Finding Multiple Derivatives: The General Product Rule

          As some readers may have picked up already, in my last post, Further Differentiation: Multiple Products Rule, I missed a very exciting next step that will combine it with my work on both Finding Multiple Derivatives, and the General Polynomial Expansion.  This fantastic insight is a General Product Rule for Multiple Derivatives.

        By following the reasoning in my posts on the Multiple Products Rule, and the post Finding Multiple Derivatives: The Product On Product Rule, it arises that for further differentiation of multiple products, the result will be similar to the polynomial expansion of the sum of the multiplicands to the power of the order of derivative.  If in a term a function has a certain power, that power should become the order of derivative on that function within that term rather than a power in accordance with my previous work.

      Examples

(1)          f(x) = g(x)h(x)i(x)

(2)         f'(x) = g'(x)h(x)i(x) + g(x)h'(x)i(x) + g(x)h(x)i'(x)

(3)         f''(x) = g''(x)h(x)i(x) + g'(x)h'(x)i(x) +g'(x)h(x)i'(x) + g'(x)h'(x)i(x) + g(x)h''(x)i(x) + g(x)h'(x)i'(x) + g'(x)h(x)i'(x) + g(x)h'(x)i'(x) + g(x)h(x)i''(x)

                      = g''(x)h(x)i(x) + 2g'(x)h'(x)i(x) + 2g'(x)h(x)i'(x) + g(x)h''(x)i(x) + 2g(x)h'(x)i'(x) +  + g(x)h(x)i''(x)

                    (by the Multiple Product Rule)

          Alternatively,

(1)          (g + h + i)⁰ = g⁰h⁰i⁰
(2)          (g + h + i)¹ = g¹h⁰i⁰ + g⁰h¹i⁰ + g⁰h⁰i¹
(3)          (g + h + i)² = g²h⁰i⁰ + 2g¹h¹i⁰ + 2g¹h⁰i¹ + g⁰h²i⁰ + 2g⁰h¹i¹ + g⁰h⁰i²
                     (by Polynomial Theorem)

∴    (1)    f(x) = g(x)h(x)i(x)
      (2)    f'(x) = g'(x)h(x)i(x) + g(x)h'(x)i(x) + g(x)h(x)i'(x)
      (3)   f''(x) = g''(x)h(x)i(x) + 2g'(x)h'(x)i(x) + 2g'(x)h(x)i'(x) + g(x)h''(x)i(x) + 2g(x)h'(x)i'(x) +  + g(x)h(x)i''(x)
                     (General Product Rule)

Further Differentiation: Multiple Products Rule

          In my recent post, Finding Multiple Derivatives: The Product On Product Rule, I have continued my work on further differentiation to extend it to the product rule.  For the purpose of finding a general rule for extending this to the chain rule (or function of a function rule), I have found a extension for the product rule, a 'Multiple Products Rule,' if you like.  What I mean by this is that, instead of finding the derivative of two functions multiplied together, we find the derivative for the product of multiple functions.  The well known garden variety product rule is given by:

          If f(x) = g(x)h(x),

          f'(x) = g'(x)h(x) + g(x)h'(x)

          One might expect that the general rule would be a similar sum of products where each function has its derivative multiplied by the other functions.  So, it is as I shall show now.

          Assuming true for f(x) = g(x)h(x)i(x)...,

          f'(x) = g'(x)h(x)i(x)... + g(x)h'(x)i(x)... + g(x)h(x)i'(x)... + ...
          To prove for e(x)f(x),

          Show that for d(x) = e(x)f(x),

          d'(x) = e'(x)g(x)h(x)i(x)... + e(x)g'(x)h(x)i(x)... + e(x)g(x)h'(x)i(x)... + e(x)g(x)h(x)i'(x)... + ...

          d'(x) = e'(x)f(x) + e(x)f'(x)

                  = e'(x)(g(x)h(x)i(x)...) + e(x)(g'(x)h(x)i(x)... + g(x)h'(x)i(x)... + g(x)h(x)i'(x)... + ...)

                 = e'(x)g(x)h(x)i(x)... + e(x)g'(x)h(x)i(x)... + e(x)g(x)h'(x)i(x)... + e(x)g(x)h(x)i'(x)... + ...

          ∴ If true for f(x), it is true for e(x)f(x), c(x)e(x)f(x), and so on for the product of an infinite number of functions

          Since for 1xF(x), F'(x) = 1xF'(x) + 0xF(x) = F'(x), it is true for all such products.

