Why is e^(pi i) = -1?(math.toronto.edu)
math.toronto.edu
Why is e^(pi i) = -1?
http://www.math.toronto.edu/mathnet/questionCorner/epii.html
16 comments
This is a good explanation of why the result is negative and why it has zero imaginary component - ie why it is pointing the opposite way.. But it doesn't explain why the result is unit one in magnitude / length. Can you extend it do that that?
Thanks! Yes, when you grow in a direction at right angles to yourself, you don't increase in magnitude. So the turning arrow is the same length at the end as it was at the beginning.
This is actually a slightly tricky point, because it requires the direction of growth to change as you grow. If you grow by a finite amount all at once, then you do get slightly longer. So you might expect an outward spiral.
But a point moving round in a circle is always moving at right angles to the radius connecting it to the centre.
And in the same way, an arrow whose tip is moving at right angles to its shaft isn't extending. If the length is changing, then the direction isn't 90 degrees.
To reason about this properly I think you need some sort of theory of infinitesimals and continuous motion.
This is actually a slightly tricky point, because it requires the direction of growth to change as you grow. If you grow by a finite amount all at once, then you do get slightly longer. So you might expect an outward spiral.
But a point moving round in a circle is always moving at right angles to the radius connecting it to the centre.
And in the same way, an arrow whose tip is moving at right angles to its shaft isn't extending. If the length is changing, then the direction isn't 90 degrees.
To reason about this properly I think you need some sort of theory of infinitesimals and continuous motion.
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Feynman gives the same explanation in his awesome Lectures On Physics, volume 1, the chapter titled "Algebra". One of the great works of 20th-century literature.
[deleted]
That's a great explanation. Is this how they explain it in math classes? Because if not, then it should be. At least at a superficial level.
Thanks! No, I worked it out for myself while I was thinking about what the exponential of a linear operator is. The explanation at my school was the one with Taylor Series.
By the time I got to university it was just taken as a given, so nobody explained it at all.
My explanation seems to have attracted a lot of positive votes in a few minutes. I've written a more detailed version here if anyone doesn't get it:
http://johnlawrenceaspden.blogspot.com/2009/10/ei-pi-1.html
If it gets much attention I might draw the diagrams that go with it.
I'm always bewildered how many scientific types (and even some mathematicians) find complex numbers mystical. They were originally discovered by mystic methods, but Argand showed us what was really going on, and they're no more weird than 2-dimensional vectors.
Actually a lot less weird than the Reals, which really are magical and mysterious, but which everyone seems to be quite happy with!
By the time I got to university it was just taken as a given, so nobody explained it at all.
My explanation seems to have attracted a lot of positive votes in a few minutes. I've written a more detailed version here if anyone doesn't get it:
http://johnlawrenceaspden.blogspot.com/2009/10/ei-pi-1.html
If it gets much attention I might draw the diagrams that go with it.
I'm always bewildered how many scientific types (and even some mathematicians) find complex numbers mystical. They were originally discovered by mystic methods, but Argand showed us what was really going on, and they're no more weird than 2-dimensional vectors.
Actually a lot less weird than the Reals, which really are magical and mysterious, but which everyone seems to be quite happy with!
Thats neat. Only thing is that (to me anyway) this takes the idea of the complex plane as being very fundamental as opposed to just something convenient. I'm not sure how to convince someone that 1 + i is the same as the coordinate (1, 1) without saying "thats just how we define it because things work out."
The complex plane is the fundamental thing.
It's the only way to get to the complexes without doing anything spooky! Defining i to be "the square root of minus one" is about as sane as defining it to be the square root of the colour blue.
The first people who thought about it followed that approach, and were rightly scared stiff and confused by it. Even Euler made trivial mistakes.
Argand came up with the right way of thinking about it:
Take all the tuples (x,y). (Where x and y are just integers). Define on them addition and multiplication rules.
Oh look! There's a big system of these things, and embedded within it is a sub-system which works exactly the same as the integers and their rules.
Since they're exactly the same for all practical purposes, we may as well forget about the difference and say that we'll write (a,0) as a and (0,b) as ib, and (a,b) as a+ib
And although non of the pairs in that subsystem square to be (-1,0), also known as -1, there are two things that do! So now we know that both (0,1) and (0, -1), otherwise known as 1i and -1i, or i and -i, are square roots of -1.
No slight of hand or magical thinking necessary.
