## 0179: "E to the Pi times I"

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telcontar42
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### Re: e to pi times i

Well, I wouldn't be so quick to call imaginary numbers fake number. They don't have the same physical representation as real numbers in that you can have 2 apples but not 2i apples. Still, a number like 2 and a number like i are both just mathematical constructs that we have defined. i is an important mathematical unit that is necessary in the representation of many real world systems. For example, in quantum mechanics, the momentum operator is i(h-bar)(d/dx). Momentum is certainly a real quantity and it involves i.

Also, you can get real numbers from just imaginary numbers. i*i=-1

rho421
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### Re: e to pi times i

Galle wrote:Do the Taylor expansion for e^(x*i). You will find that terms that contain an i form the Taylor expansion for sin, and the terms without an i form the expansion por cos. So, you are left with e^(x*i)=cos(x)+i*sin(x). Replace x with Pi then sin becomes 0, so you are left with a real number.

Oh, that makes so much sense now. (Calculus FTW) I always just kind of took it as fact; never stopped to wonder why.

Matterwave1
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### Re: e to pi times i

The Taylor series expansion method doesn't satisfy me too well...you've expanded the functions around 0, and so the expansion should only be valid for points close to 0 shouldn't it? pi is not close to zero...

Archena
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### Re: e to pi times i

Tronald wrote:i Dont think imaginary numbers Can = Real ones http://xkcd.com/179/

But to be pedantic, nobody is saying that imaginary numbers can equal real ones, and obviously they cannot. e^ipi is an algebraic expression whose value is a real number. So the equation just implies that real numbers can equal real numbers.

oh well that makes sense. so a fake number with the help of real numbers can = a real number but fake numbers them selfless can not = real number, Right?
.

In exactly the same way as two integers can create an real, but non-integer, number - e.g. 3/4.
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scarletmanuka
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### Re: e to pi times i

Matterwave1 wrote:The Taylor series expansion method doesn't satisfy me too well...you've expanded the functions around 0, and so the expansion should only be valid for points close to 0 shouldn't it? pi is not close to zero...

In general, yes, a Taylor series expansion will only be valid when you are close enough to the point about which you're expanding. (Whether [imath]\pi[/imath] would be close enough depends on the individual expansion.) In this case, however, "close enough to" is equivalent to "any finite distance from". In other words, this expansion is valid for any value. That factorial term on the denominator is handy for that.

qinwamascot
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### Re: e to pi times i

Matterwave1 wrote:The Taylor series expansion method doesn't satisfy me too well...you've expanded the functions around 0, and so the expansion should only be valid for points close to 0 shouldn't it? pi is not close to zero...

The radius of convergence for e^x is infinite, so the Taylor series converges exactly to the function value at any finite input. A Taylor series is only an approximation if finitely many terms are used, otherwise it's an expansion that is exactly equal to the function value for all points within the radius of curvature, which as I previously stated is infinite in this case.

also, I'd argue that pi is a lot closer to 0 than a googol, or graham's number, is. It's all relative. It's of course also a lot farther away from 0 than .000001, for example.
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enginerd22
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### Re: e to pi times i

I agree, I've always found that Taylor series proof unsatisfying, even though it is mathematically rigorous (because a function is equal to it's Taylor series if the number of terms is infinite and there aren't any discontinuities).

I've thought of one more. The solution to differential equations is unique, and one can show that linear combinations of
{e^ix, e^-ix} and {sin(x),cos(x)} are each solutions to d^2 f(x) / dx^2 + f(x) = 0. Pick f(0)= 0, f ' (0)= 1, and you can derive the formulas for sin, f (0)= 1, f '(0)= 0 gives you cosine.

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### Re: e to pi times i

It would really be hard to explain it without using calculus.

