Wikipedia:Reference desk/Archives/Mathematics/2012 October 3

= October 3 =

Log- Linear Plot given the coordinates?
When logy is graphed as a function of x, a straight line results. Graph straight lines, each given by two points, on a log linear plot, and determine the functional relationship (The original x-y coordinates are given)

(x1,y1) = (0,5) and (x2,y2) = (3,1)

The correct answer is y= 5 * (0.58)^x Could someone tell me how to arrive at this answer? thank you! — Preceding unsigned comment added by 24.86.170.175 (talk) 00:41, 3 October 2012 (UTC)


 * Let's see. You start with:

y = a(b)x


 * You plug in x1 for x and y1 for y:

5 = a(b)0 = a(1) = a


 * So, a = 5:

y = 5(b)x


 * Now plug in x2 for x and y2 for y:

1 = 5(b)3


 * Divide both sides by 5:

0.2 = b3


 * Take the cube root of both sides:

0.5848 = b


 * Also, since you are dealing with logarithms, I suppose I should put that last "finding the cube root" step in terms of logs. Here you find the antilog of log(0.2)/3, or the antilog of -.233, to get 0.5848. StuRat (talk) 01:16, 3 October 2012 (UTC)

This process worked with this question but when i tried applying it to other questions it does not work any longer. For example, the same question, except with coordinates (-2,3) and (1,1). I first substituted the first set of coordinates into the equation to get 3b^2= a Then i substituted the second set of coordinates to get a = b. After a bit of algebra my answer was y = 1/3 * (1/3)^x. However, this is incorrect. I have spent some time trying to figure this out but I still do not understand. — Preceding unsigned comment added by 24.86.170.175 (talk) 01:40, 3 October 2012 (UTC)


 * The second given point tells you ab=1, not a=b. —Tamfang (talk) 02:26, 3 October 2012 (UTC)

Oh! that was a very silly mistake. Thank you to the both of you for helping — Preceding unsigned comment added by 24.86.170.175 (talk) 02:40, 3 October 2012 (UTC)


 * You're quite welcome. StuRat (talk) 02:57, 3 October 2012 (UTC)

Generating function of a particular sequence
I have a sequence in n which is $$\sum_{g=0}^n \binom{g + m -1}{m-1} z^{n-g}$$ where m is a natural number and z is some indeterminate, and I'm trying to find a rational generating function for this sequence. In other words, a rational function f(x), such that when expressed as a power series,

$$f(x) = \sum_{n=0}^\infty \sum_{g=0}^n \binom{g + m -1}{m-1} z^{n-g} x^n.$$

With z = 1, this is just $$f(x) = 1/(1-x)^{m+1}$$, but the z term complicates things.

If it's helpful, one way to think about the sequence is that the nth element is equal to the sum of all monomials of degree n in m+1 variables with one variable evaluating to z and the rest to 1. Another way is that it counts the number of monomials of degree ≤ n in m variables but weighting the sum based on their degree. I'm not totally sure that the sequence does have a rational generating function, but I have some reasons to believe it should.

Anyone have any ideas? Rckrone (talk) 04:51, 3 October 2012 (UTC)
 * Ok now I'm feeling a bit silly. I spent a while on this earlier and got nowhere, and then 30 min after I post it here the answer occurs to me.  It should just be $$f(x) = \frac{1}{(1-x)^m(1-zx)}$$, right? Rckrone (talk) 05:24, 3 October 2012 (UTC)

Model theory
Does every (first order) proposition having a model, constitute a part - of a finite class of (first/second order) propositions that has a unique model (up to isomorphism) for the individuals? HOOTmag (talk) 16:23, 3 October 2012 (UTC)

Why trig doesn't work?
let say I have this triangle. Why I didn't get the correct answer when I used inversed of cosine or inversed of tangent or inversed of sine? I saw the answer key used cosine law. Why only cosine law gives the correct answer?Pendragon5 (talk) 22:42, 3 October 2012 (UTC)
 * This is not a right triangle, so none of the side-length ratios will be equal to the sine, cosine or tangent of the angle. However the law of cosines can be applied to any triangle (not just right triangles) so you can use it to compute cos θ. Rckrone (talk) 23:34, 3 October 2012 (UTC)