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May 14

Perimeter estimation

Do someone know a very accurate estimation for the perimeter of an ellipse? For example evaluation for (a=5;b=1). And what is the error % of this estimation? How do you make the evaluation? Wich formula?TASDELEN (talk) 18:51, 14 May 2010 (UTC)[reply]

See circumference or ellipse. Dmcq (talk) 20:37, 14 May 2010 (UTC)[reply]

The circumference on an ellipse with semiaxes a, b is

\int _{0}^{2\pi }{\sqrt {a^{2}\cos ^{2}t+b^{2}\sin ^{2}t}}\ dt

. This can be evaluated to any desired accuracy with numerical integration techniques. For

a=5,\ b=1

this is

21.010044539689000944699164588473738912894812339134152623096835657214254472447144946328558302691971167999951485...

, correct to the number of digits given. I could just as easily give 100,000 accurate digits. -- Meni Rosenfeld (talk) 17:53, 15 May 2010 (UTC)[reply]

The ellipse article gives this as a good approximation:

C\approx \pi \left(a+b\right)\left(1+{\frac {3\left({\frac {a-b}{a+b}}\right)^{2}}{10+{\sqrt {4-3\left({\frac {a-b}{a+b}}\right)^{2}}}}}\right);\!\,

For

a=5,\ b=1

this is 21.010027, so you can judge how good the approximation is from Meni's figure.→86.132.162.141 (talk) 19:26, 15 May 2010 (UTC)[reply]

Thanks a lot.By estimation I understand Meni's evaluation and not mathematica tool's estimation.He used Ramanujan's formula and find C=21,010026651559300 (at 15 digit).

So,the error %=(Lexact-Lestimated)/Lexact=0,000000851482682.May you be interested whith a new formula which gives error%=-000000417870650,a %50 less error than Ramanujan's estimation?TASDELEN (talk) 23:18, 15 May 2010 (UTC)[reply]

In situations where accuracy is important, you wouldn't use such formulas but rather advanced techniques which are scalable to any accuracy. If you must have an elementary closed form, you can find expressions which are arbitrarily accurate, but the more accurate expressions are also more complicated. So finding an expression which is marginally more accurate than another is not interesting. If the new expression is more accurate and also simpler than the commonly known estimates, it might be somewhat interesting. -- Meni Rosenfeld (talk) 08:58, 16 May 2010 (UTC)[reply]

Meni, I am impressed. How did you do that? Bo Jacoby (talk) 04:31, 16 May 2010 (UTC).[reply]

Mathematica. Specifically,

NIntegrate[(25 Cos[t]^2 + Sin[t]^2)^(1/2), {t, 0, 2 Pi}, WorkingPrecision->150]

For higher precision, the following may be more robust:

Integrate[(25 Cos[t]^2 + Sin[t]^2)^(1/2), t]
(% /. {t -> 2Pi}) - (% /. {t -> 0})
N[%, 100000];

-- Meni Rosenfeld (talk) 08:33, 16 May 2010 (UTC)[reply]

Gentlemen,I am using Thales theorem to find the perimeter of an ellipse.An exact formula but implicit and cracking Thales theorem I may reach to a nice and practically high accuracy for astroids (x/a)^r+(y/b)^r=1.(r=2 is an ellipse).Not "simpler",but more accurate than all known approximations.TASDELEN (talk) 09:37, 16 May 2010 (UTC)[reply]

May 15

2cosx+2cos(2x)=0

Consider the function $f(x)=2\sin x+\sin 2x$
Locate and classify the function's stationary points.

