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Dan
Dozens Disciple
Dan
Dozens Disciple
Joined: Aug 8 2005, 02:45 PM

Jun 6 2017, 06:05 AM #1

It's been a long time since high school, but as I recall, trigonometric concepts were introduced in the following order:
  • Pythagorean Theorem
  • Definitions of sin, cos, & tan based on a right triangle.
  • The Unit Circle
  • Law of Sines, Law of Cosines, and various trigonometric identities.
  • Derivatives
  • Taylor series
But what if we start at the end? Define the trig functions in terms of Taylor series, prove some theorems using them, and eventually connect it to geometry?

Define:
  • S(x) = x - x3/3! + x5/5! - x7/7! + x9/9! - ...
  • C(x) = 1 - x2/2! + x4/4! - x6/6! + x8/8! - ...
  • E(x) = 1 + x + x2/2! + x3/3! + x4/4! + ...
A few obvious results follow:
  • S(0) = 0
  • C(0) = 1
  • E(0) = 1
  • S(-x) = -S(x)
  • C(-x) = C(x)
What about derivatives? Using the Power Rule:

S'(x) = d/dx (x - x3/3! + x5/5! - x7/7! + x9/9! - ...)
= 1 - 3x2/3! + 5x4/5! - 7x6/7! + 9x8/9! - ...
= 1 - x2/2! + x4/4! - x6/6! + x8/8! -
= C(x)

C'(x) = d/dx (1 - x2/2! + x4/4! - x6/6! + x8/8! - ...)
= 0 - 2x/2! + 4x3/4! - 6x5/6! + 8x7/8! - ...
= -x + x3/3! - x5/5! + x7/7! - ...
= -(x - x3/3! + x5/5! - x7/7! + ...)
= -S(x)

E'(x) = d/dx (1 + x + x2/2! + x3/3! + x4/4! + ...)
= 0 + 1 + 2x/2! + 3x2/3! + 4x3/4! + 5x4/5! + ...
= 1 + x + x2/2! + x3/3! + x4/4! + ...
= E(x)

Now, consider the function f(x) = S(x)2 + C(x)2. Using the Power Rule + Chain Rule, its derivative is:

f'(x) = d/dx (S(x)2 + C(x)2)
= 2 S(x) S'(x) + 2 C(x) C'(x)
= 2 S(x) C(x) + 2 C(x) (-S(x))
= 2 S(x) C(x) - 2 S(x) C(x)
= 0

Thus, f is a constant function. To find the value of this constant, simply plug in an arbitrary x. Zero will do nicely.

f(0) = S(0)2 + C(0)2
= 02 + 12
= 0 + 1
= 1

So S(x)2 + C(x)2 = 1 for any x. This might come in handy later, but for now I don't know what significance it may have.
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Dan
Dozens Disciple
Dan
Dozens Disciple
Joined: Aug 8 2005, 02:45 PM

Jun 6 2017, 06:20 AM #2

So far, I guess I've been implying that x is a real number, but the concept of an infinite polynomial series works just at well with complex numbers. For example, consider the function:

E(x) = 1 + x + x2/2! + x3/3! + x4/4! + x5/5! + x6/6! + x7/7! + x8/8! + x9/9! + ...

Let t be a real number and let x = it (a pure imaginary number).

E(it) = 1 + it + (it)2/2! + (it)3/3! + (it)4/4! + (it)5/5! + (it)6/6! + (it)7/7! + (it)8/8! + (it)9/9! + ...
= 1 + it + i2t2/2! + i3t3/3! + i4t4/4! + i5t5/5! + i6t6/6! + i7t7/7! + i8t8/8! + i9t9/9! + ...
= 1 + it + (-1)t2/2! + (-i)t3/3! + 1t4/4! + it5/5! + (-1)t6/6! + (-i)t7/7! + 1t8/8! + it9/9! + ...
= (1 - t2/2! + t4/4! - t6/6! + t8/8! - ...) + i(t - 3/3! + t5/5! - t7/7! + t9/9! - ...)
= C(t) + i S(t)

Interesting coincidence there, linking together the three functions defined in the original post.

Also:

E(iu) E(iv) = (C(u) + i S(u)) (C(v) + i S(v))
= C(u) C(v) + i C(u) S(v) + i C(v) S(u) - S(u) S(v)
= C(u) C(v) - S(u) S(v) + i (C(u) S(v) + C(v) S(u))
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Double sharp
Dozens Demigod
Double sharp
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Joined: Sep 19 2015, 11:02 AM

Jun 6 2017, 08:01 AM #3

Burkill's An Introduction to Mathematical Analysis introduces the exponential and circular functions in exactly this way, starting from power series. Hardy in Pure Mathematics starts by defining log and arctan as the antiderivatives of 1/x and 1/(1+x^2) respectively.

