Problems with Maxima and Minima

         Influenced by the continuous result of a square in exercises that ask for a maximum area, given a perimeter or minimum perimeter, given the area, I have determied a general equation for calculating these. The questions require that each side of a rectangle be represented in terms of one side given the length x.  This is solved as follows for a general perimeter (Formula 1) and area (Formula 2).

Formula 1

         Given the Perimeter of a rectangle P, find the Maximum Area enclosed by the rectangle.

                   Let the length and width of the rectangle be x and y respectively

                   P = 2x + 2y

                   y = P / 2 - x                                      ---(1)

                  A = x(P / 2 - x)

                 A' = P / 2 - 2x

                 Maximum value is at the stationary point (A' = 0)

                   x = P / 4                                            ---(2)

                   y = P / 2 - P / 4                (from (1) and (2))

                      = P / 4

                ∴ Maximum Area is given by a square with sides P / 4  (A = P² / 16)

         Note: the stationary point could also be found using the axes of symmetry, x = -b / 2a, of a parabola

Formula 2

          Given the Area of a rectangle A, find the Minimum Perimeter that encloses this area.

                     Let the length and width of the rectangle be x and y respectively.

                     A = xy

                      y = A / x

                     P = 2A / x + 2x

                    P' = -2A / x² + 2

                    Maximum value is at the stationary point (P' = 0)

                    2x² = 2A

                        x = √A

                        y = A / √A

                           = √A

                   ∴ Maximum Perimeter is given by a square with sides equal to √A (P = 4√A)

Curiouser and Curiouser

         Continuing my work on paradoxes, I did some research and found a paradox that I couldn't explain.  From this I obtained a general form for the paradox in which, when multiplication is reduced to a series of additions, the process of differentiation breaks down and we obtain two different answers.  To elaborate, I provide the general form for any polynomial in x here:

          C.xⁿ + ... = C. x^{n - 1}(a + a + ... + r) + ...  (where r is the remainder of x / a)  ---(1)

          d(C.xⁿ + ...) / dx = C.nx^{n - 1} + ...                                                                                ---(2)

            d(C.xⁿ + ...) / dx = C(n - 1) x^{n - 2}(a + a + ... + r) + C.x^{n - 1}(0) + ...    (from (1))

                                            = C(n - 1)x^{n - 2}(x) + 0 + ...

                                            = C(n - 1)x^{n - 1} + ...                                                                           ---(3)

                                         n = n - 1                                                (from (2) & (3))     ---(4)

          The logic behind this is that xⁿ is a shorthand expression for x.x.x, x times, which is shorthand for x + x + ..., x times x times x times ... and lastly, any number x is shorthand for a + a + a + ... + r (where r is the remainder when x is divided by a), e.g. We could say the number '23' is short hand for 4 + 4 + 4 + 4 + 4 + 3.

        As an illustration, the original paradox was given as:

                            x² = x + x + x + ...    (x times)

                   dx² / dx = 2x

                   dx² / dx = 1 + 1 + 1 + ...

                                 = x

                            2x = x      (dividing by x)

                              2 = 1

      Here C = 1, n = 2, a = 1, and r = 0.  The final expression is:

           1.2x^{2 - 1} = 1(2 - 1)x^{2 - 1} 

                  2x = x, and so on

          OR from (4)

                    2 = 2 - 1

                    2 = 1

Graphing the Difference

        On my previous post, "Approximation of Points of Intersection," I dealt with the use of finding the points of intersection of two graphs to plot the function h(x), where h(x) = f(x) - g(x).  I have decided now to include this method in a seperate post in order to expand upon the process for the general function h(x), where h(x) = f(x) + g(x).

Original

         The principle of h(x) representing the signed difference between f(x) and g(x) can also be applied to graphing.  Rather than adding the ordinates of the graphs of f(x) and -g(x), simply find the difference, if f(x) is above g(x) then h(x) is positive, otherwise h(x) is negative.

          For example graphing x³ - x², first graph x³ and x² as normal.