As it happens, we can do the same thing with the reals, to form something that embeds the reals. But the reals really are dark and mysterious and need to be brought about by a kind of magic.
It's the only way to get to the complexes without doing anything spooky! Defining i to be "the square root of minus one" is about as sane as defining it to be the square root of the colour blue.
The first people who thought about it followed that approach, and were rightly scared stiff and confused by it. Even Euler made trivial mistakes.
Argand came up with the right way of thinking about it:
Take all the tuples (x,y). (Where x and y are just integers). Define on them addition and multiplication rules.
Oh look! There's a big system of these things, and embedded within it is a sub-system which works exactly the same as the integers and their rules.
Since they're exactly the same for all practical purposes, we may as well forget about the difference and say that we'll write (a,0) as a and (0,b) as ib, and (a,b) as a+ib
And although non of the pairs in that subsystem square to be (-1,0), also known as -1, there are two things that do! So now we know that both (0,1) and (0, -1), otherwise known as 1i and -1i, or i and -i, are square roots of -1.
No slight of hand or magical thinking necessary.
As it happens, we can do the same thing with the reals, to form something that embeds the reals. But the reals really are dark and mysterious and need to be brought about by a kind of magic.
Bravo. You have a gift for explanation. This is exactly the sort of explanation helpful for someone (like myself) who has been through all the foundations at one point or another, but sometimes forgets the big picture.
For anyone that was momentarily confused by johnaspden's explanation of how (0,1) squared became (-1,0), there is excellent explanation of it here:
http://www.math.toronto.edu/mathnet/answers/imagexist.html
I'll gratuitously paste the relevant part here, but the summary is that to multiply two tuples, you simply take the cross product, as you would a binomial.
---
Remember that any collection of objects for which there is a definition of what the objects are and when two objects are equal, there is a rule for how to add two objects, there is a rule for how to multiply two objects, and these rules obey familiar arithmetic laws like commutativity, associativity, and distributivity, is, by definition, a number system.
These properties are all satisfied by complex numbers.
We have a definition of when two complex numbers are to be considered equal: they are equal if and only if they are the same pair of real numbers.
We have a rule for adding two complex numbers (which, remember, are nothing more than pairs of real numbers):
Therefore, complex numbers form a number system.
Within this number system, is there an object which, when squared, gives -1? Yes. It is the pair (0,1). When you square it using the above rule of multiplication, you get
However, complex numbers of the form (a,0) behave identically to the way ordinary real numbers a behave. They add and multiply in exactly the same way that ordinary real numbers do:
http://www.math.toronto.edu/mathnet/answers/imagexist.html
I'll gratuitously paste the relevant part here, but the summary is that to multiply two tuples, you simply take the cross product, as you would a binomial.
---
Remember that any collection of objects for which there is a definition of what the objects are and when two objects are equal, there is a rule for how to add two objects, there is a rule for how to multiply two objects, and these rules obey familiar arithmetic laws like commutativity, associativity, and distributivity, is, by definition, a number system.
These properties are all satisfied by complex numbers.
We have a definition of when two complex numbers are to be considered equal: they are equal if and only if they are the same pair of real numbers.
We have a rule for adding two complex numbers (which, remember, are nothing more than pairs of real numbers):
(a,b) + (c,d) = (a+c, b+d)
(a,b)(c,d) = (ac-bd, ad+bc)
Therefore, complex numbers form a number system.
Within this number system, is there an object which, when squared, gives -1? Yes. It is the pair (0,1). When you square it using the above rule of multiplication, you get
(0,1)(0,1) = ( (0)(0) - (1)(1), (0)(1)+(1)(0) ) = (-1,0).
However, complex numbers of the form (a,0) behave identically to the way ordinary real numbers a behave. They add and multiply in exactly the same way that ordinary real numbers do:
(a,0) + (b,0) = (a+b,0)
(a,0)(b,0) = (ab,0)
http://betterexplained.com/articles/a-visual-intuitive-guide...
http://betterexplained.com/articles/intuitive-arithmetic-wit...
The first link was posted here a while ago, I think.
http://betterexplained.com/articles/intuitive-arithmetic-wit...
The first link was posted here a while ago, I think.
Glad the articles were useful for you :).
I really like Euler's Formula, but feel many explanations focus on the raw bits (showing the series are equal). The physics explanation "90-degree rotation" is a good one. Here's how I explained it to my mom:
"The formula is saying there's two ways to get to a place on a circle. One is to go across (cos) and up (sin), and the other is to go out (1.0) and rotate (x radians)."