Another reasonable approach is to define sin and cos as solutions to the differential equation y" = -y, with suitable initial conditions.
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wendy.krieger
wendy.krieger

Jun 6 2017, 09:27 AM #4

The approach i use for geometry mirrors number theory.

You start with the polytopes and lattices (which are like the integers).

One then heads off to the isoseries functions, which govern the ratio of chords in all polygons t(n+1) = at(n)-t(n-1). This allows one to construct all of the chords of the polygons. When 2 is part of the series, at t(0) then t(n) = a^^n, the isopower.

The sin and cos series are the chords and shortchords.

Evaluated for integer a, this makes the iso-series.

From here, we do a simple geometric proof that the chords form an isoseries in a, and that it is possible to derive the solutions of all polygons, and such are linearly dependent.

The progression to multiplication of complex numbers = geometric plane, and that the product of points (by rotating and dialation of 0,1 to 0,P), equates to the point having the coordinate c+is = rþ. One proves the general case that i²=-1 geometrically. The logrithm of a point is then (log r, þ).

This leads to the cyclotomic numbers as the span of powers of x, where x^n = -1. From this, we can prove the general case of the isoseries is a kind of series of 2, 10.1, 100.01, etc.

Isomorphism is then discussed.

From this we head over to the Schläfli series and dynkin series.

Polytopes and the stott expansion are then brought into the picture.

As an aside, we develop hyperbolic geometry and the principle of circle drawing.

Various proofs about adding length, the right angle theorm and others are devised. The right-angle theorm is 2(2+h²)=(2+a²)(2+b²), which reduces to h²=a²+b² as these go to vanishing.

The general conformal model is introduced.
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Kodegadulo
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Kodegadulo
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Joined: Sep 10 2011, 11:27 PM

Jun 6 2017, 11:56 AM #5

Dan @ Jun 6 2017, 06:05 AM wrote:Define:
  • </li>
  • S(x) = x - x3/3! + x5/5! - x7/7! + x9/9! - ...
  • C(x) = 1 - x2/2! + x4/4! - x6/6! + x8/8! - ...
  • E(x) = 1 + x + x2/2! + x3/3! + x4/4! + ...
Or equivalently, using \(\LaTeX\):

\[S(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}\]
\[C(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}\]
\[E(x) = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}\]
S'(x) = d/dx (x - x3/3! + x5/5! - x7/7! + x9/9! - ...)
= 1 - 3x2/3! + 5x4/5! - 7x6/7! + 9x8/9! - ...
= 1 - x2/2! + x4/4! - x6/6! + x8/8! -
= C(x)

C'(x) = d/dx (1 - x2/2! + x4/4! - x6/6! + x8/8! - ...)
= 0 - 2x/2! + 4x3/4! - 6x5/6! + 8x7/8! - ...
= -x + x3/3! - x5/5! + x7/7! - ...
= -(x - x3/3! + x5/5! - x7/7! + ...)
= -S(x)

E'(x) = d/dx (1 + x + x2/2! + x3/3! + x4/4! + ...)
= 0 + 1 + 2x/2! + 3x2/3! + 4x3/4! + 5x4/5! + ...
= 1 + x + x2/2! + x3/3! + x4/4! + ...
= E(x)
\[S'(x) = \sum_{n=0}^{\infty} (-1)^n \frac{(2n+1) x^{2n}}{(2n+1)!} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} = C(x)\]

\[C'(x) = \sum_{n=0}^{\infty} (-1)^n \frac{(2n)x^{2n-1}}{(2n)!} = \sum_{n=1}^{\infty} (-1)^n \frac{x^{2n-1}}{(2n-1)!} =
- \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} = -S(x)\]

\[E'(x) = \sum_{n=0}^{\infty} \frac{nx^{n-1}}{n!} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!} = E(x)\]

(I hope I got that right... :) )
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Dan
Dozens Disciple
Dan
Dozens Disciple
Joined: Aug 8 2005, 02:45 PM

Jun 7 2017, 04:47 AM #6

Let a be a constant. An alternate Taylor series for E(x), centered around a, is:

E(x) = E(a) + E'(a)(x-a) + E''(a)(x-a)2/2! + E'''(a)(x-a)3/3! + E''''(a)(x-a)4/4! + E'''''(a)(x-a)5/5! + E''''''(a)(x-a)6/6! + ...

But E is its own derivative, so we can just remove all those prime symbols.

E(x) = E(a) + E(a)(x-a) + E(a)(x-a)2/2! + E(a)(x-a)3/3! + E(a)(x-a)4/4! + E(a)(x-a)5/5! + E(a)(x-a)6/6! + ...
= E(a) (1 + (x-a) + (x-a)2/2! + (x-a)3/3! + (x-a)4/4! + (x-a)5/5! + (x-a)6/6! + ...)
= E(a) E(x - a)

Let u = a and v = x - a. Then u + v = x, so:
E(u + v) = E(u) E(v)

Let's plug in some imaginary numbers. Let x and y be real. We know from an earlier theorem that E(i(x + y)) = C(x + y) + i S(x + y).

