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

= 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 &minus; tan A = 1/30, etc. Michael Hardy (talk) 01:25, 11 December 2008 (UTC)

....OK, let's take it a bit further: let &fnof;(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, &fnof; 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) = 1x = e2&pi;ix = cos(2&pi;x)+i sin(2&pi;x) 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 1x is the constant 1x = 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

 * $$ \sec(\theta_1 + \cdots + \theta_n) = \frac{\sec\theta_1 \cdots \sec\theta_n}{e_0 - e_2 + e_4 - \cdots} $$
 * $$ \sec(\theta_1 + \cdots + \theta_n) = \frac{\sec\theta_1 \cdots \sec\theta_n}{e_0 - e_2 + e_4 - \cdots} $$

where ek is the kth-degree elementary symmetric polynomial in the n variables xi = tan &theta;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)