Trigonometric Limits

           Recently in 2 Unit Maths we have looked at some special limits involving Trig Ratios.  Although these are quite simple to evaluate, I have proven a result that could potentially make the process a lot easier.  To cut a long story short, halfway through the exercise in the text book I came up with this:

                              lim asinbx = lim (c .asinbx).ab

                              ^x>0    cx        ^x>0 (ab   cx    )  c
                                                = lim (sinbx).ab

                                                    ^x>0 (  bx  )  c

                                                = 1x (ab/c)

                                                = ab/c

          So, we can simply take the product of the two constants on the top and divide by the constant on the bottom!

Finding Multiple Derivatives: The Product on Product Rule

          In the manner of my last post, Irrational Roots, I have been looking over some of my previous work. I have even been able to extend some of it such as that on Finding Multiple Derivatives.  What I had been missing in the naiivety was a simple pattern that I dared not approach for fear of complexity.  Until now I hadn't the confidence to even explore the possibility of a 'Product on Product Rule,' as I'm choosing to name it now.

          My irrational fear was that anything that was more complex the first time would only compound in difficulty over multiple stages.  This was not the case and it actually was quite a bit simpler since I had already immersed myself in Binomial Theorem in my work on the General Polynomial Expansion, which meant that I was quite familiar with the pattern that emerged.  Firstly I shall give you a look at the pattern that comes about.

          f(x) = g(x)h(x)

          f'(x) = g'(x)h(x) + g(x)h'(x)

          f''(x) = g''(x)h(x) + g'(x)h'(x) + g'(x)h'(x) + g(x)h''(x)

                  = g''(x)h(x) + 2g'(x)h'(x) + g(x) h''(x)

          f'''(x) = g'''(x)h(x) + g''(x)h'(x) + 2g''(x)h'(x) + 2g'(x)h''(x) + g'(x)h''(x) + g(x)h'''(x)

                   = g'''(x)h(x) + 3g''(x)h'(x) + 3g'(x)h''(x) + g(x)h'''(x)

          As my introduction has probably given away, these arrangements are similar to binomial expansions.  In fact, they are exactly correlated, as I shall show now.  If we were to conduct a similar operation to the binomial expansions for (g'(x) + h'(x))ⁿ, where we treat the power on a term as the number of derivatives, we would get the following:

          (g'(x) + h'(x))⁰ = ⁰C₀g(x)h(x)
                                  = g(x)h(x)      (notice in the expansion the second term starts with a power of 0 and the first term ends with a power of 0)
          (g'(x) + h'(x))¹ = ¹C₀g'(x)h(x) + ¹C₁g(x)h'(x)
                                  = g'(x)h(x) + g(x)h'(x)
          (g'(x) + h'(x))² = ²C₀g''(x)h(x) + ²C₁g'(x)h'(x) + ²C₂g(x) h''(x)
                                 = g''(x)h(x) + 2g'(x)h'(x) + g(x) h''(x)
          (g'(x) + h'(x))³ = ³C₀g'''(x)h(x) + ³C₁g''(x)h'(x) + ³C₂g'(x)h''(x) + ³C₃g(x)h'''(x)
                                  = g'''(x)h(x) + 3g''(x)h'(x) + 3g'(x)h''(x) + g(x)h'''(x)

          The explanation for this phenomenon is much the same as that for the actual binomial expansions.  When the product is applied once, there are all possibilities of combinations between each of the functions, with the sum of the number of derivatives on each in each term being 1.  When this is differentiated the possibilities of each term are broken up, thus giving all possibilities with the sum of the number of derivatives being two.  Furthermore, the number of times a combination appears is found by ‘choosing’ that number for the derivative from the total number of the derivative for the same reason it is done when doing a binomial expansion.

Irrational Roots

          This result bears much similarity to my treatment of the Irrationality of Fermat’s Last Theorem, however is much more powerful due to its greater range, efficiency and applicability.  It also is a method for the proof of irrationality.  Although it is a simple argument it is quite far reaching.  I’ll make you wait in anticipation no longer, only working in R:

                              x ∈ Q
                              x ∉ J
                              x = m/p     (where m and p share no prime factors)
                              xⁿ =­­ mⁿ/pⁿ
                              Since the prime factors of mⁿ will be the same as those for m (because of unique factorization mⁿ can be made up by multiplying the prime factors of m by themselves – e.g 12 = 2x2x3, 144 = 2x2x2x2x3x3), and the same can be said for pⁿ and p,
                              xⁿ ∉ J
                              ∴ The nth root of a number cannot be a rational fraction, since the nth power of a fraction is always another fraction.  So, if an integer is not the nth power of another integer, its nth root is irrational.

