Wikipedia:Reference desk/Archives/Mathematics/2008 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)

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)

: $\backslash sec\; A\; -\; \backslash tan\; A\; =\; \backslash frac\{\backslash sec\; A\; -\; \backslash tan\; A\}\{1\}\backslash ,\backslash cdot\backslash ,\; \backslash frac\{\backslash sec\; A\; +\; \backslash tan\; A\}\{\backslash sec\; A\; +\; \backslash tan\; A\}$

= \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

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

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

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

and

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

we get

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

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

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

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

: $f(A+B)\; =\; f(A)f(B)\backslash ,$

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:

:

:

: $\backslash sec(\backslash theta\_1\; +\; \backslash cdots\; +\; \backslash theta\_n)\; =\; \backslash frac\{\backslash sec\backslash theta\_1\; \backslash cdots\; \backslash sec\backslash theta\_n\}\{e\_0\; -\; e\_2\; +\; e\_4\; -\; \backslash cdots\}$

:

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)

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)