${\frac {df(x)}{dx}}=2\cos x+2\cos 2x$
$0=2\cos x+2\cos 2x$
$0=\cos x+\cos 2x$
$0=\cos x+\cos ^{2}x-\sin ^{2}x$
$0=\cos x(1+\cos x-\sin x\cdot \tan x)$
$\cos x=0$
$x={\frac {\pi }{2}}$
doesn't work because tan is undefined at pi/2 --Alphador (talk) 10:17, 15 May 2010 (UTC)[reply]

Pi/2 is a potentially false solution since in the case of x=pi/2 dividing by cos x is dividing by zero. So this solution has to be tested. Setting x=Pi/2 into 2cos x+2cos 2x=2cos Pi/2+2cos pi=-2Taemyr (talk) 11:17, 15 May 2010 (UTC)[reply]

Convert

0=\cos x+\cos ^{2}x-\sin ^{2}x

into a quadratic in terms of

\cos x

. --COVIZAPIBETEFOKY (talk) 11:55, 15 May 2010 (UTC)[reply]

0=\cos x+\cos ^{2}x-(1-\cos ^{2}x)

0=2\cos ^{2}x+\cos x-1

\cos x={\frac {-1\pm {\sqrt {1^{2}-4\cdot 2\cdot (-1)}}}{2\cdot 2}}={\frac {1}{2}},-1

x={\frac {\pi }{3}},\pi

—Preceding unsigned comment added by 220.253.221.60 (talk) 12:14, 15 May 2010 (UTC)[reply]

The derivative of sin(2x) is not cos(2x). 76.229.218.70 (talk) 18:35, 15 May 2010 (UTC)[reply]

It is

2\cos 2x

, as the OP has correctly written. -- Meni Rosenfeld (talk) 18:42, 15 May 2010 (UTC)[reply]

-\pi /3

is another solution, and all solution are of course

+2k\pi

. In fact, all solutions can be written succinctly as

(2k+1)\pi /3\;\!

. -- Meni Rosenfeld (talk) 18:42, 15 May 2010 (UTC)[reply]

May 17

ridiculous indefinite integral question

Evaluate $\int \cot x\cdot \ln {|\sin x|}dx$ --115.178.29.142 (talk) 00:59, 17 May 2010 (UTC)[reply]

Change

\cot x

to

{\frac {\cos x}{\sin x}}

and solve by substitution, would be my guess at a glance. I'll look back when I'm not in a hurry. 76.228.199.229 (talk) 01:05, 17 May 2010 (UTC)[reply]

In fact, you don't even have to go that far. Hint: the antiderivative of cotangent is

\ln |\sin x|+C

. 76.228.199.229 (talk) 01:18, 17 May 2010 (UTC)[reply]

Geometry problem

Hey guys. I came across a really difficult (for me) olympiad question and I'm clueless as to how to solve it. Here it is.

Put O as the circumcenter of triangle ABC, and take an arbitrary point D on BC. Put E as the intersection of two circles; a circle with AD as its diameter, and the nine-point circle of ΔABC. Prove that angle ADE = angle ODC (E is not on BC)

I've tried everything from looking for similar triangles, finding cyclic quadrilaterals, to assigning every length using the law of sines. But I keep getting stuck no matter what I try. Can anybody help me?? thanks. Johnnyboi7 (talk) 12:47, 17 May 2010 (UTC)[reply]

This gave me a lot of trouble, and the way I ended up doing it isn't too pretty but here goes. Let X, Y, Z be the midpoints of BC, AC, AB respectively, N be the nine-point center, F the midpoint of AD (and so the center of that first circle) and G the base of the altitude from A. NX is the nine-point radius, and AO is the circumradius which is twice the nine-point radius. Since N is halfway between the altitude and the perpendicular bisector to BC, NX and AO are parallel. The two circles intersect at G and E, so angle NFG = angle NFE. Call this angle θ. Reflect N over the line YZ and call that point N' (and note that A is the reflection of G). N' is the midpoint of AO. So then AFN' and ADO are similar triangles and so angle ADO = angle AFN' = θ. angle EDG is inscribed in circle F, and has corresponding central angle EFG, so angle EDG = 180 - θ, which means angle CDE is also θ. Since angle CDE = angle ADO, angle ADE = angle ODC. Rckrone (talk) 18:38, 19 May 2010 (UTC)[reply]

