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Euler's formula approximation

Anonymous Sat 12 Mar 17:54:00 2022 No.14298407 View Reply Original Report

Quoted By:

An even better approximation using a different base:

$2^{1.442696\pi i} + 1 = 0$

Prove me wrong.

What is the algeabraic meaning of raising a real number to the power of an imaginary number?

How does Matlab calculate the result?

Anonymous

Anonymous Sat 12 Mar 2022 18:07:36 No.14298446 Report

Quoted By: >>14298464 >>14298466

$ \log \pi^{ei} = i e \log \pi \neq i (2k+1) \pi, \forall k \in \mathbb Z $
Thus $\pi^{ei} \neq -1$ .

Anonymous

Anonymous Sat 12 Mar 2022 18:08:32 No.14298448 Report

Quoted By: >>14298462

(2^(1.442696 * ? * i)) + 1 =
2.18103313 × 10-12 - 2.08854685 × 10-6 i

Anonymous

Anonymous Sat 12 Mar 2022 18:09:45 No.14298450 Report

Quoted By: >>14299681

Reminder that being amazed at this is equivalent to being amazed that 2 + 2 = 4

Reminder that anyone who claims to be amazed is most likely a midwit who doesn't eve understand math and pretends to look smart

Anonymous

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Anonymous Sat 12 Mar 2022 18:14:33 No.14298462 Report

Quoted By:

>>14298448
This looks wrong.
How is this calculated anyway?

Anonymous

Anonymous Sat 12 Mar 2022 18:15:12 No.14298464 Report

Quoted By: >>14298466

>>14298446
The last inequality is due to the fact that the function $x \mapsto \frac{x}{\log x}$ is minimized at $x = e$ , therefore
$ e \log \pi = \frac{e}{\log e} \log \pi < \frac{\pi}{\log \pi} \log \pi = \pi. <script type="math/tex; mode=display">$

Anonymous

Anonymous Sat 12 Mar 2022 18:16:06 No.14298465 Report

Quoted By: >>14298468 >>14300454

this is incredibly wrong
I don't even know where to begin other than by telling you that pi to the power of e is 23.
This isn't an accurate approximation at all.
The reason Euler's equation works is due to e^(ipi) being equivalent to the Taylor series of the sum of i*Sine and Cosine. Pi has no such relation. Since those are trigonometric angles functions, that's where theta (which is pi as it is written in Euler's identity) comes in.

Anonymous

Anonymous Sat 12 Mar 2022 18:16:19 No.14298466 Report

Quoted By:

>>14298446
>>14298464
Sorry, typo
$ e \log \pi = \frac{e}{\log e} \log \pi < \frac{\pi}{\log \pi} \log \pi = \pi. $

Anonymous

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Anonymous Sat 12 Mar 2022 18:18:20 No.14298468 Report

Quoted By: >>14298692

>>14298465
it is fairly accurate.
I wish I knew how it's calculated.

Anonymous

Anonymous Sat 12 Mar 2022 20:02:37 No.14298692 Report

Quoted By: >>14298705 >>14298739 >>14298774

>>14298468
To solve this, I'll just focus on $pi^{e*i}$ I'll worry about the +1 at the end.
First thing, using euler's relation, is to exponentiate it. This gives:
$e^{e*i*ln(?)}$
Now you have to actually look at what's going on here. The number is going to be pulled around the complex plane. e is just going to increase the rate at which that happens, the radius of the number won't be different and will still equal one.
$e^{i*ln(?)} = e*(cos(ln(?)) + i(e*sin(ln(?)))$
To make it easier to understand, I'll put this in decimal form.
This comes out to around $-0.999553+0.02989*i$
Notice that this is the number your calculator came out to.
The magnitude of this equation will be 1, think of e as just multiplying the amount it has been "pulled" in a circle around the complex plane.
$?^{i*e} = -0.99553+0.2989i \neq -1$
Therefore, $?^{e*i} + 1 \neq 1$

Anonymous

Anonymous Sat 12 Mar 2022 20:09:01 No.14298705 Report

Quoted By: >>14298739

>>14298692
Small correction:
$e^{e*i*ln(?)} = e?(cos(ln(?))+i(e?sin(ln(?)))$
I accidentally left e out of the left hand side there.