This is a beautiful result which ties together radians, exponential growth, and complex numbers -- I have a post drafted that I want to share on this :).
I really like Euler's Formula, but feel many explanations focus on the raw bits (showing the series are equal). The physics explanation "90-degree rotation" is a good one. Here's how I explained it to my mom:
"The formula is saying there's two ways to get to a place on a circle. One is to go across (cos) and up (sin), and the other is to go out (1.0) and rotate (x radians)."
This is a beautiful result which ties together radians, exponential growth, and complex numbers -- I have a post drafted that I want to share on this :).
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Where does e come into this specifically? That is, why does 5^(i*pi) have an imaginary part?
I'll give it a shot. For me, I think about e in terms of compound interest.
I.e. you have a bank account and put in a dollar. Its a great bank, they give you 100% interest. So, after a year, you have (1 + 1) = 2 dollars.
But wait, instead, they compound twice. That is, every 6 months, you get 50% interest. So now you have (1 + 1/2)^2 = 2.25 dollars.
Instead of compounding twice, they compound 4 times. So now you have (1 + 1/4)^4 = 2.44 dollars. Each time you get interest, its given in proportion to what you have in the account at the time. So, if they are compounding 4 times:
1 --> (1.25 * 1) = 1.25 --> (1.25 * 1.25) = 1.56 --> (1.25 * 1.56) = 1.95 --> (1.25 * 1.95) = 2.44.
(Please note: Rounding error. I just typed in to two places after the decimal point).
Why 2 or 4? Lets just call it compounding n times. As n gets larger, you get more money, but it doesn't go on forever...
For n times, the formula for how much money you have is (1 + 1/n)^n. What if you want to continuously compound? I.e. at every instant, growth is proportional to the current value? Well, then you would want something like (1 + 1/infinity)^infinity. But, we can't really do that, so instead we say:
lim (n --> inf) of (1 + 1/n)^n
And thats what we call e. Start putting in larger and larger values of n and you'll see that it converges to 2.718... The point is that e is pretty much defined as the number you get when you take a number starting at 1 and have it grow in proportion to itself continuously. Instead, if you use 5 instead of e, your growing too fast. In terms of johnaspden's explanation, you'd move too far around the circle, past (-1, 0).
Also, johnaspden, your explanations are truly great, really helped me understand Euler's formula.
Edit: Small change.
I.e. you have a bank account and put in a dollar. Its a great bank, they give you 100% interest. So, after a year, you have (1 + 1) = 2 dollars.
But wait, instead, they compound twice. That is, every 6 months, you get 50% interest. So now you have (1 + 1/2)^2 = 2.25 dollars.
Instead of compounding twice, they compound 4 times. So now you have (1 + 1/4)^4 = 2.44 dollars. Each time you get interest, its given in proportion to what you have in the account at the time. So, if they are compounding 4 times:
1 --> (1.25 * 1) = 1.25 --> (1.25 * 1.25) = 1.56 --> (1.25 * 1.56) = 1.95 --> (1.25 * 1.95) = 2.44.
(Please note: Rounding error. I just typed in to two places after the decimal point).
Why 2 or 4? Lets just call it compounding n times. As n gets larger, you get more money, but it doesn't go on forever...
For n times, the formula for how much money you have is (1 + 1/n)^n. What if you want to continuously compound? I.e. at every instant, growth is proportional to the current value? Well, then you would want something like (1 + 1/infinity)^infinity. But, we can't really do that, so instead we say:
lim (n --> inf) of (1 + 1/n)^n
And thats what we call e. Start putting in larger and larger values of n and you'll see that it converges to 2.718... The point is that e is pretty much defined as the number you get when you take a number starting at 1 and have it grow in proportion to itself continuously. Instead, if you use 5 instead of e, your growing too fast. In terms of johnaspden's explanation, you'd move too far around the circle, past (-1, 0).
Also, johnaspden, your explanations are truly great, really helped me understand Euler's formula.
Edit: Small change.
Pi is Wrong: http://www.math.utah.edu/~palais/pi.pdf
It should be e^(pi i) = 1, but pi was unfortunately defined at half the appropriate value in the 17th century.
It should be e^(pi i) = 1, but pi was unfortunately defined at half the appropriate value in the 17th century.
Pi is defined as the circumference of a circle of unit diameter. This isn't somehow less correct than the alternative of making that circle one of unit radius. It causes some annoying extra factors (as the PDF points out), but the alternative would cause different complications. The area of a circle, for example, would be r^2*pi/2.