Fig. 14 Graph of x³ and x²

           Now, at x = 0 and x = 1 (the point of intersection), f(x) is equal to g(x) and thus h(x)has roots at 0 and 1.  Furthermore, we can observe that between 0 and 1, the distance grows and decreases slightly, below 0 the distance increases dramatically, and above 1 the distance increases at a slightly lesser rate.  Graphing this information we get.

Fig. 15 Graph of x³ - x²

End Of Original

           We can further apply this principle by representing f(x) + g(x) as f(x) - (-g(x)).  As an extension of the original example:

          To graph h(x) = f(x) + g(x), first graph x³ and -x² as normal, then use the process described above to obtain the graph for h(x).  From this we see that the graph is a mirror image of that of the original example.

Approximation of Points of Intersection

      It is often difficult to solve simultaneous equations when finding points of intersection.  Thus we can use the method of halving the interval used for approximating roots to find these.  For example:

      If we want to find the point of intersection of f(x) and g(x), where f(x) = sinx and g(x) = (x-1)²,

      First we obtain (from the graphs) an initial approximate range for the x-value of the point of intersection)

Fig. 13 Sine Curve and Parabola

          There are two points of intersection between x = 0 & x = 0.5 and x = 1.5 & x = 2.  For the purpose of this demonstration I will focus on the former.  To find the x co-ordinate, we halve the interval and find the diference (f(x) - g(x)), comparing this to the results for the original approximations.

         For this example,

         f(0) - g(0) = sin0 - (0 - 1)²

                          = 0 - 1

                          = -1

         ∴ at x = 0 sinx is below (x - 1)² and the graphs are 1 unit apart

         f(0.5) - g(0.5) = sin0.5 - (0.5 - 1)²

                                ≈ 0.48 - 0.25

                                = 0.23

         ∴ at x = 0.5 sinx is above (x - 1)² and the graphs are approximately 0.23 units apart

                   (The above results can be verified by the graph)

         Halving the interval gives x = 0.25

         f(0.25) - g(0.25) = sin0.25 - (0.25 - 1)²

                                    ≈ 0.25 - 0.56

                                    = -0.31

         ∴ at x = 0.25 sinx is below (x - 1)² and the graphs are approximately 0.31 units apart

         The point of intersection is between x = 0.5 and x = 0.25, and closer to x = 0.5

         After a second application we reach x = 0.375 as the best considered approximation

         To find the y co-ordinate, it is reasonable to say, that because the graphs are so close, the average of the values of the two functions with the given x co-ordinate will give a rough estimation for the y co-ordinate.

          So, to find the y co-ordinate as an example,

          (f(0.375) + g(0.375))/2 = (sin0.375 + (0.375 - 1)²) / 2

                                                ≈ (0.38 + 0.39) / 2

                                                = 0.385

                   (for comparison, GeoGebra gives the point as (0.39, 0.38))

          It soon becomes obvious that this process is equivalent to decreasing f(x) by g(x) and finding the root, i.e. graphing the signed distance of g(x) from f(x) and finding the point at which the distance is 0 and thus the x-intercept of the graph.

          Using this information, we can now use Newton's method of approximation, using the identity h(x) = f(x) - g(x), to find the root of h(x), and thus the point of intersection of f(x) and g(x).  Obviously the order of the functions does not matter, the only change will be the sign of h(x), not the magnitude, and this will be completely opposite so as to cancel out any effect it might have.  The range of h(x) will be restricted to the smaller of the two ranges of f(x) and g(x), however this is of little importance as the point of intersection will not occur outside this range.

          Using our first example, and an initial approximation of x = 0.5 (since this was the closest)

          h(x) = sinx - (x-1)²

          h(0.5) ≈ 0.23

          h'(x) = cosx - 2(x - 1)

          h'(0.5) ≈ 0.88 - (-1)

                     = 1.88

                  x = 0.5 - 0.23 / 1.88

                     ≈ 0.38

         This gives a y co-ordinate of 0.38, which is reasonably close - and in only one step!