Anonymous

Anonymous Sat 12 Mar 2022 20:22:00 No.14298739 Report

Quoted By: >>14299553

>>14298705
>>14298692
And on that note I realize I wrote another error.
e is in theta so it should be inside the trigonometric functions. Not on the outside like I wrote.
$= (cos(ln(?)*e)+i(sin(ln(?)*e))$
Sorry if that confused anyone.

Anonymous

Anonymous Sat 12 Mar 2022 20:34:43 No.14298774 Report

Quoted By: >>14298780 >>14298966

>>14298692
>First thing, using euler's relation, is to exponentiate it
Why? Just so we have something familiar to work with, since we now can think in terms of cos and sin, essentially the polar coordinates ?

>pulled around the complex plane.
do you mean it is going to rotate around the unit circle? i get that.

>?0.99553+0.2989i
it doesn't exactly equal -1 but it is pretty damn close. Trigonometrically, we just underrotated a little bit: the X coordinate is pretty much -1 and the Y-coord is 0.29.
What I don't understand is the significate of the base of this exponential function. If we replace e with some arbitrary base, we still have the unit circle, and the vector rotating around that unit circle, with a certain initial phase and frequency, which doesn't matter. So we have x^iy where x and y are arbitrary. And we just spin our vector, like a roulette, and check where it stops on the unit circle. I don't understand the significance of the base being e, other than using the Taylor series to see that e^ix = cos (x) + i sin (x). But what if the base is something other than e?? We will get some kind of a complex number a + ib
except it won't be cos and sin. Possibly. I don't know. I am confused.
How does matlab numerically calculate the result of a^ib where a and b are arbitrary real numbers?

Anonymous

Anonymous Sat 12 Mar 2022 20:38:10 No.14298780 Report

Quoted By: >>14298790

>>14298774
>I don't understand the significance of the base being e, other than using the Taylor series to see that e^ix = cos(x) + i sin(x)
That's the reason this relation exists.
>But what if the base is something other than e?
Then it won't be equivalent to cos(x)+isin(x).

Anonymous

Anonymous Sat 12 Mar 2022 20:44:20 No.14298790 Report

Quoted By: >>14298792

>>14298780
But the resulting complex number is still some a+ib
and both a and b can be expressed in terms of cos and sin in the polar coordinates. So we still spin our vector on the unit circle.

Anonymous

Anonymous Sat 12 Mar 2022 20:47:01 No.14298792 Report

Quoted By:

>>14298790
You are deeply confused.

Anonymous

Anonymous Sat 12 Mar 2022 21:49:27 No.14298966 Report

Quoted By: >>14299172

>>14298774
>I don't understand the significance of the base being e
the derivative of e^x is e^x
the derivative of a^x is ln(a) * a^x
the derivative of e^ix is i * e^ix, from which we can conclude that it moves around the unit circle at a rate of 1
the derivative of a^ix is i * ln(a) * a^ix, from which we can conclude that it moves around the unit circle at a rate of ln(a)

Anonymous

Anonymous Sat 12 Mar 2022 23:47:04 No.14299172 Report

Quoted By: >>14299467

>>14298966
>moves around the unit circle at a rate
thats the frequency, right?

that makes sense. i get that. so the base of exponent determines the frequency. so if we change the base from e to a we can compensate for the changed frequency by tweaking the exponent b of a^ib so we could in theory get the same result for a given a and find b such that a^ib = -1, right? in that case we could have a^ib = cos (something) + i sin (something) and that something must have ln(something) in it?