The Enigma Of The Wine In The Water

          This problem comes from a book called Enigma (Fabrice Mazza and Sylvain Lhullier), that was gifted to my stepfather (a maths teacher) recently.  A particularly deluding puzzle is one entitled 'Water In The Wine.'  Although I think it is well known, I've never actually tried to solve it before.  The problem start:

                              "You have two completely identical tankards; one contains 150 ml wine, the other 150 ml water.  You take a spoonful of liquid from the tankard of water, empty it into the tankard of wine and mix it in well.  Then, using the same spoon, you take a spoonful of this mixture and empty it back into the first tankard of water.  Thus each tankard once again contains 150ml liquid.  Is there more water in the wine, or more wine in the water?"

          All most everyone would follow their intuitive instinct, which might say something like 'Surely a full teaspoon of water was put in the wine, yet less wine was taken back to the water?  Of course the aim of this puzzle is to trick the puzzler in exactly this manner, and as perhaps some clever chicken reading this has already spotted, the intuitive statement is true.  But it's not the whole truth.  You see a full teaspoon of water was poured into the wine, but it didn't stay there, since, when less wine was taken, the rest of the teaspoon was taken up by some of the water.  In Enigma, the solutions a the back explains the phenomenon such:

                              "Suppose the tankard contains 80% water and 20% wine at the end of the process.  The missing 20% of water can only be in the other tankard just as the missing 80% wine must also be there..."

          Of course, the power of intuition is mighty indeed, and these explanations were still not enough to conquer the minds of my family members.  So I devised a few demonstrations in order to attempt to use the art of persuasion to sway their attitudes.  Fortunately I had an impressionable youth handy, and acquiring there assistance in insistence was not difficult.  The first method of demonstration was a geometrical one.

Fig. 32 Geometry of the Water in the Wine

          This diagram represents the two vessels after the water from the first to the second.  The first vessel contains 150 ml - 1 tsp of water, and the second contains 150 ml wine + 1 tsp water.  The division on the right is the tsp of liquid that will be poured back into the first vessel.  Since it is equal in volume to the amount of water in the vessel and the small portion of water in the top right corner is common to the two, the volume of water left in the vessel and the volume of wine taken to be put in the first vessel must be equal.

          While I may have convinced some, there are still a few who hold onto the idea that my logic is missing something.  Such was the reaction of my subjects who continued their heated objection to my so obviously flawed logic.  My next approach seemed a little more powerful though, and it was this that converted my younger sibling to the cause.  It is a diagrammatic approach that gets to the roots of the problem by representing the liquids as a group of smaller volumes.  These can be likened to the molecules that make up the substances, which adds a sense of fundamental truth to an already convincing argument, if I do say so myself (which I do!).

Fig. 33 Particles of the of Water in the Wine

           Again, this diagram shows the result of the first transfer.  Since an equal amount is taken back as was put in, we must take three of the particles from vessel two and put them into the first.  If we take three wine particles there will be three of each the wrong vessel.  If the three water particles are taken back, the starting state is reached in which no water is in the wine, and no wine is in the water.  More likely, some of each is taken.  If two wine particles are taken with one water particle, there will be two particles of each left in the wrong vessel.  In fact, the less the number of wine particles taken, the more water is taken back and therefore the less that remains in the wine.  This has the added advantage of showing that no matter how much stirring occurs, there will still be equal amounts in the wrong vessels.

          Despite a couple of protestations my subjects were fairly convinced at this stage (with some aid from my younger sibling).  To push my advantage I adapted this concept to rigorously prove the result beyond all doubt.  How you ask?  With ... (drumroll)... Algebra!  First we establish the volumes

                               Let the volume of Vessel 1 be V ml, and the volume of Vessel 2 be W ml

          Now, we perform the transfers.

                              Take p ml of water from Vessel 1 and pour it into Vessel 2.

                              Now Vessel 1 contains V - p ml water and Vessel 2 contains W ml wine and p ml water

                              If we take, in a teaspoon, q ml wine and p - q ml water from Vessel 2, we have taken a total of q + (p - q) = p ml.

                              Now Vessel 2 contains W - q ml wine and p - (p - q) = q ml water.

                              Pouring the contents of the teaspoon into Vessel 1 gives (V - p) + (p - q) = V - q ml water and q ml wine.

                              ∴ There is q ml wine in the water = q ml water in the wine.

Tic Tac Triangle

           My next topic is a complicated game of tic-tac-toe, with the cool twist that its played in an infinite triangle, and its a game not between two wits but between a humble mathematician and the ultimate opponent, mathematics itself (dah dah dah! [foreboding musical number] Okay, bit of an anticlimax, sorry).  A-a-a-ny-who, back to what I was saying, the subject matter this time round is a problem in which a certain rule is applied a number of times to a set of naughts and crosses, the problem being to determine the end result without working out the next line in the sequence.