But another way to look at this same expression is:

E(i(x + y)) = E(ix) E(iy)
= C(x) C(y) - S(x) S(y) + i (C(x) S(y) + S(x) C(y))

So C(x + y) + i S(x + y) = C(x) C(y) - S(x) S(y) + i (C(x) S(y) + S(x) C(y)).

Equating the real and imaginary parts separately gives:
  • C(x + y) = C(x) C(y) - S(x) S(y)
  • S(x + y) = C(x) S(y) + S(x) C(y)
When y = x, then:
  • C(2x) = C(x)2 - S(x)2
  • S(2x) = 2 C(x) S(x)
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Double sharp
Dozens Demigod
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Joined: Sep 19 2015, 11:02 AM

Jun 7 2017, 10:09 AM #7

I'll be honest and say that I never actually bothered memorising the product formulae for cos A cos B, sin A sin B, and sin A cos B; it's much easier to start from the complex exponential and work backwards, like you are doing. ^_^
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Dan
Dozens Disciple
Dan
Dozens Disciple
Joined: Aug 8 2005, 02:45 PM

Jun 11 2017, 02:02 AM #8

Now, what about the roots of these infinite polynomials (where their values are equal to zero)?

E doesn't seem to have any. It definitely doesn't have any positive roots, due to the lack of sign changes in its expansion. And whatever negative roots it has seem to disappear when I add another term to the Taylor series.

But it seems like C has a root somewhere between 1;6 and 1;7. And S obviously has a root at 0, but it seems to have another one between 3;1 and 3;2.

Based on the theorem from my original post that S(x)2 + C(x)2 = 1, whenever either S(x) or C(x) is 0, the other must be either ±1. So let's consider all 4 possibilities, and take a look at how S and C behave near these roots.

Case 1: S(a) = 0, C(a) = +1

Then, from my previous results:
  • C(a + x) = C(a) C(x) - S(a) S(x) = 1 C(x) - 0 S(x) = C(x)
  • S(a + x) = C(a) S(x) + S(a) C(x) = 1 S(x) + 0 C(x) = S(x)
This gives the interesting result that if a root a &#8800; 0 indeed exists, then both functions are periodic, and have the same period.

Let &#964; denote this period, if it indeed exists. Since S(0) = 0 and C(0) = 1, then S(&#964;) = 0 and C(&#964;) = 1 also.

Case 2: S(a) = 0, C(a) = -1
  • C(a + x) = C(a) C(x) - S(a) S(x) = (-1) C(x) - 0 S(x) = -C(x)
  • S(a + x) = C(a) S(x) + S(a) C(x) = (-1) S(x) + 0 C(x) = -S(x)
Case 3: S(0) = +1, C(a) = 0
  • C(a + x) = C(a) C(x) - S(a) S(x) = 0 C(x) - 1 S(x) = -S(x)
  • S(a + x) = C(a) S(x) + S(a) C(x) = 0 S(x) + 1 C(x) = C(x)
Case 4: S(0) = -1, C(a) = 0
  • C(a + x) = C(a) C(x) - S(a) S(x) = 0 C(x) - (-1) S(x) = S(x)
  • S(a + x) = C(a) S(x) + S(a) C(x) = 0 S(x) + (-1) C(x) = -C(x)
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Dan
Dozens Disciple
Dan
Dozens Disciple
Joined: Aug 8 2005, 02:45 PM

Jun 11 2017, 03:20 AM #9

Recall that for any x:
  • C(2x) = C(x)2 - S(x)2
  • S(2x) = 2 C(x) S(x)
Now, what if x = &#964;/2? Then:
  • C(2x) = C(&#964;) = C(0) = 1
  • S(2x) = S(&#964;) = S(0) = 0
Plug these back into the original equations. Can we solve for C(x) and S(x)?
  • 1 = C(x)2 - S(x)2
  • 0 = 2 C(x) S(x)
From the second equation, C(x) = 0 or S(x) = 0 (or both). What happens when we plug them into the first equation?

If C(x) = 0, then we get S(x)2 = -1, which means that S(x) isn't a real number. So, let's try S(x) = 0. Then, C(x)2 = 1, so C(x) is either +1 or -1. But if it's +1, then we picked the wrong value for &#964;, because that makes the functions &#964;/2-periodic. So, C(x) = -1.

In summary:
  • C(&#964;/2) = -1
  • S(&#964;/2) = 0
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Double sharp
Dozens Demigod
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Joined: Sep 19 2015, 11:02 AM

Jun 11 2017, 04:10 AM #10

I would personally have gone for defining &#960; as the smallest positive root of S(x). ^_^

Also, here is a relevant blog post by Timothy Gowers. (And if anyone is concerned about differentiating power series, here is another useful post from his blog.)
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