DISCUSSION

The following is the derivation of the weight-loss formula:

w_i = weight in A.M. on day i

c_i = calories consumed on day i

K w_{i =} calories consumed on day i to maintain body weight and

K = K₁₅₄ ( a + b w_i ), which allows for the fact that the required

caloric intake may depend on the bodyweight in a linear way.

For a 154 lb man, K₁₅₄ = 2700 cal / 154 lb – day (Ref. 1)

then, K = 2700 cal / 154 lb – day = K₁₅₄ ( a + b (154) ).

So, of course, when K = K₁₅₄ , a + 154 b = 1, exactly.

We propose:

(w_{i + 1} – w_{i )} D / ∆ t = ( c_i – K w_i ) where D is a constant to be found

in the literature, and ∆ t = t _{i + 1} – t _i = one day, and one pound,

w_{i + 1} – w_i = 1.

In other words, the weight gain is is one pound and the calorie

difference between day i and day i + 1 is c_i – K w_i . Then c_n – Kw_n = D

where D is the calories to gain or lose one lb. body weight on some day n.

D = 3500 cal. / lb. (Ref. 2)

For a period of fasting, we assume c_i = ( 1 – f ) K w_i where f is the

fraction of the week fasted. In other words, the calorie loss is averaged

over the whole week. When f = 0 body weight is unchanged and when

fasting the whole week, ( f = 1 ), the daily weight loss is roughly K w_i / D.

During the periods of the fasts, f is taken to be a constant during the fasts.

Gathering terms, we have the master equation:

(w_{i + 1} – w_i ) _/ ∆ t = – f K₁₅₄ ( a + b w_i ) w_i / D and [1]

taking the limit as i approaches i + 1 we have a differential equation:

d w / d t = – f K₁₅₄.( a + b w ) w / D [2]

Rearranging,

d w / (( a + b w ) w ) = – ( f K₁₅₄ / D ) d t [3]

Integrating, we have two equations in two unknowns:

– ( 1/ a) ln ( ( a + b w ) / w ) = – ( f K₁₅₄ / D ) t + C [4]

where C is a constant of integration. From the data,

the first equation:

( ( 3 / 7) a ( 2700) / ( 154 x 3500 ) ) ( 56 ) = [5]

= ln ( ( a + 173.5 ( b ) ) / 173.5 ) – ln ( ( a + 201 ( b ) ) / 201 )

and the second equation:

( ( 1 / 7) a ( 2700) / ( 154 x 3500 ) ) ( 112 ) = [6]

= ln ( ( a + 162.7 ( b ) ) / 162.7 ) – ln ( ( a + 173.5 ( b ) ) / 173.5 )

Adding the equations:

( ( 3 / 7) a ( 2700) / ( 154 x 3500 ) ( 56 ) + ( ( 1 / 7) a ( 2700) / ( 154 x 3500 ) ) ( 112 ) =

= ln ( ( a + 162.7 ( b ) ) / 162.7 ) – ln ( ( a + 201 ( b ) ) / 201 ) [7]

Now, ( 2700 / ( 154 x 3500 x 7 ) ) ( 3 x 56 + 112 ) a = 0.200371 a [8]

So,

( a + 162.7 b )

ln ( —————— ) + ln ( 201 / 162.7 ) = 0.200371 a [9]

( a + 201 b )

a + 154 b = 1 or a = 1 – 154 b [10]

So,

( 1 – 154 b + 162.7 b )

ln ( —————————- ) + ln ( 201 / 162.7 ) = 0.200371 a [11]

( 1 – 154 b + 201 b )

( 1 + 8.7 b )

ln ( —————) + ln ( 201 / 162.7 ) = 0.200371 ( 1 – 154 b ) [12]

( 1 + 47 b )

( 1 + 8.7 b )

ln ( —————) + 30.857134 b = 0.200371 – ln ( 201 / 162.7 ) [13]

( 1 + 47 b ) = 0.200371 – 0.2113948 = – 0.0110238

( 1 + 8.7 b )

ln ( —————) + 0.0110238 = – 30.857134 b [14]

( 1 + 47 b )

( 1 + 8.7 b )

[ln ( —————) + 0.0110238] / – 30.857134 = b [15]

( 1 + 47 b )

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How do you get b? Newton approximation.

