Re: [math-fun] Intuitively simple trig functions
Mike wrote: << Is there a geometric way to understand the Taylor series for sin and cos? The closest I've been able to find is a combinatorical explanation (below), but it doesn't seem to help much.
I'd love to know such a way to understand their Taylor series. I recently gave a talk to a bunch of smart youngsters about complex numbers, and was unable to find a truly graceful way to explain why exp(ix) = cos(x) + i sin(x), (without deriving their Taylor series). So if anyone knows a way to see this, I'd love to know it. But of course that would require giving exp a meaning on the imaginaries. Mike continued: << The paper "Objects of Categories as Complex Numbers" by Marcelo Fiore and Tom Leinster and says that under certain conditions, objects can behave as though they had complex cardinalities. It turns out that data types of typed programming languages satisfy the conditions; the data structure T = T^2 + T + 1 of trees with 2 (ordered), 1, or 0 children, behaves like "i". It is not the case that T^4 is isomorphic to the one-element set, but T^5 = T. Since it's a countably infinite set, this may seem obvious, but there's a way of constructing the isomorphism using only distributivity and the defining isomorphism above, so the cardinality is also preserved. . . . . . .
I don't really understand this, but it sounds fascinating. --Dan ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx
On Sat, Jun 12, 2010 at 1:04 PM, Dan Asimov <dasimov@earthlink.net> wrote:
I recently gave a talk to a bunch of smart youngsters about complex numbers, and was unable to find a truly graceful way to explain why
exp(ix) = cos(x) + i sin(x),
(without deriving their Taylor series).
So if anyone knows a way to see this, I'd love to know it.
But of course that would require giving exp a meaning on the imaginaries.
You need some physics, so that may lose some youngsters, but I find the first chapter of Tristan Needham's Visual Complex Analysis to give a very nice way of thinking about complex numbers addition, multiplication, and exponentiation, and it naturally yields this identity. --Joshua
On Jun 12, 2010, at 4:04 PM, Dan Asimov wrote:
I recently gave a talk to a bunch of smart youngsters about complex numbers, and was unable to find a truly graceful way to explain why
exp(ix) = cos(x) + i sin(x),
(without deriving their Taylor series).
So if anyone knows a way to see this, I'd love to know it.
I've sometimes explained e^(i x) by a (admittedly strange) economic metaphor for imaginary interest. Imagine a cooperative bank where you deposit or borrow money, with interest paid in gold = imaginary money. There is a property tax on gold at the same rate, e.g. $1 earn .01 gold pieces/year, and each gold piece is taxed at $.01 / year. You can take a short position in gold, with the same terms. This is just to dress up the differential equation f'(t) = i f(t) in a more concrete form. You can then plot what happens over time in the (money, gold) plane, and it's easy to see the solutions of the differential equation go around in a circle, hence sine and cosine. Alternatively: f'(t)=i f(t) => f''(t) = -f(t). This latter equation works independently on the real and imaginary parts, and is one of the reasonable ways to define the cosine and sine functions. Bill
What age is a "youngster", and how "smart"? Do they already know what sine and cosine are? When I was ten my older brother asked me: "If i is the square root of -1, what is the square root of i?". I knew about the complex plane, so just started plotting various values of a+bi and their squares. An hour later I told him the answer, as near as I could tell, was about 0.7 + 0.7*i. So I think this is easy if you break it up into two pieces: Start with showing where the powers of i are. Then let A=(i+1)/sqrt(2) and plot the locations of A, A^2, A^3, A^4. If necessary you could do the same thing with A=(sqrt(3)+i)/2. The pattern is easy to notice and generalize. That gets you halfway there (establishing the sine and cosine part). I understand the other half (generalizing the exponential function to imaginary, and therefore complex, arguments) through vector calculus and the geometric interpretation of complex derivative. So, I'm no help there (-: On Sat, Jun 12, 2010 at 16:04, Dan Asimov <dasimov@earthlink.net> wrote:
I'd love to know such a way to understand their Taylor series.
I recently gave a talk to a bunch of smart youngsters about complex numbers, and was unable to find a truly graceful way to explain why
exp(ix) = cos(x) + i sin(x),
(without deriving their Taylor series).
So if anyone knows a way to see this, I'd love to know it.
But of course that would require giving exp a meaning on the imaginaries.
