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Continuous Compounding – The Curious Case of e

Let’s look at a shortcut to calculate continuous compound interest using Euler’s number e . The popular formula for compound interest is: Where: t =number of years, A = final amount after t years, r =interest rate per annum.       The formula assumes that the interest rate is compounded once a year. If interest rates are compounded more frequently, the formula is: Where n is the yearly compounding frequency. Say we want to find out the compound rate of return, r , given the following: P =100000, A =8000000, t =25 years, with compounding frequency set to 1 per year. To calculate r , we can use (2) with n =1: Rearranging the terms, we get: To simplify the calculation, we may also use logarithms on the expression for (2): Therefore, r =0.1915 or 19.15% per year. In other words, if we compounded 100000 at the rate of 19.15%, once a year, for 25 years, we would end up with 8000000. Let’s increase the compounding frequenc...
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Solved Problem 4 - Expressions for the Distances from the Vertices of the Incentral Triangle to the Incenter

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RESULT For any  Δ ABC , with: sides a , b , and c angles A , B , and C internal angle bisectors AD , BE , and CF incenter I , and inradius r Proof 1 By the angle bisector theorem: Using the formula for the length of the angle bisector: Proof 2 We can use basic trigonometry to prove the result as described below. Let us drop a perpendicular IP from I onto AB or AC. It is trivial that IP = r. Hence, we get the following expression for AI : We know that: We know the popular formulae for a triangle’s inradius and area, respectively : Thus, using the obtained expression for AI, we get: Using the formula for the length of the angle bisector: Taking  √( b+c-a )  common: The expressions for IE and IF follow similarly.
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Solved Problem 3 - Expression for Inradius of the Incentral Triangle (of an Isoceles Triangle)

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RESULT For an isosceles triangle  Δ ABC  with:          sides a, b, and c , and b = c = ak, where k   is any positive real greater than 1/2          area  Δ          internal angle bisectors AD , BE , and CF , and          inradius r Proof  If X is the point of intersection of AD and FE (refer to above diagram): XD = AD - AX Applying the angle bisector theorem: Using the above in XD , we get: From the result for the side lengths of  Δ DEF : By the sine rule in  Δ ABC : Returning to our expression for DE = DF and rewriting in terms of k : FE is parallel to BC, so   ∠ EFC =  ∠ FCB =  ∠ ACF. Hence  Δ EFC is isosceles. We have already seen that: We want an expression in terms of only r and k , so let us cancel out a by rewriting s as a (1+2 k )/2: Taking...
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Solved Problem 2 - Expressions for the Side Lengths of the Incentral Triangle

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RESULT For any  Δ ABC , with: sides a , b , and c angles A , B , and C internal angle bisectors AD , BE , and CF incenter I , and inradius r Proof Applying the angle bisector theorem:  Applying the cosine rule in  Δ CDE: The above expression looks a bit messy, but we can simplify it using the cosine rule in  Δ ABC : This is much better, but we can improve it further. Let us use the projection formulae: The expressions for EF and FD follow similarly.
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Solved Problem 1 - Generalized Basel Problem for "N" Nested Infinite Sums

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RESULT Proof Let us start with a Maclaurin series and work our way to the generalized result. Dividing both sides of the Maclaurin series of sinx by x we get: Since the zeroes of sinx are integral multiples of  π,   we can use the Weierstrass factorization theorem to represent the above function as a product involving its zeroes: Expanding the above product: We want  π  in the numerator in the RHS of the final result, so let us factor out the reciprocals of the even powers of  π   in the above expression: Rewriting the above expression in the general form: Equating the above expression to our original expression for sinx/x : Equating the corresponding coefficients of powers of x in the LHS and RHS above and multiplying both sides by powers of , we get the Basel problem generalized for n nested infinite sums:
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Combinatorics Problems

COMBINATORICS PROBLEMS 1.    Since I’ve been an avid 2x2, 3x3, 4x4, and Megaminx speed-cuber for quite some time, this problem was easy to think of:  for natural  K  and  X,   find how many possible states a  K x K  Rubik’s cube can be in after  X  moves. 2.    Devise a formula that gives the number of  N -sided polygons that can be formed with all integral sides less than a given integer   X .
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Calculus Problems

CALCULUS PROBLEMS 1.     A function  f(x)  when modified gives  g(x) . Find all  f(x)  such that the same modification applied to  f'(x)   will give  g'(x)     . 2.    Find all  f(x)  that have the same roots as  f'(x) . 3.    For a given area  A , find the shape with maximum/minimum perimeter  P max / P min . 4.    For a given perimeter  P , find the shape with minimum area  A min   (apart from a rectangle).
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Number Theory and Algebra Problems

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NUMBER THEORY AND ALGEBRA PROBLEMS 1.  2.     Prove/disprove that: a.       b.           c.                        3.    Given a square of integral side length   x , find the maximum number of squares with prime side lengths that can fit into it, as a function of  x: S p (x) . 4.    Find all fractions of the form   a/b , where  a  and  b  are primes, consisting only of prime digits. 5.    Find a string of  X  consecutive composite numbers that  cannot  be expressed as the sum of ( X +1)! and  k , where k is a natural from 2 to  X+1.   6.   Consider naturals  a ,  b , and  c  to be less than a given value  N .  a, b,  and ...
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