And I wouldn't refer to complex, or "imaginary," numbers as fake. They have numerous applications in the real world, most significantly with anything that has to do with waves. This is because "e" raised to an imaginary number behaves like a sinusoidal function and it's a lot easier to deal with exponentials than sinusoidal functions.
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BrainMagMo
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### Re: e to pi times i

telcontar42 wrote:Well, I wouldn't be so quick to call imaginary numbers fake number. They don't have the same physical representation as real numbers in that you can have 2 apples but not 2i apples. Still, a number like 2 and a number like i are both just mathematical constructs that we have defined. i is an important mathematical unit that is necessary in the representation of many real world systems. For example, in quantum mechanics, the momentum operator is i(h-bar)(d/dx). Momentum is certainly a real quantity and it involves i.

Also, you can get real numbers from just imaginary numbers. i*i=-1

Further, you can have 2i meters.
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SocialSceneRepairman
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### Re: e to pi times i

1/149896274?

eot
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### Re: "e to pi times i" Discussion

Is this a joke within a joke?
[imath]\sqrt{-1}[/imath] is undefined

Trican
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### Re: "e to pi times i" Discussion

Not necessarily, sqrt(-1) doesn't mean that it would come out to be a real number. It turns into an imaginary number whenever there is a negative under a square root. In this case it would be 1i, or just simply, i. That is why there is a discussion on it, how an imaginary number like that can turn out to be a whole number.

eot
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### Re: "e to pi times i" Discussion

No, you can't define it like that
Here's why:

[imath]1 = \sqrt{1} = \sqrt{-1 * -1} = \sqrt{-1} * \sqrt{-1} = i * i = -1[/imath]

i is defined as [imath]i^2 = -1[/imath]

GBog
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### Re: "e to pi times i" Discussion

Or you could just say that
$\sqrt{ab} = \sqrt{a} \sqrt{b}$ is not a general identity, but only holds when at least one of a and b is real and non-negative.

Extending the domain of a function often means giving up some structure.

eot
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### Re: "e to pi times i" Discussion

...which, I'm sure you'll agree, is quite clumsy and inelegant
tell me why you'd want to define [imath]i = \sqrt{-1}[/imath]

dragondave
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### Re: "e to pi times i" Discussion

Why should we define [imath]i = \sqrt{-1}[/imath]?

There is nothing special about [imath]i[/imath]; [imath]j[/imath] is also used in the literature, and there isn't anything preventing the use of [imath]\sqrt{-1}[/imath] itself, other than it is a pain to write repeatedly.

When trying to draw fractals it's often useful to not think about the fact that [imath]i[/imath] is not a real number. Treat it entirely as an unknown: as if we had [imath]a=b+6[/imath] with no additional information, we can still say that [imath]2a=2b+12[/imath], for example. If you treat [imath]i[/imath] like this, but with the curious property that [imath]i^2=-1[/imath], it is possible to make pretty pictures (e.g. the M set) or do 'complex' math (technical term, meaning: "math with real and imaginary numbers") with no additional mathematical knowledge. It is because of the fact that most maths works perfectly well with this change, and that these can be useful in describing rotations (Imagine a vertical sine wave from left to right, a two-dimensional construct, in a three-dimensional space. You can discuss a wave in the same direction, but moving forward and backwards, using complex numbers.)

For example, the M-set (Mandelbrot set) is generated like this: Pick a number (eg: -0.1). Square it. Add the original number, and square the result. Repeat this enough times so you know whether the number explodes towards infinity or doesn't (hint: if it gets bigger than two, it explodes). Colour that point on the number line either black or white, depending. Repeat for every point on the number line. (You end up with a short black line; from about -2 to 1/3, from memory.) Now we can do this with complex numbers, too, which makes it a two-dimensional grid.

So instead of a number, x, we now have a complex number, x+yi.
We can square this in the traditional way: [imath]x^2+y^2 i^2+2xyi[/imath].
Since we know that [imath]i^2[/imath] is -1, we can substitute that in: [imath](x^2-y^2)+(2xy)i[/imath]
This remains a complex number, where the real part is [imath]x^2-y^2[/imath] and the imaginary part is [imath]2xy[/imath]
We can use the mappings x=>[imath]x^2-y^2[/imath] and y=>[imath]2xy[/imath] to square complex numbers without needing to think about what [imath]i[/imath] actually *is*.