          The sequence in question begins with a row of any number of X's and O's.  The next line is then constructed with one less of these, each being the result of the two above them.  If a symbol's genesis is two parents of the same denomination, the child is an O (i.e. O + O --> O, X + X --> O).  If the parents are different, the child is an X (i.e. O + X --> X, X + O --> X).  Each new line is thus produced until the last, with one child, is reached.  Thus a typical arrangement might be any of the following:

                                OOX     XOXXO       XOOXXOXOXOOXO

                                 OX        XXOX          XOXOXXXXXOXX

                                   X          OXX             XXXXOOOOXXO

                                                  XO               OOOXOOOXOX

                                                    X                 OOXXOOXXX

                                                                         OXOXOXOO

                                                                          XXXXXXO

                                                                           OOOOOX

                                                                            OOOOX

                                                                             OOOX

                                                                               OOX

                                                                                 OX

                                                                                   X

            The question of what the end result will be for a given first row can be overcome with a little logical reasoning and an extra dose of cunning.  Firstly, we can take a look at what will happen with an initial row of three.  Here are the eight possibilities:

           XXX     XXO     XOX     XOO    OXX    OXO      OOX     OOO

            OO        OX        XX        XO       XO       XX         OX       OO

              O          X           O           X          X         O            X          O

          As you can see, if the two outer symbols are the same, the result is an O, and otherwise it is an X.  This is because if they are the same they will react in the same way with the centre parent, producing two children that are the same and those will then produce an O.  If they are different, they will react differently with the centre and by the same reasoning produce an X.  So wherever there appears a row of three, the child two rows down will be exactly as if the two outer parents of the triplet directly parented the child.  Now we can look at the arrangement with five in the first row.  Let's pick an arbritary arrangement:

                                    \XO\X/XO/

                                      \XX/\OX/

                                       \O/X\X/

          "I think I'll stop there."  Only because this is all we need.  You'll notice I have marked in two overlapping sets of three.   Here the two outer parents of a set of three are determined by the two outer parents and the centre parent of a set of five.  By the same reasoning as for the set of three, we can show that the end result of a set of five will depend only on the outer symbols as parents.  In the same way we can split a set of nine up as a set of five over the top of a set of five, and a set of thirteen will be a set of seven over a set of seven and so on for all odd numbers of the form 1 + 4k (where k is a positive integer).  If the set begins with an even number of the form 2(2k + 1), we only need to work out the two outer children of the next line, and then the end result will be clear.  For an odd number of the form 3 + 4k, an arrangement can be found that will end with something like this:

                                                     \XOXO\OXX/OXOO/

          In this way two parents can be determined by inspecting the child of the first symbol and the symbol one place to the right of the middle, and the child of the last symbol, and the symbol one place to the left of the middle.  Now, for the case of a set of even number of the form 4(k + 1), the outer parents of the row down, and the child of the two symbols on the right of the middle line, and the child of the two symbols on the left of the middle line, and then the end result can be achieved as above.

How Many Squares On A Chessboard?

Fig. 29 How many squares can you find?

          This problem is much like the previous one (see How Many Triangles in a Triangle?), in that a relationship must be proven for the number of geometrical figures that can be constructed from the arrangement given, as a function of the order.  In this problem, a range of different sized chessboards are given, with the conventional board being of order eight (8 sides).

Fig. 30 Unconventional Chessboards

          The text investigates the properties of these first few, then asks how many squares could be found on the one with eight squares on each side.  It then leaves as an exercise for the reader to find a general rule that could be applied.  It can be verified easily for these first few that for a function N(O = n), where N is the number of squares and O is the order of the figure, that N(O = n) = (n/6)(n + 1)(2n + 1).  I have found two different proofs for this relationship, but there may well be many more.

          Firstly we take a look at the square of order n (I'll use n = 3 to illustrate), and assume that we have already counted those for that of order n - 1 which appears in the bottom left, so we are only looking at the top and right edges (and those parts that are overlapped by the larger sized squares).  Now we can see that there will be one square of side length n (there is only one three by three square in that of order three.).  Looking at the next largest square, with side length n - 1, we can fit along the edge three of these (since it shifts one space from the top left corner and then reaches the corner and must shift down a space for the next one), in the 3x3 square, three 2x2 squares can be fit along the edge like this.

