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March 9

More map projection questions

I think I have some understanding of how certain projections can be constructed geometrically, but I just want to check my reasoning.

For the two-point equidistant projection, my guess is that one way it can be constructed pointwise is by plotting the two control points A and B a distance 1 apart on paper, measuring the distance AB between them on the globe, and then measuring distances between those two points and some other point C on a globe. To plot C, extend a compass to a length of AC/AB and draw circle about A, then extend a compass to a length of BC/AB and draw a circle about B. The point C lies at one of the intersections of the two circles. Obviously, this is not the full story as the circles will usually intersect at two points, but is this the right basic idea?

For the Lambert cylindrical equal-area projection, I believe that you simply wrap a cylinder round the Earth and install a bar light source along the diameter of the Earth and parallel to the cylinder, and plot in accordance with the shadow cast on the cylinder. Is this right? Or does the bar light source have to be infinite in length? Thanks.--Leon (talk) 08:44, 9 March 2016 (UTC)[reply]

For cylindrical equal area, you want the rays to leave the axis of the cylinder at a right angle; if you can do that with a physical lamp, it need only be as long as the diameter. —Tamfang (talk) 08:03, 11 March 2016 (UTC)[reply]

proving function inequality

Hello,

I've tried to prove that a certain expression is tightly bound by n^2, and the problem was reduced to showing that (ln(n))^7 < sqrt(n) from a certain n and on. I thought (ln(n))^7 is always bigger than sqrt(n), but wolfram alpha says it's only true for n < ~3.5 * 10^24. How can one find such bound or even know that the inequality is true from a certain point? thanks! 31.154.176.33 (talk) 21:34, 9 March 2016 (UTC)[reply]

I assume we're talking about only positive values of n, right? Well, n↦n^1/7 is a strictly increasing function, so you can apply it to both sides of an inequality without changing the solution set:

\left(\log n\right)^{7}<{\sqrt {n}}

\log n<{\sqrt {n}}^{\frac {1}{7}}=n^{\frac {1}{14}}

Similarly, you can exponentiate both sides:

n<e^{n^{\frac {1}{14}}}

See where you can get from there. --Trovatore (talk) 22:18, 9 March 2016 (UTC)[reply]

Actually Alpha says that the inequality (ln(n))^7 < sqrt(n)) is true outside an interval given in terms of the Lambert W function, but approximately n>3.35526×10^24 or n<2.94531. (You have to press the "Approximate values" button to get actual numbers.) Note that for n=1, (ln(n))^7=0<1<sqrt(n) so the statement that "(ln(n))^7 is always bigger than sqrt(n)" is easily shown to be false. One approach to solving this (without Alpha's help) is to write n=u¹⁴. This reduces the inequality to 14 logu < u or u-14logu > 0. A bit of calculus shows that f(u)=u-14logu approaches ∞ near u=0, decreases to 14(1-log14)<0 at u=14, and increases to ∞ as u increases to ∞. So the graph of f has a U-shape and crosses the x-axis twice, once at some u between 0 and 14, and again at some u>14. It's fairly easy to find these values numerically without resorting to Lambert functions, then you can raise these values to 14th power to get the solution in terms of x. --RDBury (talk) 01:07, 10 March 2016 (UTC)[reply]

thanks! 31.154.176.33 (talk) 06:05, 10 March 2016 (UTC)[reply]