Chapter 10

5. Suppose AB is taken equal to 1, and one wants to multiply BD by BC.  Join AC and draw DE parallel to CA.  Show that BE is the product of BD and BC.  Similarly, given two lengths BE and BD, construct the quotient length p. 278

Two parallel lines cutting a line have equal angles on the alternate sides, so ÐBAC @ ÐBDE, and ÐBED @ ÐBCA.  Since ÐBAC = ÐDAE, these are similar triangles.  Thus corresponding pairs of sides are in the same ratio: AB/BC as BD/BE; cross-multiplying gives AB*BE = BC*BD.  Since AB = 1, BC*BD = BE.  Dividing both sides by BD gives BE/BD = BC.

24. The best-known of Descartes's statements is, "I think, therefore I am," from the Discourse on Method. The context is Descartes's resolve to accept only those ideas that are self-evidently true.   There is a well-known joke based on this statement: Descartes goes into a restaurant.  The waiter asks him, "Would you like tonight's special?" He replies, "I think not," and disappears.  Comment on the logical validity of this joke.

Mainly, it assumes the converse of "I think, therefore I am" – "I am not, therefore I think not" has the same truth value, but "I think not, therefore I am not" does not necessarily.  Also, of course, what's being negated is not "think" but "will have," so it's mainly a grammar joke.

26. What advances in technique and/or understanding enabled Girard to state that every algebraic equation has as many solutions as its degree?  Why were Cardano and Viete unable to make such an assertion?

Girard had the advantages of algebraic notation, as Cardano did not, and he considered negative solutions, as Viete did not.  Most important,  complex numbers were beginning to be understood.  The truth of the  Fundamental Theorem of Algebra requires complex roots.

Chapter 11

1. Show that Fermat's two methods of determining a maximum or minimum of a polynomial p(x) are both equivalent to solving p'(x)=0.

(p(x1) – p(x2))/(x1-x2), if we set x2-x1=e, is (p(x+e) – p(x))/e, which is the same as (p(x+h)-p(x))/h.  Setting x1=x2 and thus e equal to 0 is the equivalent of taking the limit of that ratio as h goes to 0, which is exactly the derivative of the function p(x), p'(x).  Because p(x1) is set equal to p(x2) at the beginning, the ratio of the change in function value to the change in x value is set equal to 0, so p'(x) = 0.

2. Use one of Fermat's methods to find the maximum of bx – x3.  How would Fermat decide which of the two solutions to choose as his maximum?

bx1 – x13 = bx2 - x23  then b(x1 – x2) = x13 – x23  dividing by (x1-x2)  gives

b = x12 + x1x + x22  Setting x1 = x2, b = 3x2,  x = ±Ã(b/3) .

This is a volume problem so only positive values make sense,  bx – x3 is > 0 for x > 0 and   < b*b.  The  maximum would occur here.

4. Use Fermat's tangent method to determine the relation between the abscissa x of a point B and the subtangent t that gives the tangent line to y = x3.

f(x+e)/f(x) = (t+e)/t

(x+e)3/x3 = (t+e)/t

tx3 + 3tx2e + 3txe2 + te3 = tx3 + ex3

3tx2e + 3txe2 + te3 = ex3

3tx2 + 3txe + te2 = x3

setting e = 0 gives 3tx2 = x3, t = x/3

16. Determine the area under the curve y = pxk from x = 0 to x = x0 by dividing the interval [0,x0] into an infinite set of subintervals, beginning from the right with the points a0 = x0, a1 = (n/m)x0, ..., where n < m, and proceeding as in Fermat's derivation of the area under the hyperbola.

 

Following Fermat's method, the area of the first circumscribed rectangle, from (n/m)x0 to x0, is R1 = (x0-(n/m)x0)y0 = (1-(n/m))x0px0k = (1-(n/m))px0k+1

The area of the next rectangle is

R2 = ((n/m)x0-(n/m)2x0)p((n/m)x0)k = (n/m)(1-(n/m))x0(n/m)kpx0k = (n/m)k+1(1-(n/m))px0k+1

     = (n/m)k+1R1

and the area of R3 is (n/m)2(k+1)R1

So the area of all the rectangles is

R = R1 + (n/m)k+1R1 + (n/m)2(k+1)R1 + ... = R1[1+(n/m)k+1+(n/m)2(k+1)+...]

or, using the formula for the sum of a geometric series,

R = (1/(1-(n/m)k+1))R1 = (1/(1-(n/m)k+1))*(1-n/m)px0k+1

= 1/((n/m)+(n/m)2+...+(n/m)k+1)*px0k+1.

 The last equation can be obtained by long division of (1 – n/m) k+1  by (m/n – 1).

Then  allowing n/m to approach 1, the value becomes

px0k+1/(k+1) = (1/(k+1))x0y0.

 

23. Square Newton's power series for (1-x2)1/2 and show that the resultant power series is equal to 1 – x2.  (You need to convince yourself that every coefficient beyond that for x2 is equal to 0.)

(1 + (1/2)x2 – (1/8)x4 + (1/16)x6 – (5/128)x8 + ... )2

=(1+(1/2)x2 – (1/8)x4 + (1/16)x6 – (5/128)x8 + ... )

+      (1/2)x2 + (1/4)x4 – (1/16)x6 + (1/32)x8 – (5/256)x10 + ...)

+                    -(1/8)x4 – (1/16)x6  + (1/64)x8 – (1/128)x10 + (5/1024)x12 + ... )

+                                     (1/16)x6 + (1/32)x8 – (1/128)x10 + (1/256)x12 – (5/2048)x14 + ...)

+                                                 - (5/128)x8 – (5/256)x10 + (5/1024)x12 – (5/2048)x14 + ...)

etc...

  1 +        x2 +        0x4         + 0x6        + 0x8+ ....

=1+x2