Anonymous

Anonymous Sun 13 Mar 2022 03:05:26 No.14299467 Report

Quoted By: >>14299553

>>14299172
$a^{ib}$
$= \left(e^{\ln(a)}\right)^{ib}$
$= e^{ib \ln(a)}$
$= \cos(b \ln(a)) + i \sin(b \ln(a))$

Anonymous

Anonymous Sun 13 Mar 2022 03:55:53 No.14299553 Report

Quoted By: >>14299660 >>14299664

>>14299467
Ah so it is the same as this:
>>14298739
and when a=e we get the Euler's formula.
I think I get it. But. Why do we represent a as e^ln(a)? To convert to the Euler's formula? What if we didn't know the Euler's formula?
Is that because we have some trigonometric intuition (the rotation around the unit circle)?

Anonymous

Anonymous Sun 13 Mar 2022 04:57:23 No.14299660 Report

Quoted By: >>14299702

>>14299553
>Is that because we have some trigonometric intuition
no its because a^x=e^(xln(a)) dumbass

Anonymous

Anonymous Sun 13 Mar 2022 05:01:51 No.14299664 Report

Quoted By: >>14299749 >>14300272 >>14300461

>>14299553
>To convert to the Euler's formula?
Yes.
>What if we didn't know the Euler's formula?
Then we'd have to work it out.

We're trying to find out real functions f and g such that
$e^{ix} = f(x) + i g(x)$ .
Take its derivative
$f'(x) + i g'(x) = i e^{ix} = i (f(x) + i g(x)) = - g(x) + i f'(x)$
and we see
$f'(x) = -g(x)$
$g'(x) = f(x)$ .
From $e^{i0} = 0$ , we have $f(0) = 1$ and $g(0) = 0$ .
The solutions for f and g are cosine and sine.

In more geometric terms, consider a point at $e^{it}$ where t is the time. The velocity of that point is the derivative of the function, $i e^{it}$ , which is the vector from the origin to the point rotated 90 degrees counterclockwise. This makes the point go in a circle at unit speed.

You could do all of this with a different base but then you would have factors of $\ln(a)$ all over the place. The point would move around the circle at speed $\ln(a)$ .

Anonymous

Anonymous Sun 13 Mar 2022 05:07:35 No.14299681 Report

Quoted By: >>14300458

>>14298450
midwit detected

Anonymous

Anonymous Sun 13 Mar 2022 05:24:11 No.14299702 Report

Quoted By: >>14300272 >>14301854

And what >>14299660 says is correct; an easy way to define what imaginary and other strange exponents mean is to first define $e^x$ and then define other stuff based on that.

If you wanted to, you could avoid using e explicitly by using a definition like

$a^b = \lim_{n \to \infty} (b(\sqrt[n]{a} - 1) + 1)^n$
(where n is an integer)
(Integer exponents are presumed to be already defined. Showing that this formula is consistent with the already defined integer exponents when b is an integer is left as an exercise for the reader.)

but that's really just hiding the e because

$\lim_{n \to \infty} n(\sqrt[n]{a} - 1) = \ln(a)$
$\lim_{n \to \infty} (x/n + 1)^n = e^x$

Anonymous

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Anonymous Sun 13 Mar 2022 05:49:11 No.14299749 Report

Quoted By:

>>14299664

Anonymous

Anonymous Sun 13 Mar 2022 14:26:49 No.14300272 Report

Quoted By: >>14300461 >>14300538

>>14299664
>>14299702
so there is really no way around e and the euler's formula. why is that? is that how the world works?