Pondering Postulative Thinking (The Necessity of Postulates in Mathematics)

          The explanation given by Euclid that Axiomatic / Postulative thinking is the only method of acquiring any concrete truth is that any proof is based on either another proof or an assumption.  This in itself makes itself logically redundant, as obviously is show that either it is based on a fundamental assumption (as by its own logic, any proof can be shown to be derived from some previous assumption), in which case it may be taken to not represent a fact, or it is in fact a proof that does not require a former assumption, in which case it contradicts itself by providing evidence that it is not necessarily the truth.

           What this implies is not that this method is incorrect, but that it is not necessarily true, and there may be other methods.  Thus we cannot prove that our logic is the most suitable, only measure how closely it relates to reality, and refine it until it becomes a satisfactory model that can make future predictions on the behaviour of our environment.  Now that rant's over I'll get back to the maths!

Parallel Postulate... Why?

      Using the method I developed in "Angle Properties of Parallel Lines," we can prove that parallel lines have co-interior ∠s that are supplementary without the need for Euclid's Fifth Postulate in Book 1 of 'Elements,' or the "Parallel Postulate" as it is termed.  Given this we can then prove the ∠ sum △, using the fact that ∠s on a straight line are supplementary, and that alternate ∠s in parallel lines are equal, as in the diagram:

Fig. 11 Triangle formed within Parallel Lines

If this is acceptable, the "Parallel Postulate" can easily be proved using my 'reductio ad absurdum' proof as follows.

Fig. 12 Intersecting lines with transversal

          LR intersects L'R' at I

          Assume the co-interior ∠s between the transversal TT' and the lines LR and L'R' are supplementary

          ∠TIT' + ∠ITT' + ∠IT'T = 180 (∠ sum △TIT')

                                      ∠TIT' = 180 - 180 (∠ITT' and ∠IT'T are supp.)

                                                 = 0

          ∴ LR || L'R'

          But LR and L'R' intersect

          ∴ sum co-interior ∠s < 180 (to give ∠ sum of △ = 180) on side of intersection

          As they are supplementary in parallel lines, any two lines who's co-interior ∠s are not supplementary form a △ on the side where their sum < 180, and ∴ intersect on that side.

A Perplexing Paradox

WARNING: The author takes no responsibility for any cranium explosion that might occur due to the baffling nature of the following content...

Just joking, although this seemingly impossible occurrence may be confusing at first, I assure you there is a completely logical explanation for this phenomenon.

We begin by defining the relationship of two variables a and b as:
                                 a = b
          Now,
                            a - b = 0
             a - b + a² - ab = a² - ab
          a² + (1 - b)a - b = a(a - b)
               (a + 1)(a - b) = a(a - b)
          * Dividing by (a - b)
                            a + 1 = a
                                  1 = 0
          But it cannot be!  Therefore should we abandon the principles of algebra?

          What went wrong??

          * The Answer: We divided by a - b, which is not a defined operation as
                      we also said that a - b = 0


          To construct a paradox like this one, begin with a false expression such as:
                      2 = 4
          Now add any expression possible (including as many variables as you want)
              3a + 2 = 3a + 4     (I use expressions of a here for simplicity)
          Multiply by (a - b)   ---> remember the opposite operation, division,
                                                       is the bit that isn't allowed
          3a² - 3ab + 2a - 2b = 3a² - 3ab + 4a - 4b
          Now cancel equal expressions
          2a - 2b = 4a - 4b
          2b - 2a = 0
                   b = a
          Write these steps backwards and Voila!

          To be tricky you can also use a different ratio of a : b, in which case, multiply instead by (na - mb), where m : n = a : b (the final expression in the working should be na = mb.  The original paradox is a special case where m : n = 1 : 1.

