$ is a different equation from $a^l+b^l+c^l=0$, in the sense that for given values of a, b and c, they have different solutions. However, the family of all equations representable in the form $a^l+b^l-c^l=0$ is the same as the family of all equation representable as $a^l+b^l+c^l=0$. If we are given a particular equation $a^l+b^l-c^l=0$, what we should really do is let $d=-c$, solve $a^l+b^l+d^l=0$ and then take $c=-d$. We can abuse notation to save up on letters and use c instead of d, remembering that this c is different from the original c. We still do need to remember to reverse the transformation in the end.
2. I think the key words are "something like". I don't know if the general equation can be reduced to $y^2=x^3-x$, but the graph of any particular equation should qualitatively look like the figure.
-- Meni Rosenfeld (talk) 15:04, 17 December 2007 (UTC)
That doesn't sound right. $a^l+b^l+c^l=0$, subsituing c = (-c), $a^l+b^l+(-c)^l=0$. Substracing the 2 equations you have $c^l=(-c)^l$ which are only true when l is even integer. NYCDA (talk) 19:21, 17 December 2007 (UTC)
In the context in which this is presented in Adler's lecture from which it is taken, it is known that $l$ is odd, so that $-c^l$ is transformed by the bijective substitution $c:=-c$ into $+c^l$. --Lambiam 19:28, 17 December 2007 (UTC)
[ec]The c is different in each equation. They are not simultaneously satisfied, and hence you cannot subtract them. "$c=-c$" is just an abuse of notation, if you take it literally of course c can only be 0. -- Meni Rosenfeld (talk) 19:30, 17 December 2007 (UTC)
Sorry I should've typed substract $a^l + b^l$ from both equations. NYCDA (talk) 20:40, 17 December 2007 (UTC)
Same answer, the two equations are not satisfied simultaneously, you can't manipulate them as if they were. -- Meni Rosenfeld (talk) 21:04, 17 December 2007 (UTC)
I think there is a typo where Adler talks about graphing the equation (1); he means equation (2):
$y^2 = x(x-a^\ell)(x+b^\ell)\,,$
which is rather like $y^2 = x(x-1)(x+1) = x^3-x$.  --Lambiam 19:38, 17 December 2007 (UTC)

The paper only said the graph looks like. It does not say the 2 equation are the same. Theorem 3 states, n, a, b, c are integers. For n > 2, abc = 0. We know if n = 2 then 3, 4 and 5 will satisfy the equation as here $3^2 + 4^2 = 5^2$. But for n > 2, atleast one of the variable must be 0. The paper states, for n = 3 or 4, it is proven by Euler and Fermat. It then assumes n is a prime number and used l to denote it. So we have $a^l + b^l = c^l$ and rearranged as $a^l + b^l - c^l = 0$ but since l is prime and therefore an odd number, substing c = (-c), you have $a^l + b^l - (-c)^l = 0 = a^l + b^l + c^l = 0$ since we know it's an odd power. NYCDA (talk) 20:12, 17 December 2007 (UTC) (Edit, I changed the wording so that it appears in the right context. I meant to reply to the original poster, not someone else)

What does this have to do with the original questions? -- Meni Rosenfeld (talk) 20:22, 17 December 2007 (UTC)
He also had a question on why c was replaced with -c. I know it was subsequently answere, which unfortunatly I did not see since I was in edit mode then. NYCDA (talk) 20:32, 17 December 2007 (UTC)
Yes, but you seem to have explained why $(-c)^l=-c^l$, which isn't what the OP asked about (but rather what does the transformation $c=-c$ mean). -- Meni Rosenfeld (talk) 21:04, 17 December 2007 (UTC)
I don't understand. By rearranging $a^l + b^l = c^l$ to $a^l + b^l - c^l = 0$, how can we just decide to add $c^l$? That would just mean we would have to add $c^l$ on the other side of the equation. Unless you mean somehow $(-c)^l$ which would be negative for odd $l$ anyway. What the hell is going on? —Preceding unsigned comment added by 91.84.143.82 (talk) 14:04, 18 December 2007 (UTC)
Please ignore everything else and take a look at my very first comment. Is it still unclear? Can you describe which part you do not understand? -- Meni Rosenfeld (talk) 14:11, 18 December 2007 (UTC)
I see what they intend now, thanks. —Preceding unsigned comment added by 91.84.143.82 (talk) 14:31, 18 December 2007 (UTC)