I guess I should add that you need to choose which vertex is B and which is C in order for it to work out, otherwise you may end up with angle ADE being supplementary to angle ODC rather than equal. Also depending on how things work out, angle EDG may end up being θ rather than 180 - θ depending on whether D falls between E and G on the circle or not. That will determine whether to assign the label C to the vertex on G side of DG or to the one on the D side. Rckrone (talk) 20:41, 19 May 2010 (UTC)[reply]

Rckrone, I love you!! thank you so much for all the effort you've put into this:) I'm sorry I didn't reply sooner.

PS. Yeah I agree, the problem's a little bit shaky considering that the angle depends on where we put D. If I were the person who created this problem, I would have said "Prove that angle ADE = angle ODX (X being the midpoint of BC" "Johnnyboi7 (talk) 09:17, 23 May 2010 (UTC)[reply]

SAS instruction books

Does anyone have any suggestions as to what SAS software instruction book/guide would be the best one to buy?--160.36.39.85 (talk) 13:41, 17 May 2010 (UTC)[reply]

Anything by Andy McNab. Sorry. Terrlible joke. feel free to remove me;) Vespine (talk) 02:37, 21 May 2010 (UTC)[reply]

May 18

Differential solid angle

Why is the solid angle element d(omega) equal to dl*dm/sqrt(1-l^2-m^2), where l and m are direction cosines with respect to two orthogonal axes? --99.237.234.104 (talk) 03:57, 18 May 2010 (UTC)[reply]

wheels of constant width and variations

A mathematician might be able to give a quicker answer to this question Wikipedia:Reference_desk/Science#Square_wheels than I can. Thanks.77.86.10.27 (talk) 15:58, 18 May 2010 (UTC)[reply]

A neat and elegant proof can be found here. The method can be extended to any regular polygon by truncating the catenary at a point corresponding to the exterior angle. Dbfirs 20:46, 19 May 2010 (UTC)[reply]

Fit an exponential curve

I have two points, and the desired slope at one of the points. How can I translate this information into an equation of the form y=a*e^(b*x)+c ? It seems like it should be possible, but trying to solve the equations is just tying me in knots. 173.52.5.181 (talk) 19:11, 18 May 2010 (UTC)[reply]

The problem here is that b appears in a nonlinear way in the equations. To make this problem a little less bad, you can try to isolate the nolinearity in one eqation, instead it popping up in all of the simultaneous set of the equations. You can do that by considering the quantity:

[y2 - y1]/y1'

where y2 and y1 are the the y-coordinate of the two points and y1' is the slope. This only depends on b and you can solve for it using e.g. Newton Raphson. Once you have b, you have to deal with only linear equations for the two other variables. Count Iblis (talk) 20:06, 18 May 2010 (UTC)[reply]

You have the 3 equations

ae^{bx_{1}}+c-y_{1}=ae^{bx_{2}}+c-y_{2}=abe^{bx_{1}}-y'_{1}=0

in the 3 unknowns, $a,b,c.$ If this is confusing you may rename

(a,b,c,x_{1},y_{1},x_{2},y_{2},y'_{1})\rightarrow (x,y,z,a_{1},b_{1},a_{2},b_{2},c)

getting the 3 equations

xe^{a_{1}y}+z-b_{1}=xe^{a_{2}y}+z-b_{2}=xye^{a_{1}y}-c=0

in the 3 unknowns $x,y,z.$ A standard approach is to eliminate variables to obtain 3 equations in one variable each, and then solve these equations numerically. The unknown $z$ is eliminated by subtraction:

(xe^{a_{1}y}+z-b_{1})-(xe^{a_{2}y}+z-b_{2})=xye^{a_{1}y}-c=0

and simplification

x(e^{a_{1}y}-e^{a_{2}y})-b=xye^{a_{1}y}-c=0

where $b=b_{1}-b_{2}$ . Multiplication and subtraction

ye^{a_{1}y}(x(e^{a_{1}y}-e^{a_{2}y})-b)-(e^{a_{1}y}-e^{a_{2}y})(xye^{a_{1}y}-c)=0

eliminates $x$ after simplification:

ye^{a_{1}y}(-b)-(e^{a_{1}y}-e^{a_{2}y})(-c)=0

(b/c)ye^{a_{1}y-a_{2}y}-(e^{a_{1}y-a_{2}y}-e^{a_{2}y-a_{2}y})=0

Insert $a=a_{1}-a_{2}$ and $A=b/c.$

Aye^{ay}-(e^{ay}-1)=0.

The false solution $y=0$ is discarded

e^{ay}-{\frac {e^{ay}-1}{Ay}}=0.

Simplify by setting $w=ay$ and $B=A/a$

e^{w}-{\frac {e^{w}-1}{Bw}}=0.

The solution $w$ is found numerically and substituted into the former equations which are then easily solved. Then undo the renaming and you are finished. Bo Jacoby (talk) 12:42, 19 May 2010 (UTC).[reply]

Eigenvectors

I'm looking at this matrix:

\left[{\begin{matrix}1&1/2&1/8&1/48&1/384&\cdots \\[6pt]0&1/2&1/4&1/16&1/96&\cdots \\[6pt]0&0&1/8&1/16&1/64&\cdots \\[6pt]0&0&0&1/48&1/96&\cdots \\[6pt]0&0&0&0&1/384&\cdots \\[6pt]\vdots &\vdots &\vdots &\vdots &\vdots &\ddots \end{matrix}}\right].

It goes on forever. The pattern is this:

Below the main diagonal every entry is 0;
The entries in the first row are the reciprocals of

2,\qquad 2\cdot 4,\qquad 2\cdot 4\cdot 6,\qquad 2\cdot 4\cdot 6\cdot 8,\quad \dots \dots \,

Each column is proportional to a row of Pascal's triangle; e.g. the last column shown above is proportional to 1, 4, 6, 4, 1.
Consequently the sequence of numbers on the main diagonal is the same as the first row.
Also consequently, if each column is summed, the sums are 1, 1, 1/2, 1/6, 1/24, the terms being the reciprocals of the factorials.
Each row is a shift of a scalar multiple of the first row.
This matrix is the matrix of coefficients in the "inversion formulas" section of this article.

Since it's an upper-triangular matrix, the eigenvalues are the numbers on the main diagonal. I've found the first three corresponding eigenvectors:

\left[{\begin{matrix}1\\0\\0\\0\\0\\\vdots \end{matrix}}\right],\quad \left[{\begin{matrix}1\\-1\\0\\0\\0\\\vdots \end{matrix}}\right],\quad \left[{\begin{matrix}5\\-14\\21\\0\\0\\\vdots \end{matrix}}\right]

What more can be said about the eigenvectors? Michael Hardy (talk) 23:44, 18 May 2010 (UTC) Michael Hardy (talk) 15:15, 20 May 2010 (UTC)[reply]

In case it helps, the following eigenvectors are {205, -987, 3243, -5405} and {111049, -761404, 3950262, -13752764, 24067337}. Also, for all eigenvectors up to at least the 40th, the entries of the eigenvector have alternating signs. -- Meni Rosenfeld (talk) 04:42, 20 May 2010 (UTC)[reply]

Thank you; I'll see if the Online Encyclopedia of Integer Sequences says anything about those. And otherwise see if I can spot any patterns that shed some light. Michael Hardy (talk) 15:03, 20 May 2010 (UTC)[reply]

May 19

Dart question

"A dart, thrown at random, hits a square target. Assuming that any two parts of the target of equal area are equally likely to be hit, find the probability that the point hit is nearer to the center than to any edge." I got that the probability would be 25%, but this seems too easy for a Putnam question. Can someone confirm my answer? 173.179.59.66 (talk) 02:51, 19 May 2010 (UTC)[reply]