Anonymous

Anonymous Sun 13 Mar 2022 16:17:04 No.14300454 Report

Quoted By:

>>14298465
Every a^ib can be taylord into a sum of cos and i*sin. choosing e as a Just gets rid of anoying Log(a)'s infesting the whole series

Anonymous

Anonymous Sun 13 Mar 2022 16:18:37 No.14300458 Report

Quoted By:

>>14299681
Found the midwit

Anonymous

Anonymous Sun 13 Mar 2022 16:19:33 No.14300461 Report

Quoted By: >>14300564 >>14301887

>>14300272
Functions of the form $f(x) = a^x$ have the property that their derivative is a constant times themselves. e is the base for which this constant is 1. If you're doing anything involving derivatives of these functions, e is going to show up. And as you can see from >>14299664 the simplest way of understanding why we define imaginary exponents the way we do is that it's the choice that makes the derivative work as expected with complex numbers. We could use this idea as a definition too: $f(x) = a^x$ is the unique function that agrees with the usual definition for rational exponents and is complex differentiable. I don't know what you mean by no way around Euler's formula. That is something we can derive, as seen in >>14299664.

Anonymous

Anonymous Sun 13 Mar 2022 17:00:21 No.14300538 Report

Quoted By: >>14300564

>>14300272
imagine if you actually buckled down and learned calculus some day

Anonymous

Anonymous Sun 13 Mar 2022 17:09:21 No.14300564 Report

Quoted By: >>14300650

>>14300461
thanks anon. this thread has been incredibly useful for me even if i only understand about 90%. but i think i get the most important part, the taylor series of exp() and its connection with cos + isin

>>14300538
i've been learning calculus but i don't have a deep understanding of it. threads like this at least make me think in the right direction even if some parts may still be over my head. i will save and review these posts.

Anonymous

Anonymous Sun 13 Mar 2022 17:42:54 No.14300650 Report

Quoted By: >>14301220

>>14300564
>i've been learning calculus but i don't have a deep understanding of it
I appreciate your humility. it shows character and isn't he norm on 4chan.
>but i think i get the most important part, the taylor series of exp() and its connection with cos + isin
This formula (E) connects calculus and differential equations (the exp part) with trigonometry and geometry (the right hand side).
But you must try to appreciate the ALGEBRAIC connection as well. Given (E), it is easy to show that the set of all $z\in\mathbb{C}\ \text{ s.t. }\ z^n=1$ makes a structure called a "group" with n elements. If n=5, for example, you get $z=1,\exp(\pm j2\pi/5),\exp(\pm j4\pi/5)$ all of which satisfy $z^5=1$ . Well call this group "nth roots of unity" and note that it can also be shown to be a cyclic group. The nth roots group is a a subset of the circle group (the set of all z with |z|=1). Both these turn out to be commutative or "abelian."
These structures are very natural and it's ridiculously convenient for anyone who cares about applications to have all of this pre-encoded in (E). Algebra is just as important as the connection to trig and differential equations. I use $j$ btw

Anonymous

Anonymous Sun 13 Mar 2022 22:03:37 No.14301220 Report

Quoted By:

>>14300650
>algebraic
>groups
I need to look this up, I know the basics but it is way over my head.

>i use j btw
Do you post in /diy/ /ohm/ by chance?
I occasionally ask math related questions there about bode plots / transfer functions / phasors etc. sort of related to this topic.

Anonymous

Anonymous Mon 14 Mar 2022 03:38:43 No.14301854 Report

Quoted By: >>14301896 >>14302093

>>14299702
What does the number e signify?
Why exactly does the number e make this work?
What is the number e interacting with to do what?
What is the value of e contributing towards?
What fact of the matter is hinging on e being e?
I get what i is I get what pi suggests, what is e 's precise relation to the situation?
What is forcing the working of the idea to turn 180 degrees, to relate to the number e in any way?

Anonymous

Anonymous Mon 14 Mar 2022 03:47:11 No.14301887 Report

Quoted By: >>14302308

>>14300461
Does the number e being relavant in that turning the 180 degrees; something about an establishing measurement degree by degree? The squaring it bringing it to 360 degrees?

So the number e is some ratio of having to do with each degree?

What provoked my thinking this is looking up about e and seeing one of it's uses having to do with plotting to the days of the year, which is 365; but maybe with leap years, knocks down around 360 and their might be relvance.