## xmas

Just saw my 'old' maths teacher from school after almost 20 years, which reminded me to come and wish you a merry christmas, especially meni rosenfeld who has been a great help. And a happy new year to all the others too many to name. Bon voyage!87.102.76.184 (talk) 14:37, 17 December 2007 (UTC)

On behalf of myself (wishing a merry christmas to me in particular might not be completely appropriate, but never mind the technicalities) and others, thanks, and a merry christmas \ happy new year to you as well! -- Meni Rosenfeld (talk) 14:52, 17 December 2007 (UTC)
Definitely, thanks to everyone contributing here. Merry christmas ! --Xedi (talk) 15:18, 17 December 2007 (UTC)
Yeah, there was something quite amusing in singling out Meni for Christmas greetings. 68.183.18.54 (talk) 17:51, 17 December 2007 (UTC)
I hope it was not a case of assuming good faith.  --Lambiam 19:10, 17 December 2007 (UTC)
Merry Christmas all, I have enjoyed being of help and learning some things myself here as a "new" wikipedian (for a number of months now) A math-wiki (talk) 10:54, 18 December 2007 (UTC)

## Does inverse tetration yield a new type of numbers?

When you start with the natural numbers, the inverse operation of addition yields a new type of numbers (the integers), the inverse of multiplication yields a new type of numbers (the rationals) and the inverse of exponentiation yields a new type of numbers (the reals). Each operation, when applied to two natural numbers, yields a natural number, but the inverse operation (on certain natural numbers) has a solution only in some new number system. Does the inverse operation of tetration (super root or super log), applied to natural numbers, lead to a new number system as well, or do the reals encompass all operations like this beyond exponentiation? risk (talk) 14:43, 17 December 2007 (UTC)

I'm inclined to think that reals are more or less enough. Their wonderful feature is completeness, manifested in our case as the Intermediate value theorem. While there is apparently no commonly accepted generalization for a noninetger second argument, I'd expect any reasonable generalization to be, in a suitable region, continuous and monotonous. That means, say, that since ${}^12=2$ and ${}^22=4$, there must be some real $1 such that ${}^x2=3$, and so on.
Of course, you might wonder if non-real values are necessary for extending this beyond a "suitable region" (for example, finding x such that ${}^x2=-1$), or if non-real values are necessary for obtaining a suitable generalization in the first place. These look like extremely difficult questions, which will probably not be solved to a satisfying extent in our lifetime. -- Meni Rosenfeld (talk) 15:14, 17 December 2007 (UTC)
[ec]Actually, the first might not be so hard. slog is apparently extendible to the entire real line once we have it for a nice region; As for hyperroot, I think that for any natural 'n, the equation ${}^nx=y$ has at least one complex "solution" for almost every y, and thus new numbers are not necessary. -- Meni Rosenfeld (talk) 15:30, 17 December 2007 (UTC)
Because exponentiation is not commutative ($x^y \ne y^x$) there are actually two inverse operations - roots (inverse of $f(x)=x^y$) and logarithms (inverse of $f(x)=y^x$). Either one takes you into complex numbers if you want their domain to extend over all rationals, not just positive rationals. Since the field of complex numbers is both algebraically closed and analytically complete (with the standard metric), I don't think any finite or infinite sequence of operations or inverse operations will ever take you from inside complex numbers to something outside complex numbers. AFAIK all other number systems (such as quaternions or p-adic numbers) start ab initio by positing elements with different algebraic properties or a different metric. Gandalf61 (talk) 15:30, 17 December 2007 (UTC)
Thank you both. That clarifies a lot. Come to think of it, my little sequence wasn't really correct. The nth root of a natural number doesn't reveal the real numbers but only those that can be expressed as an nth root of a natural number. These are a subset of R just like the rationals and the integers. Presumably any successive operation, inverted and applied to natural numbers will only define new subsets of R in this way. risk (talk) 16:09, 17 December 2007 (UTC)
In a certain sense, the combined inversion of addition, multiplication, and exponentiation leads to another subset of the reals, namely the algebraic numbers. Icek (talk) 18:43, 23 December 2007 (UTC)