The area of a circle, for example, would be r^2 pi/2.
This is actually my favorite example of why pi is wrong—it's the "exception" that proves the rule. To see why, set τ = C/r = 2 pi, and then consider the following chart of common quadratic forms:
N.B. I made a previous comment along these lines, but put it in the wrong place. If you're still a "pi is wrong" skeptic, considering the improvement in radian angle measure may yet convince you:
The explanations mapping complex exponentiation to rotations are basically right. In this context, it's worth noting that the conventional choice of circle constant is off by two. We should be using tau = τ = C/r as the circle constant, rather than pi = π = C/D. Read τ as "turn" and all those radian angle measures suddenly make sense. Ninety degrees? Instead of the confusing π/2 we have 90° = τ/4 = one quarter turn. And so on: 60° = τ/6 = one sixth of a turn, 180° = τ/2 = one half turn, etc.
In these terms, Euler's formula would be recast as
e^(i τ) = 1
That is, the exponential of the imaginary unit i times the circle constant τ is unity: one full rotation.
This is actually my favorite example of why pi is wrong—it's the "exception" that proves the rule. To see why, set τ = C/r = 2 pi, and then consider the following chart of common quadratic forms:
integral of u 1/2 u^2
kinetic energy 1/2 m v^2
distance fallen 1/2 g t^2
spring energy 1/2 k x^2
triangular area 1/2 b h
circular area 1/2 τ r^2
N.B. I made a previous comment along these lines, but put it in the wrong place. If you're still a "pi is wrong" skeptic, considering the improvement in radian angle measure may yet convince you:
The explanations mapping complex exponentiation to rotations are basically right. In this context, it's worth noting that the conventional choice of circle constant is off by two. We should be using tau = τ = C/r as the circle constant, rather than pi = π = C/D. Read τ as "turn" and all those radian angle measures suddenly make sense. Ninety degrees? Instead of the confusing π/2 we have 90° = τ/4 = one quarter turn. And so on: 60° = τ/6 = one sixth of a turn, 180° = τ/2 = one half turn, etc.
In these terms, Euler's formula would be recast as
e^(i τ) = 1
That is, the exponential of the imaginary unit i times the circle constant τ is unity: one full rotation.
Hm, is using tau a standard convention?
I've thought pi is horrible... and adopted using a loopy pi as 2pi. I loop the first vertical line over the horizontal, right, and then make it the horizontal, and so on. If that makes any sense.
I'm thinking of making a LaTeX package containing it... (and also some other personal conventions)
Also, to strengthen your point, 2pi makes the radian system a lot more natural.
I've thought pi is horrible... and adopted using a loopy pi as 2pi. I loop the first vertical line over the horizontal, right, and then make it the horizontal, and so on. If that makes any sense.
I'm thinking of making a LaTeX package containing it... (and also some other personal conventions)
Also, to strengthen your point, 2pi makes the radian system a lot more natural.
I made up the usage of τ myself, partially because of its typographic similarity to π, partially because it leads naturally to the usage "τ = turn". It's too bad that π has two legs while τ has only one; it would be poetic if π were 2τ, but it wasn't to be.
N.B. I have a secret master plan to spread the use of τ, but this comment is too small to contain it. ;-)
N.B. I have a secret master plan to spread the use of τ, but this comment is too small to contain it. ;-)
I'd love to hear it. christopherolah.coATgmail.com
All of those (except the circle) have the 1/2 because they're integrals of something linear. While it's true that area and integral are closely related (the latter being a special case of the former), a circle is clearly not linear.
As a circle expands, its area grows proportional to its circumference. The circumference is proportional to the radius.
So you're getting the area by integrating a linear thing.
So you're getting the area by integrating a linear thing.
As a circle expands, its area grows proportional to its circumference. The circumference is proportional to the radius.
Exactly. In symbols, this reads dA/dr = C ∝ r. Setting τ = C/r = 6.2831853…, we have dA = C dr = τ r dr => A = ½ τ r². Q.E.D.
Exactly. In symbols, this reads dA/dr = C ∝ r. Setting τ = C/r = 6.2831853…, we have dA = C dr = τ r dr => A = ½ τ r². Q.E.D.