It was originally argued that zero was not a number. After all, one apple is not the same as one banana, so why is no apples the same as no bananas, or nothing?
Likewise, fractions (Part of an apple is different to an apple; part of an apple will rot quicker than a whole one.)
And negative numbers (How can I have minus three apples in my hands)
And irrational numbers (What do you mean, I can't write down the value of pi? It's got to end or repeat at *some* point!)
Imaginary and Complex numbers are merely further abstractions of the concept of numbers.

Rentsy
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### Re: "e to pi times i" Discussion

And everything can work fine!

[imath]i[/imath] is a real number, just not part of the set of reals. Saying that [imath]i[/imath] isn't a number is like saying [imath]\frac{1}{2}[/imath] isn't a number. "Ooo, you're using previously defined number (in this case, integers) and moving them around to define a new, richer set of numbers that includes the old but also includes these new things."

Remember: If you can define it, you can do it!

GBog
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### Re: "e to pi times i" Discussion

eot wrote:...which, I'm sure you'll agree, is quite clumsy and inelegant
tell me why you'd want to define [imath]i = \sqrt{-1}[/imath]

Not very clumsy. The equality holds up to a sign change anyway. But there is an inelegancy in the inability to extend it to other powers of complex numbers.

I don't want to define [imath]i[/imath] as [imath]\sqrt{-1}[/imath], [imath]i[/imath] is defined by [imath]i^2=-1[/imath], but if you want to define square roots of negative numbers, [imath]\sqrt{-1}=i[/imath] is a natural choice. As shown, you then loose some structure. But you can.

PebblesRox
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### Re: "e to pi times i" Discussion

Here's a really good explanation that uses visualization to explain how the equation works:

http://johnlawrenceaspden.blogspot.com/ ... -pi-1.html

There's also an interesting article and discussion at Hacker News:

http://news.ycombinator.com/item?id=911961

Alx_xlA
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### Re: "e to pi times i" Discussion

Rentsy wrote:My [imath]i[/imath] is your [imath]-i[/imath]

And everything can work fine!

[imath]i[/imath] is a real number, just not part of the set of reals. Saying that [imath]i[/imath] isn't a number is like saying [imath]\frac{1}{2}[/imath] isn't a number. "Ooo, you're using previously defined number (in this case, integers) and moving them around to define a new, richer set of numbers that includes the old but also includes these new things."

Remember: If you can define it, you can do it!

Umm... [imath]i[/imath] is not a Real Number. It is a member of the imaginary number set, which shares the complex plane with the real one.
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Cynical Idealist
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### Re: "e to pi times i" Discussion

Alx_xlA wrote:
Rentsy wrote:My [imath]i[/imath] is your [imath]-i[/imath]

And everything can work fine!

[imath]i[/imath] is a real number, just not part of the set of reals. Saying that [imath]i[/imath] isn't a number is like saying [imath]\frac{1}{2}[/imath] isn't a number. "Ooo, you're using previously defined number (in this case, integers) and moving them around to define a new, richer set of numbers that includes the old but also includes these new things."

Remember: If you can define it, you can do it!

Umm... [imath]i[/imath] is not a Real Number. It is a member of the imaginary number set, which shares the complex plane with the real one.

[imath]i[/imath] isn't a member of the Real Numbers, but it is a real number in the sense that it exists and is defined. It is the imaginary unit, but it isn't a fake number.
Stupid reuse of terms...
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### Re: "e to pi times i" Discussion

Tronald wrote:Quite frankly i dot get it becuase i Dont think imaginary numbers Can = Real ones http://xkcd.com/179/

If you're thinking, "how can a non-real number crop up in one side of the equation and evaluate to something as real as -1", then you're neglecting that e to any real power can never be negative.

