User blog comment:Tuner King 5/Help plz/@comment-24808864-20160112230745/@comment-28838092-20160116001834

Finally got it :`1

“At which x are they growing the fastest?”

The command given on page 108 in the book is N + 1 ⇒ N:1-(1-P) (366-N)/365 ⇒ P: {N,P} I adjusted it with leap day: N + 1 ⇒ N:1-(1-P) (367-N)/366 ⇒ P: {N,P}

Using this formula, we can get the secant slopes

Finding the secant lines:

P(0) = 0

P(10) = 0.1166 P(10) - P(0) = 0.1166 - 0 = 0.1106 P(20) = 0.4106 P(20) - P(10) = 0.4106 - 0.1166 = 0.294 P(30) = 0.7053 P(30) - P(20) = 0.7053 - 0.4106 = 0.2947 P(40) = 0.8905 P(40) - P(30) = 0.8905 - 0.7053 = 0.1852 P(50) = 0.9701 P(50) - P(40) = 0.9701 - 0.8905 = 0.796

The secant lines have the greatest speed between the interval of [15,25], so let's take the secant lines in this interval to get a better approximation

P(15) = 0.2522978592

P(16) = 0.2829413896 P(16) - P(15) = 0.2829413896 - 0.2522978592 = 0.030635304 P(17) = 0.3142882141 P(17) - P(16) = 0.3142882141 - 0.2829413896 = 0.0313468245 P(18) = 0.3461382151 P(18) - P(17) = 0.3461382151 - 0.3142882141 = 0.031850001 P(19) = 0.3782953521 P(19) - P(18) = 0.3782953521 - 0.3461382151 = 0.032157137 P(20) = 0.4105696371 P(20) - P(19) = 0.4105696371 - 0.3782953521 = 0.032274285 P(21) = 0.4427789465 P(21) - P(20) = 0.4427789465 - 0.4105696371 = 0.0322093094 P(22) = 0.4747506463 P(22) - P(21) = 0.4747506463 - 0.4427789465 = 0.0319716998 P(23) = 0.5063230118 P(23) - P(22) = 0.5063230118 - 0.4747506463 = 0.0318165488 P(24) = 0.5373464291 P(24) - P(23) = 0.5373464291 - 0.5063230118  = 0.0310234173

At x = 20, they are growing the fastest