-- Robert Munafo -- mrob.com
Quoting Dan Asimov <dasimov@earthlink.net>:
Mike wrote:
<< Is there a geometric way to understand the Taylor series for sin and cos? The closest I've been able to find is a combinatorical explanation (below), but it doesn't seem to help much.
I'd love to know such a way to understand their Taylor series.
I recently gave a talk to a bunch of smart youngsters about complex numbers, and was unable to find a truly graceful way to explain why
exp(ix) = cos(x) + i sin(x),
(without deriving their Taylor series).
So if anyone knows a way to see this, I'd love to know it.
But of course that would require giving exp a meaning on the imaginaries.
I don't know about complex categories (?) but finding a way to get Euler's formula has interested me for a long, long time. From calculus, the series for sine and cosine can be merged, but what if you don't know about calculus and infinite series yet. An approach I have seen is to define cis (cosine plus i times sine) and use trigonometric identities to manipulate it; it is an ersatz exponential and leaves any connection with an actual exponential unanswered. It meshes nicely with using polar coordinates in the complex plane, but doesn't have the radial factor. Anyway, I finally understood things to my satisfaction by using 2x2 real matrices to represent multiplication in the complex plane and using tghe definition of the exponential as the limit of the product (1 + (small angle) x matrix representing complex number)). The coefficients of the series come from the binomial formuls, taking into account that the matrix product must be ordered. Matching that with cis gives the desired result, with the taylor's series as a byproduct, not as an initial hypothesis. The series convergrs to a rotation by neatly fencing the rotated vector in by a sequence of contracting boxes. -hvm ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos
I think I can summarize Needham's argument without having to draw any pictures. But it's a lot better with the pictures. Imagine a particle beginning at (1,0) in the plane, a.k.a. at 1 in the complex plane. This particle moves such that its velocity is always perpendicular to the vector from the origin to its current position, and its speed is always equal to the magnitude of its distance from the origin. That is, with s(t) being its position, s'(t) = i * s(t). Thus s(t) = e^(it), up to some constants which turn out to be 1 because of the given initial conditions. But velocity perpendicular to displacement vector means circular motion: the velocity never has a radial component so the distance from the origin never changes. Furthermore, the acceleration is perpendicular to the velocity here, and thus the magnitude of the velocity never changes; it's always 1 as well. Thus the particle is in uniform circular motion around the origin, at speed 1 (distance unit per second, hence also radian per second) so s(t) = cos(t) + i sin(t). --Joshua
On Sat, Jun 12, 2010 at 1:04 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Mike wrote:
<< Is there a geometric way to understand the Taylor series for sin and cos? The closest I've been able to find is a combinatorical explanation (below), but it doesn't seem to help much.
I'd love to know such a way to understand their Taylor series.
I recently gave a talk to a bunch of smart youngsters about complex numbers, and was unable to find a truly graceful way to explain why
exp(ix) = cos(x) + i sin(x),
(without deriving their Taylor series).
So if anyone knows a way to see this, I'd love to know it.
But of course that would require giving exp a meaning on the imaginaries.
I think the Greeks would have had a hard time with Taylor series, because they insisted that all numbers have units. You can't have X cm + (X cm)^2. Can one do unitless calculations geometrically? -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
On Sun, Jun 13, 2010 at 8:56 AM, Mike Stay <metaweta@gmail.com> wrote:
I think the Greeks would have had a hard time with Taylor series, because they insisted that all numbers have units. You can't have X cm + (X cm)^2. Can one do unitless calculations geometrically?
X must refer to something such as a ratio of two lengths, not a number of centimeters. One can certainly view radians as unitless, being the ratio between an arc length and a radius. --Joshua
I've always liked the "compound interest" explanation for the Euler-DeMoivre formula: Imagine continuous compounding of $1 at 100X % interest for one year, giving $e^X. Visualize this as an expansion, beginning at the point (1,0), at the rate 100X% / year. Letting X = I, the expansion becomes a rotation. The number of years' rotation to go around the unit circle is just the length of the circle, which explains why e^2piI = 1. Rich ---- Quoting Dan Asimov <dasimov@earthlink.net>:
Mike wrote:
<< Is there a geometric way to understand the Taylor series for sin and cos? The closest I've been able to find is a combinatorical explanation (below), but it doesn't seem to help much.
I'd love to know such a way to understand their Taylor series.
I recently gave a talk to a bunch of smart youngsters about complex numbers, and was unable to find a truly graceful way to explain why
exp(ix) = cos(x) + i sin(x),
(without deriving their Taylor series).
So if anyone knows a way to see this, I'd love to know it.