In short, e^x < 0 is as impossible as x^2 < 0 for real numbers. The nifty thing about i is that you can arbitrarily define it as satisfying both of these impossible equations, and get a consistent arithmetic model.
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Eebster the Great
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### Re: "e to pi times i" Discussion

eot wrote:...which, I'm sure you'll agree, is quite clumsy and inelegant
tell me why you'd want to define [imath]i = \sqrt{-1}[/imath]

Defining [imath]i = \sqrt{-1}[/imath] is actually meaningless until we have defined [imath]\sqrt{z}[/imath] for complex z, so for that reason you do actually see i defined as [imath]i^2 = -1[/imath] (or equivalently, as one of the solutions to [imath]x^2 + 1 = 0[/imath]). However, once we define [imath]\sqrt{ }[/imath] as meaning the principal branch of the complex square root function (which is the complex extension of the principal square root function), it is indeed true that [imath]i = \sqrt{-1}[/imath]. Your claim that$1 = \sqrt{1} = \sqrt{-1 * -1} = \sqrt{-1} \sqrt{-1} = i * i = -1$ fails because either [imath]\sqrt{ }[/imath] is the principal square root function, which is not defined for negative numbers, or it is the principal branch of the complex square root function, which does not in general distribute over multiplication.

SocialSceneRepairman
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### Re: "e to pi times i" Discussion

eot wrote:...which, I'm sure you'll agree, is quite clumsy and inelegant

Not at all. In fact, hardly anyone would agree. As you get further in math, you'll find that pretty much all the rules you take for granted with reals have caveats like that. "Clumsy and inelegant" would be to define a rule without such caveats.

The reason he used the notation he did is because mathematicians and the general public typically call this number "i," but scientists almost always call it "j," because university departments like to be petty. (Scientists also use phi for azimuthal angle and theta for polar angle, where mathematicians write it the other way around.) (Okay, anyone will tell you that it's because i is current, or index, but if that were the case, scientists would use a different letter for e as well.)

Eebster the Great
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### Re: "e to pi times i" Discussion

SocialSceneRepairman wrote:The reason he used the notation he did is because mathematicians and the general public typically call this number "i," but scientists almost always call it "j," because university departments like to be petty. (Scientists also use phi for azimuthal angle and theta for polar angle, where mathematicians write it the other way around.) (Okay, anyone will tell you that it's because i is current, or index, but if that were the case, scientists would use a different letter for e as well.)

It's when the letters start getting random (I for current, J for current density, K for momentum (wtf? p wasn't good enough?), U for work, etc.) that I start becoming very annoyed and confused.

pubescent tuba
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### Re: "e to pi times i" Discussion

I really like this formula! I had never come across it before, but it makes at least some sense (to a high school student, I think that's reasonable) and it's so neat.

SocialSceneRepairman
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### Re: "e to pi times i" Discussion

pubescent tuba wrote:I really like this formula! I had never come across it before, but it makes at least some sense (to a high school student, I think that's reasonable) and it's so neat.

It's actually pretty easy to prove with even AP Calculus; once you get to Maclaurin series, it's just what some of my professors would call "plug-and-chug." The full (more useful) version of the theorem is [imath]e^{i\theta} = \cos\theta + i\sin\theta[/imath], where theta is in radians.

Even to a high school student, it's probably pretty obvious how you get from there to the formula in the comic.

pubescent tuba
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### Re: "e to pi times i" Discussion

Ja, but I hadn't come across that either. But now I have, and it makes maths even more happy for me to know such beautiful equations exist.

theta4
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### Re: "e to pi times i" Discussion

When plugging a complex number into [imath]e^{x}[/imath], you always get a complex number. It just so happens that [imath]e^{\pi i} = -1 + 0i[/imath], i.e., the imaginary part is zero, so it goes away, leaving -1.

Is it that hard to understand? Math can get crazy (like Taylor series; who woulda guessed?) so we just have to deal with it. [imath]e^{\pi i} + 1 = 0[/imath], EOF.
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SocialSceneRepairman
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### Re: "e to pi times i" Discussion

Well, real numbers are complex numbers. There's nothing less "real" about the number you get from a complex formula. In fact, complex formulas that cancel are the backbone of a lot of physics.

