Wikipedia:Reference desk/Archives/Mathematics/2007 October 3

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October 3

TI-89 Question

I want to solve an equation but restrict the domain to integers. ex. Solve(2^x+1=3^y-1,{x,y}), but i want to restrict it. How would I do that? xxpor ^yo!|_{see what i've done} 02:33, 3 October 2007 (UTC)

You wouldn't. You can't. You'd need a very, very advanced calculator to solve diophontine (sp?) equations, seeing as we can't always find a method for solving them -- that's a number theory thing, not simple math. Sorry! Gscshoyru 02:39, 3 October 2007 (UTC)

I wish to point out that 2^x and 3^y are not polynomials. This makes it even harder to solve. Looking at the equation

2^x + 1 == 3^y - 1

2^x + 2 == 3^y

Since x and y are integers. We can try y=0,1,2,3,4,...

and look for 3^y that are even since "2^x + 2" is even.

Find all 3^y that are even and "3^y - 2" that are of the form 2^x

But of course this is just brute force and not elegant. I must be stupid because I'm having trouble finding integer values of y such that 3^y is even. How odd! 202.168.50.40 03:47, 3 October 2007 (UTC)

In case the original questioner misses the point, the answer is that there are no solutions to 2^x+1 = 3^y-1 in positive integers because 2^x+1 is odd and 3^y-1 is even. No calculator required. Gandalf61 09:26, 3 October 2007 (UTC)

If you don't require the integers to be positive, however, there is one solution.  --Lambiam 09:42, 3 October 2007 (UTC)

lol @ Lambiam's approach. I love loopholes like that!

• $2^{x}+1=3^{y}-1\!$

--Mostargue 18:29, 3 October 2007 (UTC)

Also, see Diophantine equation.--Mostargue 18:34, 3 October 2007 (UTC)

Haha, I mixed up, its $x^{2}+1=x^{3}-1$. but there must be someway in general restrict the domain to integers. xxpor ^yo!|_{see what i've done} 18:36, 3 October 2007 (UTC)

Again, see diophantine equations. In the case that you've given, though, with only one variable, there are exactly three complex solutions (possibly real, possibly rational, possibly integer). With two variables, the problem becomes much more complicated. Although $y^{2}+1=x^{3}-1$ can be easily re-written in a standard form as an elliptic curve and solved using those methods. Donald Hosek 20:16, 3 October 2007 (UTC)

• $x^{2}+1=x^{3}-1$

Collect like terms:

• $x^{3}-x^{2}-2=0$

From here, you can use synthetic division to find the roots.

Just from looking at it right now, I can tell there is an odd number of roots between 1 and 2. So, either you have 2 more chances of finding getting integer roots, or none. Good luck ^_^.--Mostargue 05:36, 4 October 2007 (UTC)

Actually, that's not quite right. If the single-variable equation has any integer solutions, they can be any of $\{-2,-1,1,2\}$. The simplest approach is not to use synthetic division though. It's far easier to just plug each of these four possible values into the polynomial $x^{3}-x^{2}-2$ and see if they work out (here's a hint, they don't). I think the most likely scenario is that the OP was really looking for integral solutions to the elliptical curve $y^{2}=x^{3}-2$. Donald Hosek 21:24, 4 October 2007 (UTC)

Linear System Elimination Method

I can use the elimination method to solve equations, but I can't understand why it works, which really bothers me when I use it. My algebra book explains absolutely nothing about why anything is true (which is extremely frustrating!). Could someone please give me an explanation of why this works. Thanks in advance. Imaninjapirate^{talk to me} 03:24, 3 October 2007 (UTC)

Gaussian elimination, which I believe is the method being discussed, relies on the fact that the solution to a system of linear equations is not changed under any of the following operations:

1. Multiplying both sides of one equation by a non-zero number. (e.g. 2x = 2 has the same solution as 4x = 4).
2. Swapping two of the equations around. (e.g. 3x + 2y = 5, x - y = 1 has the same solution as x - y = 1, 3x + 2y = 5).
3. Adding a multiple of one of the equations to another. (A little trickier, but if you just think in terms of adding two equations together, then 3x + 2y = 5, x - y = 1 has the same solution as 3x + 2y = 5, (x - y) + (3x + 2y) = 4x + y = 5 + 1 = 6).

Gaussian elimination is then just a systematic method of applying these three operations to simplify the equations. Confusing Manifestation 03:56, 3 October 2007 (UTC)

We can think of a system of linear equations in many ways, each providing insights and benefits.