[deleted]
johnaspden has it right. To put it more explicitly, we can calculate the area of a circle by integrating the differential element of area dA for an infinitesimal annulus from 0 to r. Now, dA is simply the arclength (circumference C) times the thickness dr; since the circumference scales linearly with radius, this leads to the integral of a linear function as follows:
dA = C dr = τ r dr => A = ½ τ r².
dA = C dr = τ r dr => A = ½ τ r².
Yes, the alternative would cause different complications, but fewer of them. The pi-is-wrong pdf goes into it.
It was programming a lot that brought me to realize this. I got in the habit of always questioning the primitives I'm given, in the hope of simplifying the definitions built on them. It was a little boggling to notice that pi, something I'd grown up with, deserved the same kind of questioning.
It was programming a lot that brought me to realize this. I got in the habit of always questioning the primitives I'm given, in the hope of simplifying the definitions built on them. It was a little boggling to notice that pi, something I'd grown up with, deserved the same kind of questioning.
I'm not convinced that redefining Pi as 2*Pi would cause an equal amount of complications (besides confusing everyone with the change). The radius is more fundamental to a circle. The diameter is just derived from the radius. The idea of a circle with a unit radius is so widely used that it is simply called a "unit circle" with the obvious implication that the radius is what the "unit" refers to. Every time I've done math with radians, the fact that Pi is only half way around has added a subtle but noticeable disconnect. I have to force my intuition to fit the definition.
The radius is only more fundamental to the notion of a circle because of the modern definition of a circle (set of all points exactly r away from the given point). One could also use a definition like:
"Continuous set of points such that each is distance d from exactly one other, and further from none." The modern definition is convenient because it's a special case of a convenient definition of ellipse, but there's nothing terribly fundamental about it.
And yes, every time you've done math with radians, that problem has haunted you. That's because we decided to use radians instead of "diameter-ans", to keep with the above definition.
And yes, every time you've done math with radians, that problem has haunted you. That's because we decided to use radians instead of "diameter-ans", to keep with the above definition.
The article didn't have any mention of Euler, but this is typically called Euler's Formula and is considered one of the most beautiful formulas to Mathematicians.
Here's some more info if you want to find out some info, applications and such: http://en.wikipedia.org/wiki/Eulers_formula
Here's some more info if you want to find out some info, applications and such: http://en.wikipedia.org/wiki/Eulers_formula
Write this as e^(iπ) + 1 = 0 and you have my favourite equation.
It relates all the fundamental mathematical numbers: e, i, π, 1 and 0.
It relates all the fundamental mathematical numbers: e, i, π, 1 and 0.
... using the fundamental equivalence relation =, and the operations +, * & ^
You remember those half-angle formulae
sin 2A = 2 sin A cos A
cos 2A = (cos A)^2 - (sin A)^2
If you are happy using sine of small x is x and cosine of small x is 1 as your base cases, you can write sine and cosine as mutually recursive functions:
It had never occurred to me to do exponential the same way, as
(exp (complex 0 pi)) => #C(-1.0012055113842453d0 6.167817939221069d-7)
sin 2A = 2 sin A cos A
cos 2A = (cos A)^2 - (sin A)^2
If you are happy using sine of small x is x and cosine of small x is 1 as your base cases, you can write sine and cosine as mutually recursive functions:
(defun sin (z)
(if (small z)
z
(* 2
(sin (half z))
(cos (half z)))))
(defun cos (z)
(if (small z)
1
(- (square (cos (half z)))
(square (sin (half z))))))
It had never occurred to me to do exponential the same way, as
(defun exp (z)
(if (small z)
(+ 1 z)
(square (exp (half z)))))
(exp (complex 0 pi)) => #C(-1.0012055113842453d0 6.167817939221069d-7)
The explanations mapping complex exponentiation to rotations are basically right. In this context, it's worth noting that the conventional choice of circle constant is off by two. We should be using tau = τ = C/r as the circle constant, rather than pi = π = C/D. Read τ as "turn" and all those radian angle measures suddenly make sense. Ninety degrees? Instead of the confusing π/2 we have 90° = τ/4 = one quarter turn. And so on: 60° = τ/6 = one sixth of a turn, 180° = τ/2 = one half turn, etc.
In these terms, Euler's formula would be recast as
e^(i τ) = 1
That is, the exponential of the imaginary unit i times the circle constant τ is unity: one full rotation.
In these terms, Euler's formula would be recast as
e^(i τ) = 1
That is, the exponential of the imaginary unit i times the circle constant τ is unity: one full rotation.
I enjoyed Lakoff and Núñez's section on this in their book Where Mathematics Comes From[1]. The relevent pages are available at Google Books[2].
