Wikipedia:Reference desk/Archives/Mathematics/2008 December 10

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December 10

Reciprocal of a graph

What does the reciprocal of a graph actually show you? Harland1 (^t/_c) 09:00, 10 December 2008 (UTC)

I am not familiar with the term "reciprocal of a graph" - or, rather, I can think of several possible meanings. Can you give an example ? Assuming you mean graph of a function (and not graph as in graph theory), what is the reciprocal graph of y=x²+1 ? Is it the graph of y= sqrt(x-1) or y=(x²+1)^-1 or y=x^-2+1 or something else ? Or, if you mean graph as in graph theory, do you mean a graph's complement graph ? Gandalf61 (talk) 10:15, 10 December 2008 (UTC)

I mean the if you have the graph of y=x^2+6x-12 what does the graph of y=1/(x^2+6x-12) show you? Harland1 (^t/_c) 21:40, 10 December 2008 (UTC)

Well, it shows you what values y=1/(x^2+6x-12) takes... I'm not sure I understand the question... --Tango (talk) 22:00, 10 December 2008 (UTC)

Harland you are talking about a graph of the reciprocal of a function. In your example the function is y=x^2+6x-12. Cuddlyable3 (talk) 23:33, 10 December 2008 (UTC)

Well, obvious facts, assuming the original function is as nice as yours: The positive part of the graph will be turned upside down (mirrored in the line y = 1), and distorted. Likewise for the negative part (mirrored in the line y = -1). It will approach zero when the original function approaches infinity, and it will approach infinity (positive, negative or both) where the original function has a zero. I don't think there's anything deeper to it. -- Jao (talk) 00:21, 11 December 2008 (UTC)

Trigonometric identities

I need some help with a question on trig. identities.
Given that sec A + tan A = 2
I need to show that sec A - tan A = 1/2

Im really stumped on this one. I dont expect anyone to do it for me but a hint would be nice. I've done the obvious which is to write sec A as 1/cos A but after that Im not sure what to do --RMFan1 (talk) 19:31, 10 December 2008 (UTC)

Multiply the left hand sides together and it should become clear. If not recognize the factored difference of squares and use the pythagorean theorem to get that (sec A + tan A)*(sec A - tan A) = 1, so (sec A - tan A) = 1/(sec A + tan A). As a general hint, if you see X+Y=G, X-Y=H, then you should consider if (X+Y)*(X-Y)=G*H and 2*X=G+H are relevant. In this case 2+1/2 is unlikely to be useful, but XX-YY=1 is a familiar trigonometric identity. JackSchmidt (talk) 19:55, 10 December 2008 (UTC)

But remember in a formal proof, you have to work backwards (i.e start with sec^2(A) - tan^2(A) = 1, factor it into (sec(A) + tan(A)) and (sec(A) - tan(A)) = 1 and since one of these is 2, the other must be 1/2). That is a formal proof.

Topology Expert (talk) 20:02, 10 December 2008 (UTC)

$\sec A-\tan A={\frac {\sec A-\tan A}{1}}\,\cdot \,{\frac {\sec A+\tan A}{\sec A+\tan A}}={\frac {\sec ^{2}A-\tan ^{2}A}{\sec A+\tan A}}={\frac {1}{\sec A+\tan A}}.$

So, for example, if sec A + tan A = 30, then sec A − tan A = 1/30, etc. Michael Hardy (talk) 01:25, 11 December 2008 (UTC)

....OK, let's take it a bit further: let ƒ(A) = sec A + tan A. Then since secant is an even function and tangent is an odd function, we get

$f(-A)={\frac {1}{f(A)}}.$

So at least as far as that identity goes, ƒ behaves like an exponential function. How far can we take the resemblance to an exponential function? Well, since

$\sec(A+B)={\frac {\sec A\sec B}{1-\tan A\tan B}}$

and

$\tan(A+B)={\frac {\tan A+\tan B}{1-\tan A\tan B}},$

we get

$f(A+B)={\frac {\tan(A)+\left(\sec A\sec A\right)+\tan B}{1-\tan A\tan B}}.$

Now if only we could just regroup the parentheses, we'd have

${\frac {\left(\tan(A)+\sec A\right)\left(\sec A+\tan B\right)}{1-\tan A\tan B}}$

and then if we could multiply fractions by multiplying the numerators, we could say

$f(A+B)=f(A)f(B)\,$

and we'd have an exponential function. But exponential functions are not periodic, so consider this last speculation to be a work of amusing fiction. Michael Hardy (talk) 01:38, 11 December 2008 (UTC)

Exponential functions are indeed periodic. f(x) = 1^x = e^2πix = cos(2πx)+i sin(2πx) satisfies f(x+1) = f(x) . Bo Jacoby (talk) 07:45, 11 December 2008 (UTC).

True, but I was thinking of real functions of a real variable. If necessary, I think I could even defend the position that in some situations such a restriction makes sense. Michael Hardy (talk) 14:04, 11 December 2008 (UTC)

A trivial branch of the exponential function 1^x is the constant 1^x = 1. It is a periodic real function of a real variable. :-). Bo Jacoby (talk) 00:03, 15 December 2008 (UTC).

I've answered my own question, and made it a new section in the list of trigonometric identities. An excerpt:

Secants of sums of finitely many terms

$\sec(\theta _{1}+\cdots +\theta _{n})={\frac {\sec \theta _{1}\cdots \sec \theta _{n}}{e_{0}-e_{2}+e_{4}-\cdots }}$

where e_k is the kth-degree elementary symmetric polynomial in the n variables x_i = tan θ_i, i = 1, ..., n, and the number of terms in the denominator depends on n.

Michael Hardy (talk) 21:40, 14 December 2008 (UTC)

Constructing a square around a Reuleaux triangle

I am given a Reuleaux triangle and a line. Using straightedge and compass, how can I construct the square around the Reuleaux triangle (all four sides in contact with the triangle) having two sides parallel to the given line? —Bkell (talk) 20:15, 10 December 2008 (UTC)

Ah, never mind. The trick is to construct a line parallel to the given line through a vertex of the triangle, and go from there. —Bkell (talk) 20:24, 10 December 2008 (UTC)