• When we have a single linear equation in a single variable, we can always write it as ax = b. That is, a constant multiplied times a variable equals a constant. Assume a is not zero (important!); then it has a reciprocal (a multiplicative inverse) denoted by ¹⁄_a or a⁻¹. If we multiply both sides of an equation by the same quantity, the truth of the equality is preserved. Since a⁻¹a = 1 (by definition), we easily obtain x = a⁻¹b, or x = ^b⁄_a as a solution.
• In advanced mathematics, we can use matrices to follow an almost identical approach with a system of n equations in n variables, writing Ax = b. The definition of "multiplicative inverse" becomes richer, so we need to be more careful to decide if A⁻¹ exists. For large systems, elimination is almost always approached using matrices, though explicit use of an inverse is rare.

But let's look at two linear equations in two variables. Generically, the form is

${\begin{alignedat}{3}a_{11}x_{1}&+&a_{12}x_{2}&=b_{1},\\a_{21}x_{1}&+&a_{22}x_{2}&=b_{2}.\end{alignedat}}$

As before, we can multiply either equation by a constant (on both sides). We can also preserve equality by adding equal quantities to both sides. That sounds simple enough, but consider that the left side of the first equation equals its right side. Thus we can add the left side of the first equation to the left side of the second, and the right side of the first equation to the right side of the second. We call this "adding the two equations".

Assume that a₁₁ is not zero (again, important). Then it has an inverse, and we may multiply by it.

$a_{11}x_{1}+a_{12}x_{2}=b_{1}\longrightarrow x_{1}+a'_{12}x_{2}=b'_{1}\,\!$

Save this for future reference; but now multiply the new equation by −a₂₁.

$x_{1}+a'_{12}x_{2}=b'_{1}\longrightarrow -a_{21}x_{1}+a''_{12}x_{2}=b''_{1}\,\!$

We need not assume a₂₁ is nonzero, though we're wasting effort if it is. Because our next step is to add this revision of the first equation to the second. When we do, we eliminate x₁; hence the name of the method.

$a_{21}x_{1}+a_{22}x_{2}=b_{2}\longrightarrow a'_{22}x_{2}=b'_{2}\,\!$

But now our second equation has the simple form of a single linear equation in a single variable, which we already know how to solve so long as a′₂₂ is nonzero.

$a'_{22}x_{2}=b'_{2}\longrightarrow x_{2}=b''_{2}\,\!$

Having solved for x₂, we next multiply both sides of this final form of the second equation by −a′₁₂.

$x_{2}=b''_{2}\longrightarrow -a'_{12}x_{2}=b'''_{2}\,\!$

The final step is easy to anticipate; we add this equation to the saved version of the first equation.

$x_{1}+a'_{12}x_{2}=b'_{1}\longrightarrow x_{1}=b'''_{1}\,\!$

Et viola! The first equation displays the solution for x₁, as the second equation does for x₂.

Thoughout, we used only three procedures:

• Given a nonzero coefficient, find its multiplicative inverse.
• Multiply both sides of an equation by a constant.
• Add one equation to another.

For safety, we must include another step:

• Order the equations so that the first equation has the largest coefficient for x₁ (in magnitude).

The methodical approach here used for two equations in two variables extends readily to n equations in n variables. In matrix form, we call it LU decomposition with partial pivoting. (To "pivot" is to reorder the equations.) It is quite likely the single most important, and most heavily used, algorithm in all of applied mathematics. --KSmrq^T 10:20, 3 October 2007 (UTC)

sum of sum of sum... of factors

You start off with a number such as 138. Using the definition given in Perfect number, the sum of its factors is 150. The sum of the factors of 150 is 222. If you continue taking the sum of the factors, you eventually (after 168 times) get a prime number which has a sum of 1. My question is, are there any starting numbers which will never lead to 1 and continue increasing infinitely? I am excluding perfect numbers, numbers which lead to perfect numbers (25 leads to 6), and numbers such as 220 and 284 which lead to each other. 68.231.151.161 06:04, 3 October 2007 (UTC)

There are periodic sequences, such as Amicable numbers. What you are describing is an Aliquot sequence, and it is unknown whether or not there are infinite sequences (that do not repeat). - Rainwarrior 06:23, 3 October 2007 (UTC)

Wow thanks... that's exactly what I was looking for. 68.231.151.161 06:42, 3 October 2007 (UTC)

According to MathWorld, the longest periodic aliquot sequence known has period 28 and smallest member 14316. It is also interesting to note that there are fairly small numbers (like 276) whose aliquot sequences are not completely known, because they go above the range of practical factorisation. Gandalf61 09:10, 3 October 2007 (UTC)

competition

Has USA announced any competition which involves solving of 10 tough problems?If so please inform me of that. 59.96.27.63 11:50, 3 October 2007 (UTC)

There is a list of seven tough problems known as the Millenium Prize Problems, set by the Clay Mathematics Institute. One of the six has been solved. See also Unsolved problems in mathematics.  --Lambiam 12:21, 3 October 2007 (UTC)