But of course that would require giving exp a meaning on the imaginaries.
Mike continued:
<< The paper "Objects of Categories as Complex Numbers" by Marcelo Fiore and Tom Leinster and says that under certain conditions, objects can behave as though they had complex cardinalities. It turns out that data types of typed programming languages satisfy the conditions; the data structure T = T^2 + T + 1 of trees with 2 (ordered), 1, or 0 children, behaves like "i". It is not the case that T^4 is isomorphic to the one-element set, but T^5 = T. Since it's a countably infinite set, this may seem obvious, but there's a way of constructing the isomorphism using only distributivity and the defining isomorphism above, so the cardinality is also preserved. . . . . . .
I don't really understand this, but it sounds fascinating.
--Dan
________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
What's the justification for the constant speed of 1 (radian per year, or unit per year) in this approach? --Joshua On Mon, Jun 21, 2010 at 9:56 PM, <rcs@xmission.com> wrote:
I've always liked the "compound interest" explanation for the Euler-DeMoivre formula: Imagine continuous compounding of $1 at 100X % interest for one year, giving $e^X. Visualize this as an expansion, beginning at the point (1,0), at the rate 100X% / year. Letting X = I, the expansion becomes a rotation. The number of years' rotation to go around the unit circle is just the length of the circle, which explains why e^2piI = 1.
Rich
---- Quoting Dan Asimov <dasimov@earthlink.net>:
Mike wrote:
<< Is there a geometric way to understand the Taylor series for sin and cos? The closest I've been able to find is a combinatorical explanation (below), but it doesn't seem to help much.
I'd love to know such a way to understand their Taylor series.
I recently gave a talk to a bunch of smart youngsters about complex numbers, and was unable to find a truly graceful way to explain why
exp(ix) = cos(x) + i sin(x),
(without deriving their Taylor series).
So if anyone knows a way to see this, I'd love to know it.
But of course that would require giving exp a meaning on the imaginaries.
Mike continued:
<< The paper "Objects of Categories as Complex Numbers" by Marcelo Fiore and Tom Leinster and says that under certain conditions, objects can behave as though they had complex cardinalities. It turns out that data types of typed programming languages satisfy the conditions; the data structure T = T^2 + T + 1 of trees with 2 (ordered), 1, or 0 children, behaves like "i". It is not the case that T^4 is isomorphic to the one-element set, but T^5 = T. Since it's a countably infinite set, this may seem obvious, but there's a way of constructing the isomorphism using only distributivity and the defining isomorphism above, so the cardinality is also preserved. . . . . . .
I don't really understand this, but it sounds fascinating.
--Dan
________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I'm not sure what you're looking for. From a math viewpoint, you can adjust the numerical speed by adjusting the time unit, so you are free to choose 1. From a pedagogical viewpoint, a speed of 1 is simple: With no compounding, simple interest of 100% for one year, in the four directions +-1 & +-i, gives a diamond with corners at 0, 1+i, 2, 1-i. Compounding at six-month intervals gives a kite, with corners .25, .75+i, 2.25, .75-i. Decreasing the compounding interval ultimately leads to corners 1/e, cos 1 + i sin 1, e, cos 1 - i sin 1. Some amount of math detail is needed to fill in the rigor, but it's plausible that the i direction result is on the unit circle, along an arc of distance 1 from the starting point (1,0). The reason I like this is it explains why pi, defined as the length of a semicircle of radius 1, appears in the equation e^2piI = 1. I haven't seen a "natural" explanation of this. Rich --------------- Quoting Joshua Zucker <joshua.zucker@gmail.com>:
What's the justification for the constant speed of 1 (radian per year, or unit per year) in this approach?
--Joshua
On Mon, Jun 21, 2010 at 9:56 PM, <rcs@xmission.com> wrote:
I've always liked the "compound interest" explanation for the Euler-DeMoivre formula: Imagine continuous compounding of $1 at 100X % interest for one year, giving $e^X. Visualize this as an expansion, beginning at the point (1,0), at the rate 100X% / year. Letting X = I, the expansion becomes a rotation. The number of years' rotation to go around the unit circle is just the length of the circle, which explains why e^2piI = 1.
Rich
---- Quoting Dan Asimov <dasimov@earthlink.net>:
Mike wrote:
<< Is there a geometric way to understand the Taylor series for sin and cos? The closest I've been able to find is a combinatorical explanation (below), but it doesn't seem to help much.
