A Digression
by t0rajir0u, Oct 13, 2007, 7:02 PM
Apologies for not updating my blog in awhile! 
It just seemed that I had nothing interesting to talk about. However, now that I'm teaching a little class as part of the newly-formed Seattle Infinity Math Circle, I've brought up a few interesting things that might be worth posting. Here's one from last week, the Trigonometry lecture.
Exercise: Show that
.
Discussion: This is a consequence of the so-called "Sandwich Theorem," which states that if we have two functions
such that 
, and a third function 
such that 
, then 
.
Consider the points
for a small radian angle 
. It's well known that the length of arc 
is simply . The perpendicular from 
to 
has length 
, and is clearly shorter than arc . Moreover, extending to the point 
, we have a geometric view of the inequalities (for small )
As
, the Sandwich Theorem allows us to conclude the result.
Although the above is a result usually introduced in calculus (it is necessary to establish the derivatives of
), many high-school problem solvers are already familiar with it. It is necessary, however, as the last step in establishing an identity for 
first due to Vieta. (The motivation for Vieta's discovery was geometric - approximating a circle using a square, then an octagon, then a 
-gon, ad infinitum - but this algebraic presentation is nicer, in my opinion.)
The familiar double-angle identity for
can be written as
Repeated application of this identity to the factor in the RHS produces the sequence of identities
$ \begin{eqnarray*} \sin x &=& 2 \sin \frac{x}{2} \cos \frac{x}{2} \\ &=& 4 \sin \frac{x}{4} \cos \frac{x}{4} \cos \frac{x}{2} \\ &=& 8 \sin \frac{x}{8} \cos \frac{x}{8} \cos \frac{x}{4} \cos \frac{x}{2} \\ &=& ... \\ &=& 2^k \sin \frac{x}{2^k} \cos \frac{x}{2^k} ... \cos \frac{x}{2} \end{eqnarray*}$
Now,
so taking
to infinity and dividing by 
, we have
But wait! There's more! Substituting
, we have the identity
We are still not done. Repeated application of the cosine half-angle formula yields the beautiful identity
I first encountered this proof in Eli Maor's Trigonometric Delights, which I highly recommend. There's some good math in there as well as some interesting history.
I did not cover this in the lecture, but there is another infinite product for
. Euler used it in a totally nonrigorous way (and so will I; the actual proof of this product form is beyond me), but if we consider 
as a polynomial with roots 
, then we can derive the expression
Substituting as before produces the Wallis formula
Practice Problem: Prove Machin's formula,
Hint: Calculate
. Can you find other such expressions? (Other formulae of this type are known as Machin-like formulae).
It just seemed that I had nothing interesting to talk about. However, now that I'm teaching a little class as part of the newly-formed Seattle Infinity Math Circle, I've brought up a few interesting things that might be worth posting. Here's one from last week, the Trigonometry lecture.Exercise: Show that
.Discussion: This is a consequence of the so-called "Sandwich Theorem," which states that if we have two functions
such that 
, and a third function 
such that 
, then 
.Consider the points
for a small radian angle 
. It's well known that the length of arc 
is simply . The perpendicular from 
to 
has length 
, and is clearly shorter than arc . Moreover, extending to the point 
, we have a geometric view of the inequalities (for small )
As
, the Sandwich Theorem allows us to conclude the result.Although the above is a result usually introduced in calculus (it is necessary to establish the derivatives of
), many high-school problem solvers are already familiar with it. It is necessary, however, as the last step in establishing an identity for 
first due to Vieta. (The motivation for Vieta's discovery was geometric - approximating a circle using a square, then an octagon, then a 
-gon, ad infinitum - but this algebraic presentation is nicer, in my opinion.)The familiar double-angle identity for
can be written as
Repeated application of this identity to the
factor in the RHS produces the sequence of identities$ \begin{eqnarray*} \sin x &=& 2 \sin \frac{x}{2} \cos \frac{x}{2} \\ &=& 4 \sin \frac{x}{4} \cos \frac{x}{4} \cos \frac{x}{2} \\ &=& 8 \sin \frac{x}{8} \cos \frac{x}{8} \cos \frac{x}{4} \cos \frac{x}{2} \\ &=& ... \\ &=& 2^k \sin \frac{x}{2^k} \cos \frac{x}{2^k} ... \cos \frac{x}{2} \end{eqnarray*}$
Now,
so taking
to infinity and dividing by 
, we have
But wait! There's more! Substituting
, we have the identity
We are still not done. Repeated application of the cosine half-angle formula yields the beautiful identity
I first encountered this proof in Eli Maor's Trigonometric Delights, which I highly recommend. There's some good math in there as well as some interesting history.
I did not cover this in the lecture, but there is another infinite product for
. Euler used it in a totally nonrigorous way (and so will I; the actual proof of this product form is beyond me), but if we consider 
as a polynomial with roots 
, then we can derive the expression
Substituting
as before produces the Wallis formula
Practice Problem: Prove Machin's formula,
Hint: Calculate
. Can you find other such expressions? (Other formulae of this type are known as Machin-like formulae).
.
.
.
,
April 2009
March 2009