          Proof:
          Where c ≠ d, we begin with the obviously false expression:
                                 c = d
           following the previous method, where e is any expression possible
                           c + e = d + e
         (c + e)(na - mb) = (d + e)(na - mb)
cna -cmb + ena - emb = dna - dmb + ena - emb
   (c - d)na - (c - d)mb = 0
                      na - mb = 0    (this division actually is permissible as c - d ≠ 0)
                              na = mb

A Personal Pythagorean Proof

          I produced the following proof for the most famous Pythagorean Theorem (perhaps even developed by Pythagoras himself) with a little inspiration from a diagram.  The theorem states that "The square on the hypotenuse of a right triangle is equal to the sum of the squares on the two legs."  The method I used was inspired by a diagram from the Howard Eves text "An Introduction To The History Of Mathematics, Sixth Edition," which demonstrates a possible method used by the Egyptians to prove that the Pythagorean Triple {3, 4, 5}, constitutes the sides of a right triangle.  The diagram (which is actually taken from ancient Chinese geometry) is reproduced, with slight alterations, below.  It shows four congruent right triangles arranged so that the hypotenuse, c, of each form a quadrilateral and the legs, a and b, are placed so that each leg b is applied to the leg a of another congruent triangle and vice versa, so that a smaller inner quadrilateral is formed.

Fig. 10 Area Proof of Pythagorean Theorem

The first step in this proof is to establish that the outer and inner perimeters form squares.
          For the inner quadrilateral,

                    each ∠ lies on a straight ∠ and is supplementary
                          with the opposite right ∠.
                    ∴ ∠ + 90 = 180
                                ∠ = 90
                    As each △ is congruent each side is equal to a - b
                    ∴ Inner perimeter is a square.
          For the outer perimeter,
                    as the △s were taken to be congruent,
                    the ∠s at the vertices are equal to the non-right ∠s of each △
                    ∴ ∠s are complementary (∠ sum △)
                    Each side is equal to c
                    ∴ Outer perimeter is a square
          Now,
                    Area of Large Square = 4 x Area of Triangle + Area of Small Square
                                                   c² = 4ab / 2 + (a - b)²
                                                        = 2ab + a² - 2ab + b²
                                                        = a² + b<sup>2


          EDIT: After further reading, I have discovered that this proof actually appears later in the text from which I drew inspiration under the section on Hindu mathematics.  This disappoints slightly, however I will continue to search for a unique proof (27/11/11).</sup>

The Angle Sum Problem: Solved

       I have found another method of proving the ∠ sum of △ that does not utilize ∠ properties of parallel lines!  This proof however, does not rely on the ∠ sum of △ being constant as my previous work on "The Angle Sum Problem" did, rather it proves that the ∠s of a △ are supplementary for a general case, thus proving it for all cases.  Consider the following △ (the line MM' intersects the perpendicular height and two non-base sides, through their midpoints).

Fig. 9 Scalene Triangle

        By separation of horizontal and vertical vectors, we can assume midpoints lie at the same height.

                  Let the Midpoint of AD be H

                  In △AHM' and △DHM'

                  AH = HD (by definition of midpoint)

                  HM' is common

                  AHM' = 90 (given)

                  DHM' = 180 - AHM' (supp ∠s on a straight line AD)

                             = 90

                 ∴ DHM' = AHM'

                 ∴ △AHM' ≡ △DHM' (SAS)

                 △AHM ≡ △DHM (similarly)

                 △AMM' ≡ △DMM'

                 ∴ ∠MDM' = ∠MAM' (corr ∠s in congruent △s)

                 ∠MDB = ∠MBD (base ∠s in isos △)

                 ∠M'DC = ∠M'CD (similarly)

                 ∠MDM' + ∠MDB + ∠M'DC = 180 (supp ∠s on straight line
                                                                               BC)

                 ∴ ∠BAC' + ∠ABD + ∠ACB = 180 (equal ∠s as proven above)

                 ∠ sum of △ = 180

         The basis of this proof is that, if you take each corner of any △ and fold them to the point where any interior perpendicular height meets its base; the angles will fit together on the base, a straight line that must form 180 degrees.