1- http://en.wikipedia.org/wiki/Where_Mathematics_Comes_From
2- http://books.google.com/books?id=YXv6SEjTNKsC&pg=PA432&#...
1- http://en.wikipedia.org/wiki/Where_Mathematics_Comes_From
2- http://books.google.com/books?id=YXv6SEjTNKsC&pg=PA432&#...
For those of you interested in a longer look at this, check out "E, the Story of a Number" by Eli Maor.
Yes, I read when I should probably be hacking.
Yes, I read when I should probably be hacking.
I like Feynman's description, where he actually used 10^x first, just noting that 10^x for small real x was 1 + ln(10)*x (approximating from the derivative), assuming that this worked for small complex values too, and then extended to larger values by squaring.
So to broadcast my math ignorance to all:
Why does cos(pi) + i sin(pi) == -1 ??
Why does i disappear in this?
Edit: Thanks to the replies below.
Why does cos(pi) + i sin(pi) == -1 ??
Why does i disappear in this?
Edit: Thanks to the replies below.
We verify in a quick Google search:
http://www.google.com/search?hl=en&safe=off&q=sin(pi...
that sin(pi) = 0.
that sin(pi) = 0.
The formula is:
e^(ix) = cos(x) - i sin(x)
Since sin(π) == 0, i * 0 == 0 so you have cos(π) left which is -1.
So in sum:
e^(i π) = cos(π) - i sin(π) == -1 - 0 == -1
e^(ix) = cos(x) - i sin(x)
Since sin(π) == 0, i * 0 == 0 so you have cos(π) left which is -1.
So in sum:
e^(i π) = cos(π) - i sin(π) == -1 - 0 == -1
If you draw a Sine graph, you will see that sin(x) is zero at 0, pi, 2*pi, ...
If you draw a cosine graph you will see that it is -1 at pi. Therefore in the above sum the sine part just falls away.
It is easy to see that the above is true when you recall that cosine(theta) = adjacent side/(hypotenuse). Since pi=180 degrees, adjacent will be -1 hypotenuse 1 and opposite side 0.
If you draw a cosine graph you will see that it is -1 at pi. Therefore in the above sum the sine part just falls away.
It is easy to see that the above is true when you recall that cosine(theta) = adjacent side/(hypotenuse). Since pi=180 degrees, adjacent will be -1 hypotenuse 1 and opposite side 0.
sin(pi)=0
I wouldn't feel ignorant or anything. The article left out that information seemingly on purpose, presumably so that people would try to figure out where the i went.
I wouldn't feel ignorant or anything. The article left out that information seemingly on purpose, presumably so that people would try to figure out where the i went.
[deleted]
http://xkcd.com/179/
I'm sure almost everyone has seen this anyway, but why not post.
I'm sure almost everyone has seen this anyway, but why not post.
So many texts on complex analysis simply define e^{i \theta} = \cos \theta + i \sin \theta without ever explaining how. This provides a good introduction to the reason behind with only minimal recourse to Calculus.
Another good description is discussed in "Visual complex Analysis" by Needham.
Another good description is discussed in "Visual complex Analysis" by Needham.
I have known it as the celebrities formula, because if you write it this way:
e^(pi i) + 1 = 0
It contains all the "celebrities" of the math world.
It's quite cool actually.
It contains all the "celebrities" of the math world.
It's quite cool actually.
Pages like this will be so much better when we have MathML in the browser and can stop using fuzzy images for mathematical formulas.
If you control the website just use http://www.math.union.edu/~dpvc/jsMath/ and you're fine. Any users that want can install the fonts and will get a better experience.
Given how much harder MathML is to author, the jsMath solution is always going to be more popular among random math folks. Particularly since they're already used to writing TeX.
Given how much harder MathML is to author, the jsMath solution is always going to be more popular among random math folks. Particularly since they're already used to writing TeX.
A better question is how can you:
(pi/2)^2e
(pi/2)^2e
umm i m no smart ass.. but i think e^(pi i) = -1 cuz LHS = cos PI + isin PI = -1 .... :O
Exponentiation is to do with growth at a speed which is a multiple of how big you are already.
i is the multiplication which turns you through ninety degrees.
If you grow in a direction which is at right angles to yourself, you turn rather than increasing in magnitude.
Pi is how long it takes you to turn through a half circle.
So if you grow at right angles to yourself for time pi, you are pointing the opposite way.