I'd love to know such a way to understand their Taylor series.
I recently gave a talk to a bunch of smart youngsters about complex numbers, and was unable to find a truly graceful way to explain why
exp(ix) = cos(x) + i sin(x),
(without deriving their Taylor series).
So if anyone knows a way to see this, I'd love to know it.
But of course that would require giving exp a meaning on the imaginaries.
Mike continued:
<< The paper "Objects of Categories as Complex Numbers" by Marcelo Fiore and Tom Leinster and says that under certain conditions, objects can behave as though they had complex cardinalities. It turns out that data types of typed programming languages satisfy the conditions; the data structure T = T^2 + T + 1 of trees with 2 (ordered), 1, or 0 children, behaves like "i". It is not the case that T^4 is isomorphic to the one-element set, but T^5 = T. Since it's a countably infinite set, this may seem obvious, but there's a way of constructing the isomorphism using only distributivity and the defining isomorphism above, so the cardinality is also preserved. . . . . . .
I don't really understand this, but it sounds fascinating.
--Dan
________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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On Wed, Jun 23, 2010 at 5:54 PM, <rcs@xmission.com> wrote:
I'm not sure what you're looking for. From a math viewpoint, you can adjust the numerical speed by adjusting the time unit, so you are free to choose 1.
I disagree - you have already set the interest at 100% of i per time unit, compounded continuously. The question is why, after all that compounding, the speed is still 1.
From a pedagogical viewpoint, a speed of 1 is simple: With no compounding, simple interest of 100% for one year, in the four directions +-1 & +-i, gives a diamond with corners at 0, 1+i, 2, 1-i. Compounding at six-month intervals gives a kite, with corners .25, .75+i, 2.25, .75-i. Decreasing the compounding interval ultimately leads to corners 1/e, cos 1 + i sin 1, e, cos 1 - i sin 1. Some amount of math detail is needed to fill in the rigor, but it's plausible that the i direction result is on the unit circle, along an arc of distance 1 from the starting point (1,0).
I think you need to do something equivalent to a second derivative somewhere in there to show that the magnitude of the velocity vector is always 1. If there's no need for calculus, I'd like to see some of the math detail you skipped there so I can understand how we know the limit point of this sequence will be 1 unit of arc length away from the starting point.
The reason I like this is it explains why pi, defined as the length of a semicircle of radius 1, appears in the equation e^2piI = 1. I haven't seen a "natural" explanation of this.
Definitely: unit speed travel around the unit circle is what's going on here. The question I have is why 100i% interest compounded continuously results in a constant unit speed (since certainly 100% interest compounded continuously only has unit speed when you start!) Thanks, --Joshua
In the 100I% case, with an uncompounded time interval of T, the multiplier is 1 + IT. The complex norm of the multiplier is 1 + T^2. If we divide this into N intervals for compounding, the norm of one interval of T/N is 1 + T^2 / N^2. For N intervals (total time T), the norm is the Nth power of the single-interval value, ( 1 + T^2 / N^2 ) ^ N. For large N, this is roughly 1 + T^2 / N + some O(1/N^2) stuff. The effect of large N, getting close to continuous compounding, is to make the norm, for the full time interval T, approach 1. Some fine print is required to make sure that the O(1/N^2) stuff doesn't mess up the limit being 1. [1] This tends to obscure the main point, that subdividing the time interval T makes the norm closer to 1, by a factor 1/N^2, overpowering the *N effect of having N intervals. This establishes that the norm is 1 for the whole duration. A similar argument establishes that the polygonal path of the compounded values (1 + IT/N)^K, K in [0...N], is roughly the unit circle, and the limiting result is an arc of length T along the circle. My personal opinion is that the rigor doesn't help understanding, but does need to be checked. Once you accept the compound interest viewpoint, the power series for sine & cosine come from imaginary & real parts of using the binomial theorem for (1 + I T / N) ^ N, as N->infinity. I've always been puzzled that the Greeks didn't come up with logs & exp, since they knew about log spirals. Rich [1] For suitable k & x, 1 + k x < (1+x)^k < 1 / (1 - kx). ------- Quoting Joshua Zucker <joshua.zucker@gmail.com>:
On Wed, Jun 23, 2010 at 5:54 PM, <rcs@xmission.com> wrote:
I'm not sure what you're looking for. From a math viewpoint, you can adjust the numerical speed by adjusting the time unit, so you are free to choose 1.