If you're talking about national versions of the International Mathematical Olympiad, take a look at USAMO. Black Carrot 07:56, 4 October 2007 (UTC)

And at the university level, there's the Putnam. That's 12 questions, though. J Elliot 16:06, 4 October 2007 (UTC)

vivinia's theorem

I also wanted to know about vivinia's theorem. 59.96.27.63 11:50, 3 October 2007 (UTC)

Could it be that you are looking for Viviani's theorem?  --Lambiam 12:25, 3 October 2007 (UTC)

Sin2A problem

Can you prove sin2A=2sinAcosA geometrically? 59.96.27.63 11:50, 3 October 2007 (UTC)

Sure. Take an isosceles triangle with "top" angle equal to 2A. The area of the triangle, using the formula area = ½bh in which b = base and h = height, can be computed two ways, depending on which side is taken as the base. Equating the two gives you the desired result.  --Lambiam 12:16, 3 October 2007 (UTC)

You can prove most trig identities geometrically. Just rewrite the trig function as its geometric definition. Let a,b,c be the sides of a right triangle.

• $\sin A={\frac {b}{c}}$
• $\cos A={\frac {a}{c}}$
• Thus:$2\sin {A}\cos {A}=2{\frac {ab}{c^{2}}}$

Remember that because it is a right triangle, $c^{2}=a^{2}+b^{2}$ See if you can figure out the rest =)--Mostargue 17:54, 3 October 2007 (UTC)

Perhaps you, 59.96.27.63, are interested in a book called "Proofs Without Words" (well, they are actually two books). There are proofs for what you just asked and many, many more. --Taraborn 20:17, 3 October 2007 (UTC)

Comparing complex numbers

Do the inequality symbols <, >, ≤, and ≥ have any meaning under complex numbers? My first guess was to take the absolute value of the complex numbers and compare them, but that's performing a function on them, which may change the equality (x > y does not imply f(x) > f(y), which definitely doesn't hold true for the absolute value function ).--Mostargue 18:59, 3 October 2007 (UTC)

No, you can't use < and > on the complex numbers, though you can of course on their moduluses.--PaulTaylor 19:09, 3 October 2007 (UTC)

There is no standard meaning. You could introduce your own ordering, for example by defining:

u + iv ≤ x + iy  ↔  u < x OR (u = x AND v ≤ y).

Then 1 < i < 2. This is essentially the lexicographic ordering on a pair of numbers. However, for complex numbers this is not a particularly useful relation for a working mathematician. In particular, it would destroy the symmetry between i and −i: if P[i] is the conclusion of a theorem involving i, then P[−i] is also a true proposition. For example, take P[j] := (1+j)(2–j) = (3+j). Then P[i] is the proposition (1+i)(2–i) = (3+i), which is true, and therefore P[−i], the proposition (1−i)(2+i) = (3−i), is also true. This would be broken by taking P[j] := 0 < j, which is true if you substitute i for j, but false when you do this with −i.  --Lambiam 20:44, 3 October 2007 (UTC)

The most useful ordering I know of is the partial ordering by absolute value. Black Carrot 07:54, 4 October 2007 (UTC)

You can of course use the inequalities separately on the real and imaginary parts of a complex number as you would normally87.102.94.194 15:48, 4 October 2007 (UTC)

You can, naturally, but given the lack of relationship with the algebraic/analytic structure of the complex numbers, I can't immediately see why you would want to. In particular, as Lambiam pointed out above, the symmetry between i and -i means any claim that -i<i (or vice versa) is unlikely to be useful. Algebraist 16:29, 4 October 2007 (UTC)

Volume of a "bowl"

How would you find the volume of a portion of a sphere that has been "sliced" in two? Obviously, if the cut is down the center, the volume of each resulting figure will be .5 that of the sphere, but what if the cut is, say, 1/3 along the diameter of the sphere? 70.146.164.160 21:00, 3 October 2007 (UTC)

See our article on spherical caps. Gandalf61 21:37, 3 October 2007 (UTC)

Which gives the answer, but not necessarily enough information. This sounds like a Calc II homework problem type thing (although it's early in the semester for this question. The calculus approach would be to use an integral to find the result so you're looking at $\int _{a}^{b}\pi f(x)^{2}dx$ where $f(x)$ gives the radius of a slice of the sphere at vertical position x. I'll leave the determination of $f(x)$, a and b as exercises for the reader. Donald Hosek 23:17, 3 October 2007 (UTC)

In other words you need to be able to find volumes using integration.. If you don't know how to do this now would be a good time to ask..87.102.94.194 15:46, 4 October 2007 (UTC)

There are numerous pages that cover this - here is one if you need it - hopefully it should give enough info to get you started http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/applications/volume.html - you can also integrate using angles as well as length - both methods are solvable.87.102.94.194 15:46, 4 October 2007 (UTC)