( 1 + 8.7 x )

[ ln ( —————) + 0.0110238 ] / ( – 30.857134 ) = x [16]

( 1 + 47 x )

(( ln ( 1 + 47 x ) – ln ( 1 + 8.7 x ) – 0.0110238 ) / 30.857134) = x [17]

Taking the Newton approximation where x₁ = x₀ – [ f (x₀ ) / f ‘(x₀ ) ]

and starting with a good estimate of x₀ = 0.002 we get 0.002042 for b [18]

by running it through 3-4 cycles, that is x_n+1 = x_n – [ f ( x_n ) / f ‘ ( x_n ) ]

until the desired number of significant figures are obtained. Since the data

are only about three significant figures, 0.002042 suffices. Using this method

f ( x ) = [ ln ( 1 + 47 x ) – ln ( 1 + 8.7 x ) – 0.0110238 ] / 30.857134 – x

and f ‘(x₀ ) = the first derivative of f ( x ) evaluated at x = x₀. The two

Taylor series approximations gave bounds of 0.001481 and 0.004844 for

the starting value. So, a = 1 – 154 b = 1 – (154) (0.002042) = 0.6855. [19]

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TAYLOR SERIES I

Expanding the logarithmic term on the left, ln ( 1 + x ) = x (approx)

( 8.7 – 47 +  30.857134 ) b =  0.200371

– ln ( 201 / 162.7 )

( – 38.3 + 30.857134 ) b = ( 0.200371

– ln ( 201 / 162.7 ) )

b = ( 0.200371

– ln ( 201 / 162.7 ) ) / ( – 7.443 )

b = ( 0.200371 – ln ( 1.2354 ) ) / ( – 7.443 )

b = ( 0.200371 – 0.2113948 ) / ( – 7.443 )

b = 0.0110238 / 7.443 = 0.001481 (First order approximation.)

So, a = 1 – 154 b

a = 1 – ( 154 ) ( 0.001481 ) = 0.771926

Therefore,

K = K₁₅₄ ( 0.771926 + 0.001481 w )

Where again, K₁₅₄ = 2700 cal / 154 lb – day

Notice K = K₁₅₄ when w = 154

And when w = 200, ( 0.771926 + 0.001481 x 200 ) = 1.068

So there is a definite dependence of K on the weight.

Presumably, a statistical study could be done to test this.

If there were enough subjects, maybe a month’s fasting would

be enough. And probably the data could all be normalized to

fit on a common curve.

TAYLOR SERIES II

Now, to test the Taylor series up to the 2^nd order term, we have,

ln ( 1 + x ) = x – ( ½ ) x²

Going back to the previous result, 0.200371 – ln ( 1.2354 ) = – 0.0110238

0.200371 – 0.2113948 = – 0.0110238

( 1 + 8.7 b )

ln ( —————) + 30.857134 b = 0.200371 – ln ( 201 / 162.7 )

( 1 + 47 b )

8.7 b – ½ ( 8.7 ) ² b² – 47 b + ( 47 )² b² + 30.857134 b = – 0.0110238

( 8.7 – 47 ) b – ( ½ ) ( ( 8.7 )² – ( 47 )² ) b² + 30.857134 b =

= 0.200371 – ln ( 201 / 162.7 )

= – 0.0110238

Which results in the quadratic,

( 1066.655 ) b² – 7.44287 b + 0.0110238 = 0

b = ( 7.44287 ±

SQRT ( ( 7.44287 ) ² – 4 (1066.655 ) (0.0110238 ) ) )/ ( 2 x 1066.655 )

b = ( 7.44287 + SQRT ( ( 7.44287 ) ² – 47.03437 ) ) / 2133.31

b = ( 7.44287 + SQRT ( 55.39631 – 47.03437 ) ) / 2133.31

b = ( 7.44287 + SQRT ( 8.36194 ) ) / 2133.31

b = ( 7.44287 + 2.8917 )/ 2133.31

b = 10.3346 / 2133.3 = 0.004844 (Second order approximation.)

This differs from taking only the term to the first power.

References:

1. Nutritive Value of Foods U.S. Dep’t Agric. Home & Garden Bull. 72,

p. 32 (April 1981).

2) Pediatric Nutrition Handbook American Academy of Pediatrics,

p. 433 (1979).