I disagree - you have already set the interest at 100% of i per time unit, compounded continuously. The question is why, after all that compounding, the speed is still 1.
From a pedagogical viewpoint, a speed of 1 is simple: With no compounding, simple interest of 100% for one year, in the four directions +-1 & +-i, gives a diamond with corners at 0, 1+i, 2, 1-i. Compounding at six-month intervals gives a kite, with corners .25, .75+i, 2.25, .75-i. Decreasing the compounding interval ultimately leads to corners 1/e, cos 1 + i sin 1, e, cos 1 - i sin 1. Some amount of math detail is needed to fill in the rigor, but it's plausible that the i direction result is on the unit circle, along an arc of distance 1 from the starting point (1,0).
I think you need to do something equivalent to a second derivative somewhere in there to show that the magnitude of the velocity vector is always 1. If there's no need for calculus, I'd like to see some of the math detail you skipped there so I can understand how we know the limit point of this sequence will be 1 unit of arc length away from the starting point.
The reason I like this is it explains why pi, defined as the length of a semicircle of radius 1, appears in the equation e^2piI = 1. I haven't seen a "natural" explanation of this.
Definitely: unit speed travel around the unit circle is what's going on here. The question I have is why 100i% interest compounded continuously results in a constant unit speed (since certainly 100% interest compounded continuously only has unit speed when you start!)
Thanks, --Joshua
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As an aside to "compound interest", there has been a vigorous debate recently among economists about _negative_ interest rates, which could potentially be used to combat deflation & unemployment. The Fed's current target interest rate is _minus 5%_ (= 2.1% + 1.3%*inflationrate - 2.0%*unemploymentgap) [Fed Reserve Bank of San Francisco Econ. Letter 6/14/2010]. One proposal from Harvard economist Greg Mankiw: "I can now state the proposed solution: Reduce the return to holding money below zero. Imagine that the Fed were to announce that it would pick a digit from 0 to 9 out of a hat. All currency with a serial number ending in that digit would no longer be legal tender. Suddenly, the expected return to holding currency would become negative 10 percent." http://seekingalpha.com/article/211590-what-s-ben-gonna-do-now Other proposals for "hot potato(e) money": http://www.clevelandfed.org/Research/commentary/2008/0408.cfm Keynes semi-seriously proposed that the govt fill up bottles with currency & bury them in mine shafts to be dug up by the public (by analogy with digging/panning for gold). -- I haven't yet heard of any economists proposing _imaginary_ or _complex_ interest rates, but the night is still young. At 09:56 PM 6/21/2010, rcs@xmission.com wrote:
I've always liked the "compound interest" explanation for the Euler-DeMoivre formula: Imagine continuous compounding of $1 at 100X % interest for one year, giving $e^X. Visualize this as an expansion, beginning at the point (1,0), at the rate 100X% / year. Letting X = I, the expansion becomes a rotation. The number of years' rotation to go around the unit circle is just the length of the circle, which explains why e^2piI = 1.
Rich
---- Quoting Dan Asimov <dasimov@earthlink.net>:
Mike wrote:
<< Is there a geometric way to understand the Taylor series for sin and cos? The closest I've been able to find is a combinatorical explanation (below), but it doesn't seem to help much.
I'd love to know such a way to understand their Taylor series.
I recently gave a talk to a bunch of smart youngsters about complex numbers, and was unable to find a truly graceful way to explain why
exp(ix) = cos(x) + i sin(x),
(without deriving their Taylor series).
So if anyone knows a way to see this, I'd love to know it.
But of course that would require giving exp a meaning on the imaginaries.
Mike continued:
<< The paper "Objects of Categories as Complex Numbers" by Marcelo Fiore and Tom Leinster and says that under certain conditions, objects can behave as though they had complex cardinalities. It turns out that data types of typed programming languages satisfy the conditions; the data structure T = T^2 + T + 1 of trees with 2 (ordered), 1, or 0 children, behaves like "i". It is not the case that T^4 is isomorphic to the one-element set, but T^5 = T. Since it's a countably infinite set, this may seem obvious, but there's a way of constructing the isomorphism using only distributivity and the defining isomorphism above, so the cardinality is also preserved. . . . . . .
I don't really understand this, but it sounds fascinating.
--Dan
participants (8)
-
Bill Thurston -
Dan Asimov -
Henry Baker -
Joshua Zucker -
mcintosh＠servidor.unam.mx -
Mike Stay -
rcs＠xmission.com -
Robert Munafo