I did a quick back of the enveloppe calculation and I find 2/3. By symmetry, you can look at the following area. If you put the four corners of the square at x = ±1/2 and y = ±1/2, then consider the area within the square between the positive x-axis and the line y = x. This area is 1/8. The distance to the closest edge will then always be the distance to the edge at x = 1/2 which is 1/2 - x. The distance to the center is

sqrt(x^2+y^2). You can then easily see that the region that is closer to the center is from x = 0 to x = 1/4 and y = 0 from y = sqrt(1/4-x). Count Iblis (talk) 03:18, 19 May 2010 (UTC)[reply]

I see where I messed up, thanks! 173.179.59.66 (talk) 04:01, 19 May 2010 (UTC)[reply]

I don't think it is that simple, unless these thoughts are unnecessarily over complicated... Looking at iblis's triangle with corners at (0,0) (0.5,0) and (0.5,0.5). To help think about it, Draw an arc radius 0.1 and a vertical line at x=0.9 - these points are all 0.1 from the centre / edge respectively. Again arc radius 0.2 and line at x=0.8. Fnally arc 0.25 and vertical line at x=0.25. These touch at (0.25,0) and this si an equidistant point. The space remaining is a "triangle" with the vertical line, the curved arc and the original diagonal edge as its three sizes. The locus of equidistant points will curve from (0.25,0) to (0.5*(1-sqrt(2),0.5*(1-sqrt(2)), but that locus isn't a straight line nor do I think it a circular arc. It should be possible to determine x as a function of y and then integrate... -- SGBailey (talk) 13:45, 19 May 2010 (UTC)[reply]

I made a mistake in my posting yesterday a bit after bedtime. When I was in bed I saw the mistake I made. You need to do as SGBailey says but I think you then end up integrating y = x from x = 0 to 0.5*(1-sqrt(2) and then from there to x = 1/4 the function

y = sqrt[(1/2-x)^2 - x^2] = sqrt(1/4 - x)

But you need to carefully check all this. Count Iblis (talk) 14:09, 19 May 2010 (UTC)[reply]

To be equally close to the center and to the top edge, is to be on a parabola whose focus is the center and whose directrix is the top edge. Similarly for the other edges. Thus the target region is bounded by arcs of four parabolas. Michael Hardy (talk) 16:30, 19 May 2010 (UTC)[reply]

Yeah this is what you want. If you let the bottom edge be y=0 and the center be (0,1) the parabola for that edge is y = x²/4 + 1/2 running from 2-√6 to √6-2. Rckrone (talk) 19:06, 19 May 2010 (UTC)[reply]

Constant continuous maps between a topology on the natural numbers and the reals

Hi all,

I'm wondering how to show that if $\tau$ is a topology on the natural numbers 1, 2,... such that the open sets are $\mathbb {N} ,\,\phi$ , and all sets of the form $\{1,\,2,\,...,\,n\},\,n\in \mathbb {N}$ , then any continuous map from $(\mathbb {N} ,\,\tau )$ to the Euclidean topology on $\mathbb {R}$ is constant.

If I can, I'd like to show this directly without relying on much more than the definitions (e.g. without quoting results, 'a homeomorphism preserves connectedness' etc). I'm always very happy visualizing these sorts of problems in a metric space with a concept of distance, but my intuition falls short when we fix our open sets by a specified topology instead.

Could anyone suggest the nicest way to go about a problem like this?