Eebster the Great
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### Re: "e to pi times i" Discussion

SocialSceneRepairman wrote:Well, real numbers are complex numbers. There's nothing less "real" about the number you get from a complex formula. In fact, complex formulas that cancel are the backbone of a lot of physics.

Sure, but as far as I know, these are generally used as mathematical tools to analyze phenomena that could in principle be described using purely real numbers. For example, wave equations are complex, but only the real part of the result really matters in the end. e^x is just a lot easier to work with than sin x.

polymer
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### Re: "e to pi times i" Discussion

Eebster the Great wrote:Sure, but as far as I know, these are generally used as mathematical tools to analyze phenomena that could in principle be described using purely real numbers. For example, wave equations are complex, but only the real part of the result really matters in the end. e^x is just a lot easier to work with than sin x.

In the magical land of quantum mechanics stuff is actually "complex." Not just a really useful mathematical tool.

jtonca
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### Re: "e to pi times i" Discussion

There’s a neat way to show Euler’s identity computationally, using the binomial theorem, without any calculus.

Start with [imath](1+1/N)^N[/imath], with N a positive integer. Expand with the binomial theorem and take the limit of each term as N[imath]\to\infty[/imath] and you get 1+1+1/2!+1/3!+1/4!+... which converges to 2.71828... So [imath](1+1/N)^N[/imath] = Euler’s constant, e, as N[imath]\to\infty[/imath].

Next, it is easy to show, as N[imath]\to\infty[/imath], that, for any real number x, [imath](1+x/N)^N = e^x[/imath] (because for N>>x, [imath](1+1/N)^x=(1+x/N)[/imath], so [imath](1+x/N)^N=((1+1/N)^x)^N =((1+1/N)^N)^x = e^x[/imath]). And we have familiar properties such as [imath]e^x * e^y = (1+x/N)^N * (1+y/N)^N = (1+x/N+y/N)^N = (1+(x+y)/N)^N = e^{(x+y)}[/imath] .

So we’ve come up with a way of expressing [imath]e^x[/imath], without using any calculus, as an expression [imath](1+x/N)^N[/imath] (as N[imath]\to\infty[/imath]) that only involves familiar addition, division and multiplication. Because we already know how to do arithmetic on complex numbers, we define [imath]e^z[/imath], where z is a complex number, the same way: [imath](1+z/N)^N[/imath] as N[imath]\to\infty[/imath].

The only drawback for calculation is the N[imath]\to\infty[/imath] issue, but the calculus/Taylor series approach needs an infinite number of terms, too. We’ll say for a given N, we are approximating, and successively larger N’s given better approximations, like more terms on a Taylor series.

We note that squaring a complex number is an easy operation (if [imath]z=a+bi[/imath], then [imath]z^2 = (a^2-b^2)+2abi[/imath]). Since there are no restrictions on how N[imath]\to\infty[/imath], choose multiples of 2; e.g. if N=256=[imath]2^8[/imath], then [imath]e^z[/imath] ~ [imath](1+z/256)^{256}[/imath], or [imath](1+z/256)[/imath] squared 8 times (since every time you square an expression, you double the exponent).

To evaluate [imath]e^{\pi i}[/imath], we need to compute [imath](1+\pi i/N)^N[/imath]. E.g, if we set N=256, then we need to square 1.000+0.012i eight times, which gives -1.020+0.000i, which is not a bad estimate. If we set N=65536, then we need to square 1.00000+0.00005i sixteen times, which gives -1.00008+0.00000i, which is pretty close to -1, and you can do better and better with bigger N.

Granted, the math above is not completely rigorous (though it can be made so), but this approach gives a good sense for how Euler’s identity can be true, without using any calculus.

Even better, as you keep squaring the numbers, plot the successive values in the complex plane, you’ll see that they all successive rotations around the unit circle, which gives a great sense of how an exponential function becomes an oscillating function when complex numbers are involved (which you don’t get from plugging numbers into a Taylor series).

