Karl's Calculus Tutor - Solution to Exercise 5.3-1

Box 5.3-1: Solution to Exercise 5.3-1KCT logo

© 1997 by Karl Hahn

Step 1: Determine the function, f(x), that gives zero when x is the number you are after. Remember that the x we want is the number which, when raised to the twelfth power gives 2. That means you want:

   x12  =  2                                                      eq. 5.3-1-1
Clearly if you subtract 2 from both sides of that, you get a function that gives zero when x is the number you're after:
   x12 - 2  =  0                                                  eq. 5.3-1-2
So we let
   f(x)  =  x12 - 2                                               eq. 5.3-1-3
and use Newton-Raphson to find a zero of f(x).

Step 2: Determine the iteration formula for applying Newton-Raphson to equation 5.3-1-3. Remember the general formula for Newton-Raphson? It was given in equation 5.3-13, which is repeated here.

                 f(xn)
   xn+1  =  xn -                                                  eq. 5.3-13
                 f'(xn)
We know what f(x) is (equation 5.3-1-3), so to complete the iteration formula, we only have to determine what f'(x) is, which is easy. Just take the derivative of equation 5.3-1-3.
   f'(x)  =  12x11                                                eq. 5.3-1-4
So, substituting the expressions for f(x) and f'(x) into 5.3-13, we have
                 xn12 - 2
   xn+1  =  xn -                                                   eq. 5.3-1-5
                  12xn11
And that is your Newton-Raphson iteration formula.

Step 3: Get out your calculator. You'll need one that has a key on it that can raise a number to a power. The problem says to use 1.0 as your first guess, so x0 = 1.0. Now you simply apply 5.3-1-5 to x0 to get x1, and apply it to x1 to get x2, and so on. Here is the table I get using a 15-digit calculator:

   x0  =  1.0
   x1  =  1.08333333333333
   x2  =  1.06215357203892
   x3  =  1.05950026265384
   x4  =  1.05946310152991
   x5  =  1.05946309435930
When I raise x5 to the twelfth power, I get something so close to 2 that the difference couldn't possibly matter to anybody.

So the ratio of the frequencies of two adjacent notes on a piano is 1.05946309435930, to fifteen figures. Recall that Pythagoras discovered that the ratio between two notes should be the ratio of small integers in order to harmonize. So how do you get small integers out of 1.05946309435930?

Here are the ratios Pythagoras discovered when applied to the notes on a piano:

   C    #C     D    #D    E     F    #F     G    #G     A    #A     B     C
  1:1  16:15  9:8   6:5  5:4   4:3   7:5   3:2   8:5   5:3   7:4  15:8   2:1
If x = 1.05946309435930, then on the well tempered scale, we will have ratios
   C   x0  = 1.00000000000000
  #C   x1  = 1.05946309435930
   D   x2  = 1.12246204830937
  #D   x3  = 1.18920711500272
   E   x4  = 1.25992104989487
   F   x5  = 1.33483985417003
  #F   x6  = 1.41421356237310
   G   x7  = 1.49830707687668
  #G   x8  = 1.58740105196820
   A   x9  = 1.68179283050743
  #A   x10 = 1.78179743628068
   B   x11 = 1.88774862536339
   C   x12 = 2.00000000000000
None of these except the two C's are ratios of small integers. But if you take your calculator and divide out the Pythagorean ratios given above into decimals, you will find that they are all close to the well tempered numbers -- close enough that it takes a sensitive ear to tell the difference. Some piano tuners will, if you ask them, tune your piano to the Pythagorean ratios. If you are a piano player and care to try it, you will notice that when you play something familiar, it will sound different in a subtle and peculiar way -- some would say brighter, some would say harsher. Also you will find that the difference in sound depends upon what key you are playing in. And that was the original intent of the well tempered scale. It makes all keys equal, not in pitch, but in other subtle qualities.

You can find a brief biography of Pythagoras by clicking here.


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