Thankyou all very much! Typeships17 (talk) 18:53, 19 May 2010 (UTC)[reply]

Any two nonempty open sets in your topology have nonempty intersection. This gives a quick proof. Algebraist 18:59, 19 May 2010 (UTC)[reply]

Argh, what I wouldn't give to be as good as you are at Mathematics! I do envy your grasp of what seems to be just about everything under the mathematical sun. Thankyou, that's been a great help :) Typeships17 (talk) 23:01, 19 May 2010 (UTC)[reply]

More generally, any continuous function from a hyperconnected space to a Hausdorff space must be continuous (the proof uses the same idea as Algebraist notes). PS T 01:53, 20 May 2010 (UTC)[reply]

In fact, you may find it interesting to know that the topological space you mention is what is known as a Toronto space. A Toronto space is a topological space that is homeomorphic to every proper subspace of the same cardinality (it may be a good exercise to check that the topological space you mention is indeed a Toronto space). It is also true (though perhaps not immediately obvious) that the only countable Hausdorff Toronto space is the discrete space (up to homeomorphism, of course). The Toronto problem asks whether every uncountable, Hausdorff Toronto space must be discrete (and, as far as I know, this problem still remains open). PS T 02:13, 20 May 2010 (UTC)[reply]

I must confess it is exam season for me at the moment so I probably won't have time to read up on that too much right now, but I'll be sure to check it out in a couple weeks when I'm free again - thanks! Typeships17 (talk) 03:21, 20 May 2010 (UTC)[reply]

May 20

Maclaurin series

What will the Maclaurin series for $i^{i}$ be? Here $i={\sqrt {-1}}$ —Preceding unsigned comment added by Πrate (talk • contribs) 12:55, 20 May 2010 (UTC)[reply]

i^i has to be defined by choosing some branch for the logarithm. If you put i = exp(i pi/2), you get i^i = exp(-pi/2). So, you could insert -pi/2 in the Maclaurin series for exp(x). Count Iblis (talk) 14:06, 20 May 2010 (UTC)[reply]

Partial Surface Area of a Sphere

I am looking for a way to find the surface area of a portion of a sphere given two angles and a radius. The two angles would be perpindicular to each other. For example one angle would represent an azimuth and the other would represent elevation, and the radius would be the distance.

I know that arc length (L) can be calcuated by the radius multiplied by the angle (L=rθ).

To get a complete sphere SA=4πr²

The limits for the two angles would be 0 < θ < 2π for azimuth and 0 < φ < π for elevation.

Is there a similar equation for surface area of a sphere as there is for the arc length of a circle? —Preceding unsigned comment added by 138.163.106.71 (talk) 20:07, 20 May 2010 (UTC)[reply]

I don't see how you can define area with only two angles... Unless I'm reading your question wrong? 76.229.164.175 (talk) 21:32, 20 May 2010 (UTC)[reply]

Is what you are describing a "rectangle" on the sphere, - more details please.

As for areas on spheres - yes - Solid angle is one place to start - area = solid angle x radius x radius

also see Steradian

The most common formula for solid angle is that for an area defined by three planes (forming a tetrahedron) - the solid angle is a+b+c-pi where a,b,c are the angles made on the tangential surface to the sphere by the planes.

The simplest solid angle is probably that of a Lune_(mathematics)#Spherical_geometry

I'm guessing you are describing a lune truncated at the top by the elevation angle ? If so you can get the area by subtracting the solid angles of the triangular parts at top and bottom from that of a lune, and then multplying by r^2 to get the area.

Or perhaps you mean a "segment of an orange" with one of the top points cut off.77.86.115.45 (talk) 22:04, 20 May 2010 (UTC)[reply]

Applications of calculus in lower maths

What are some was that calculus can be applied to figure out lower-maths problems? By lower maths, I mean algebra, geometry, trig, and precalculus, not addition and subtraction, of course :) These ways should be quicker or about the same speed as doing it conventionally--for example it would not make much sense to use the derivative formula to find the slope of a linear function! 76.229.164.175 (talk) 21:25, 20 May 2010 (UTC)[reply]

If you know only trigonometry, you could wonder for what value of x the function sin(x) + sin(3x) has its largest value. The answer can be found quickly by using calculus. That's just one example. Michael Hardy (talk) 23:41, 20 May 2010 (UTC)[reply]

Agreed, finding the minimum or maximum of a function is perhaps the simplest common application of calculus. And, going the other way, there's also finding the area under a curve. The relationships between position, velocity, and acceleration (and a curve's equation, slope, and curvature) are common physical manifestations of the derivative/integral relationships. StuRat (talk) 02:00, 21 May 2010 (UTC)[reply]

Definition of the (units of time)...that make up the components of one second.