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### Reproduction in xkcd: vol 0

This comic is reproduced in the xkcd book, the formula "corrected" to erroneously assert:

$e^{\pi\sqrt{-1}} = 1$

appearing with a note from Randall explaining he received forty emails telling him the sign was wrong in the online edition (it wasn't). Shame. Thank you cocky fervid pedants who barely understand the complex numbers but are too sure of themselves to check before emailing "corrections".

Of course, the tens of thousands of readers who know maths and recognised the comic to be correct never wrote in to assure xkcd. Following feedback is dangerous, you make a selection bias. Upset folk are far more likely to write in to complain than pleased folk are to give praise.

geekmom
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### Re: "e to pi times i" Discussion

My favorite xkcd ever. Trying to figure out how to share it with AP Calc students without getting fired...

This was my reaction when my diff. eq's prof worked a definite integral of the gamma function and pulled a multiple of pi out as the solution.... REALLY??? wha??

Math is just so cool.

SocialSceneRepairman
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### Re: Reproduction in xkcd: vol 0

mattme wrote:This comic is reproduced in the xkcd book, the formula "corrected" to erroneously assert:

$e^{\pi\sqrt{-1}} = 1$

appearing with a note from Randall explaining he received forty emails telling him the sign was wrong in the online edition (it wasn't). Shame. Thank you cocky fervid pedants who barely understand the complex numbers but are too sure of themselves to check before emailing "corrections".

Of course, the tens of thousands of readers who know maths and recognised the comic to be correct never wrote in to assure xkcd. Following feedback is dangerous, you make a selection bias. Upset folk are far more likely to write in to complain than pleased folk are to give praise.

...really? That's such a basic identity to fuck up, I'd have to see it to believe it.

amieli
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### Re:

vf10a wrote:On the use of j - I did my A -levels last year and the (I think) most widely used math exam board in the UK now teaches j instead of i. Why this is I do not know (at least in EE there is a obvious reason).

If you're refering to MEI which I think is the only board that uses j but I'm not sure, the people who set it up are engineers, hence they believe j should be used, but they allow the use of i in exams anyway, which is what I do

Zalde Ocga
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### Re: 0179: "E to the Pi times I"

In the scroll over information, he says, "I have never been totally satisfied by the explanations for why e to the ix gives you a sinusoidal wave." The only problem is that I can't make a sinusoidal graph from it.
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### Re: 0179: "E to the Pi times I"

Zalde Ocga wrote:In the scroll over information, he says, "I have never been totally satisfied by the explanations for why e to the ix gives you a sinusoidal wave." The only problem is that I can't make a sinusoidal graph from it.

Well, according to Euler's formula, [imath]e^{ix}=\cos{x}+i\sin{x}[/imath], which is a sinusoidal wave with period [imath]2\pi i[/imath]. Alternatively, you can use Euler's formula along with a couple of basic trig identities to show that [imath]\cos{x} = \frac{1}{2} (e^{ix} + e^{-ix})[/imath] and [imath]\cos{x} = \frac{1}{2i}(e^{ix} - e^{-ix})[/imath].
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Eebster the Great
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Joined: Mon Nov 10, 2008 12:58 am UTC
Location: Cleveland, Ohio

### Re: 0179: "E to the Pi times I"

ConMan wrote:
Zalde Ocga wrote:In the scroll over information, he says, "I have never been totally satisfied by the explanations for why e to the ix gives you a sinusoidal wave." The only problem is that I can't make a sinusoidal graph from it.

Well, according to Euler's formula, [imath]e^{ix}=\cos{x}+i\sin{x}[/imath], which is a sinusoidal wave with period [imath]2\pi i[/imath]. Alternatively, you can use Euler's formula along with a couple of basic trig identities to show that [imath]\cos{x} = \frac{1}{2} (e^{ix} + e^{-ix})[/imath] and [imath]\cos{x} = \frac{1}{2i}(e^{ix} - e^{-ix})[/imath].

Obviously Randall is aware of the formal proofs and how they work (they are very simple), but the fact doesn't seem intuitive to him. There was a SMBC comic sort of like that, but with a multiplication trick using your fingers.