Fig. 31 2x2 squares on the edge of 3x3 square

          We can continue this through, forming a sum of a series of odd numbers up until we reach 1x1 squares of which there are 2n - 1, since there are n on the top row and n - 1 on the right, since the top right square was counted in the top.  So now we have this series:

                               N(O = n) - N(O = n - 1) = 1 + 3 + ... + (2n - 1)    -- (1)

           It can be shown by induction that (1) is equal to the square of n.  So for each value of n the function gives a square plus the function before, which was its square plus the function before it, and so on.  Since N(O = 1) = 1², we can say that N(O = 2) = 2² + 1², and so on up through all values of n.  Now we have reached the formula:

                               N(O = n) = n² + (n - 1)² + ... + 2² + 1²     -- (2)

          To reach the formula given at the beginning we can use an inductive approach as follows.

                    Assuming that N(O = n = k) = (k/6)(k + 1)(2k + 1),

                    N(O = n = k + 1) = (k/6)(k + 1)(2k + 1) + (k + 1)²

                                                = [(k + 1)/6][k(2k + 1) + 6(k + 1)]

                                                = [(k + 1)/6](2k² + k + 6k + 6)

                                                = [(k + 1)/6](2k² + 7k + 6)

                                                = [(k + 1)/6](2k² + 4k + 3k + 6)

                                                = [(k + 1)/6][2k(k + 2) 3(k + 2)]

                                                = [(k + 1)/6](k + 2)(2k + 3)

                                                = [(k + 1)/6][(k + 1) + 1][2(k + 1) + 1]

                     So if it is true for O = n = k it is also true for O = n = k + 1, and in turn O = n = (k + 1) + 1 = k + 2, and so on for all integer values of O = n > k.

                     But N(O = 1) = 1, and (1/6)(1 + 1)(2 x 1 + 1) = (2 x 3)/6 = 1.  Therefore it is true for all O = n greater than or equal to 1.

         The second method for this proof involves the derivation of (2).  Once again we assume that all the squares in the square of order n - 1 that appears in the bottom left have been counted.  Now, we notice that along the top row there will be 1 nxn square, 2(n - 1)x(n - 1) squares, and so on until we notice n 1x1 squares.  On the right edge we now have one less of each square (since those that fit in the top right corner have been counted).  So on the edge we have the triangular number of n (T_n) and the triangular number of n - 1 (T_{n - 1}).

         By induction it can be shown that the triangular number for a given value x, is given by T_x = (x² + x) / 2.  So now we have:

                             N(O = n) - N(O = n -1) = (x² + x) / 2 + [(x - 1)² + (x - 1)] / 2

                                                                   = (x² + x + x² - 2x + 1 + x - 1) / 2

                                                                   = (2x²) / 2

                                                                   = x²

          This is equivalent to (2) and the proof is completed in the same way from there.

How Many Triangles in a Triangle?

Fig. 27 Checkered Triangles

          This diagram appeared in the, afore mentioned, puzzle book (see The Art Of Mathematical Origami).  The problem?  Prove that in this series of triangles, there appear as many triangles as the cube of the order.  In other words, if we examined triangle n, we would find that it contained n³ unique triangles within in it (eg. 2³ = 8, 3³ = 27 are easily verified).  I especially like this proof as it shows how unifying mathematics can reap benefits.  This particular proof uses Combinations, used in probability theory and binomial expansions, to generate a result in geometry.  Although not specifically shown, I assume the triangle that corresponds to n = 1, would look like this.

Fig. 28 Triangle of the First Order

          To prove the relation given, firstly, we recognize that any triangle created must contain either of the points A, or C, since all other points are either not connected or lie on the same line as each other.  Now, we tackle each case separately.

          For those triangles that contain A, there are n lines produced from C that are intersected by the n lines produced from A, and also inherently by C, giving a total of n + 1 intersection points.  Three points must be chosen from this selection of points in order to make up a triangle.  The number of arrangements possible is given by the number of lines (n) multiplied by the number of possibilities on each line.  Assuming A is already chosen, this leaves nⁿ⁺¹C₂.

          The situation is similar for those triangles containing C, however those triangles containing A have already been accounted for and thus there is one less point available to be chosen on each line (as A lies on all of these lines).  Using the same logic as that followed for A, we can see that the number of possibilities for triangles containing B, and not A, is nⁿC₂.  Since we have covered all possible triangles, the total number of possibilities is the sum of those expressions already derived.

        Total N^o ∆s = n(ⁿ⁺¹C₂ + ⁿC₂)

                            = n[(n + 1)!/2!(n – 1)! + n!/2!(n – 2)!]

                            = n[n(n + 1) + n(n – 1)]/2

                            = n(2n²/2)

                            = n³

          Isn’t that beautiful?