A second is a unit of time. There are 60 seconds(unit of time) in one minute. What is the name or number of the components of one second of time? I realize there are parts ie: 10ths 100ths 1000ths (etc) of a second.! When asked to add 9.6 and 8.4 and 7.8 and 1.9 seconds...and then... how do you add the fractional parts,and what do you carry over? ie 6+4+8+9=27. do you say .7 and carry over the 20 as an equal to one second.? does .27 equal 2.7 seconds.?

At what point does the ttl fractional units make up one second.? is .10 equal to one second?

Thank You —Preceding unsigned comment added by 67.142.130.40 (talk) 23:05, 20 May 2010 (UTC)[reply]

To evaluate 9.6 + 8.4 + 7.8 + 1.9 you can proceed just as if you had 96 + 84 + 78 + 19 except that the decimal point would remain just before the last digit. That doesn't depend on whether they're seconds or liters or dollars or anything else.

I had to cogitate a bit before figuring out what your last question means. Here's my guess as to what you meant: Suppose you have, for example, 6.2 + 9.8, and you go add the 2 in 6.2 to the 8 in 9.8. You have 2 + 8 = 10. That's not .10. Rather, you put the 0 in the place right after the decimal point and carry the 1, getting 1 + 6 + 9 = 16, so the answer is 16.0 seconds, which is the same as 16 seconds. Michael Hardy (talk) 23:47, 20 May 2010 (UTC)[reply]

I don't think that was what he was saying, Michael. He might have thought that when computing seconds from minutes you would use base 60! Now, it's ridiculous to use that high of a base, but... 76.229.164.175 (talk) 02:19, 21 May 2010 (UTC)[reply]

Also, there isn't any common unit of time smaller than a second in the English system, although a moment is sometimes humorously used to mean 1/60th of a second. In metric, we can go to milliseconds (1/1000th) or even microseconds (a millionth). StuRat (talk) 01:51, 21 May 2010 (UTC)[reply]

May 21

Linear recurrence

Find a closed form for f(n)=3f(n-1)-2f(n-2), f(0)=2, f(1)=1.

Here's how I solved it, but the equation at the end produces the wrong value for f(2); it should be -1, but the equation gives -7.

f(n)=3f(n-1)-2f(n-2)

x^n=3x^{n-1}-2x^{n-2}

x^2-3x+2=0

x=-1,-2

f(n)=ax^n+by^n

f(n)=a(-1)^n+b(-2)^n

2=a+b

1=-a-2b

a=5, b=-3

f(n)=5(-1)^n-3(-2)^n

--70.129.186.243 (talk) 01:12, 21 May 2010 (UTC)[reply]

That's an interesting way to solve these linear recurrences. Anyway, your problem is in this step

x^2-3x+2=0

x=-1,-2

Those are not the solutions to that quadratic equation. With the correct roots, I think you'll get the final answer with no trouble.98.235.80.144 (talk) 02:15, 21 May 2010 (UTC)[reply]

It looks like you misfactored, 70. How are the solutions negative if the linear term of the quadratic is negative? 76.229.164.175 (talk) 02:23, 21 May 2010 (UTC)[reply]

behaviour of rotating hardboiled egg

Could a kind mathematician (possibly one who has read the Keith Moffat Yutaka Shimomura paper give an answer at Wikipedia:Reference_desk/Science#Spinning_Tesla_egg_and_stability - is the Moffat/Shinomura paper right? Does friction off centre of rotation cause the egg to try to roll as it rotates - and then trace a path on the surface of the egg, or what?? Is the eulur equation idea described there correct.. etc . Thanks. ~~