## Calculating deviance in flightpaths

I am very much bollocks when it comes to maths, but I would love any help to solve the current problem I have: I can fly from A to B on earth, but if I follow my compass, my bearings will be off after a certain amount of time because of natural deviance (not related to winds or anything). What I would love is a formula with which to calculate this deviance. From that formula, I could then draw a flightpath that negated the deviance. There are lots of factors one can include or exclude, or make certain reservations that translates to different latitudes. This touches upon trigonometry, but only barely before it makes a daring leap to more advanced euclidean geometry. Any help would be appreciated, especially if anyone should know this topic's name and corresponding Wiki article! 81.93.102.185 (talk) 19:45, 17 December 2007 (UTC)

I'm sorry, could you expand a little more on deviance? Are you thinking of magnetic deviation? If you are able to follow the bearing, you are describing a loxodrome. In any case, the article spherical trigonometry may be of your interest. Pallida  Mors 20:22, 17 December 2007 (UTC)
(edit conflict) Non-euclidean geometry even. Since we are on a sphere, the laws of Euclidean geometry don't apply perfectly. Let attempt to paraphrase you, to see if I understood. You are at point A,and you need to get to point B. Using your compass some way you point you plane in a certain direction, using some calculation, and then fire up the engine taking your plane in a straight line along the earth. Your method of finding the initial direction to point your plane in causes your flight path to deviate, and not hit B exactly, and you want a way of calculating where to point your plane, so that you'll hit B precisely without adjusting. Is this correct?
Presumably, you're using some method that works in the plane (ie. a map) but not on the surface of a sphere (ie. a globe), unless you're traveling really small distances. A curved space, like a sphere, doesn't quite follow euclidean rules. The shortest route between two points in a curved space is not called a straight line, but a geodesic. The shortest route between A and B in this case, is a segment of a great circle, rather than a 'straight line' (ie. a single heading). According to the great circle article, you can't travel along this route without continually changing your heading. risk (talk) 20:39, 17 December 2007 (UTC)
If you are using a compass, another problem would be due to magnetic north not being the north pole. NYCDA (talk) 20:44, 17 December 2007 (UTC)
I believe Risk pretty much nailed this. So yes, that is the case. As Pallida and NYCDA both have pointed out, there is also the magnetic aspect. The latter directly messes with the compass. Any solving of the problem with straight lines on curved spaces, would have to make use of a compass at some point - but if you feel that a solution can be provided more easily without taking into consideration the magnetic deviance, I am willing to make use of Polaris! 81.93.102.185 (talk) 22:13, 17 December 2007 (UTC)
In that case, you're looking for the loxodrome through both A and B. That is, a line of fixed bearing. Like I said, this is not the shortest route, but it will get you to B with out any course corrections. If you draw a straight line on a mercator projection, that's the loxodrome. Calculating the loxodrome given two points is (apparently) a little tricky and involves solving a nonlinear system of two variables. There are also infinitely many solutions (plenty of bearings will take you around the globe a couple of times until you it your target).
Your best bet, I think is to study the mercator projection. If you can use cartesian coordinates of a mercator projection for A and B, it's trivial to find the straight line between them. Since the mercator projection is also conformal (ie. it preserves angles), the angle of this line to the vertical axis of the map is equal to the angle of your bearing to true north. Additionally, if you know your latitude, you should be able to compensate for the difference between magnetic and true north. Magnetic declination will tell you more about that. risk (talk) 00:20, 18 December 2007 (UTC)
I believe risk is right (I mean, the previous poster). You can build a right triangle, one cathetus being the difference in longitudes and the other being the difference in meridian parts (i. e. the difference in y coordinates as given by the article Mercator projection. Once you have gotten that triangle, just find the tangent of the angle formed by the hypotenuse and the horizontal reference: that is the bearing from one place to the other under a loxodrome. Pallida  Mors 03:03, 18 December 2007 (UTC)