Something like a ratio of a step up, and s fraction of a step over.

Does this also relate to a spiral, drawing a circle through time?

Anonymous

Anonymous Mon 14 Mar 2022 03:51:01 No.14301896 Report

Quoted By: >>14302308

>>14301854
>turn 180 degrees
i does the turning, not e

Anonymous

Anonymous Mon 14 Mar 2022 05:39:49 No.14302093 Report

Quoted By:

>>14301854
$a^x$ is its own derivative when a = e.

Anonymous

Anonymous Mon 14 Mar 2022 08:21:31 No.14302308 Report

Quoted By: >>14302385

>>14301896
What is forcing (the working of the idea to turn 180 degrees)

to relate to the number e in any way?

Was the question, but there were others, any thoughts on them? >>14301887
That adding on too

Anonymous

Anonymous Mon 14 Mar 2022 08:26:38 No.14302312 Report

Quoted By:

Why was 360 degrees, 180, 90, chosen how did that come bout?

Anonymous

Anonymous Mon 14 Mar 2022 09:16:44 No.14302385 Report

Quoted By: >>14303304 >>14303371

>>14302308
e just makes the point $e^{it}$ go around the unit circle at speed 1 instead of some other speed. So that at $t = \pi$ it will be at $\pi$ radians and not somewhere else.

Anonymous

Anonymous Mon 14 Mar 2022 17:42:19 No.14303304 Report

Quoted By: >>14303470

>>14302385
What does speed mean in this context? Angular velocity? Radians per second? Or just per unit of time? I know frequency/period but I've never thought about speed.

Anonymous

Anonymous Mon 14 Mar 2022 18:05:27 No.14303371 Report

Quoted By: >>14303470

>>14302385
"frequency" is basically speed, right, revolutions per unit of time, for example 1Hz is one revolution per second which is the same as 2pi radions per second. But if we talk about speed, isn't circular motion accelerated, so sin and cos are both motion with acceleration, so it is not constant speed?

Anonymous

Anonymous Mon 14 Mar 2022 18:43:28 No.14303470 Report

Quoted By: >>14304309

>>14303304
Well, if $e^{it}$ is position on the complex plane measured in units of meters, and t is time in seconds, then $i e^{it}$ would be the velocity in meters/second and $|i e^{it}| = 1$ would be speed in meters/second.

>>14303371
>sin and cos are both motion with acceleration, so it is not constant speed?
velocity is changing, speed (magnitude of velocity) is constant

Anonymous

Anonymous Tue 15 Mar 2022 00:34:28 No.14304309 Report

Quoted By: >>14304400 >>14304608

>>14303470
What is the number e doing exactly, 2.7somethingsomethingsomething

Is it due to a consideration of every degree turned? The idea of 90 or 180 degrees can't just be stated to be turned?

You can't just say ? +1 = 0 or whatever ?^2 if it's that.

The e implies a starting point on the numberline?

And so it describes the arc curve starting at the number on the x axis, needing to advance further than they number, as s curve; as the bottom of this symbol parentheses arc

** ) ** is closer to 0,0 than the middle of it.

So something about 2.7whateverwhatever

Is dictating some ratio of curve per degree left from the x axis? Or pi already does that.

In this context pi doesn't mean, 3.14...
? means, turn 90 degrees, or is it 180 I don't remember;

So you're wanting the equation to equal 0, which implies a 360 degree turn;

What does the + 1 in this context mean;

If ? means rotate 180, what would ? plus 1 mean?

Rotate 1 more 180?

Anonymous

Anonymous Tue 15 Mar 2022 01:24:04 No.14304400 Report

Quoted By: >>14304664

>>14304309
>What does the + 1 in this context mean
it means "add one"

Anonymous

Anonymous Tue 15 Mar 2022 01:41:35 No.14304434 Report

Quoted By: >>14304628

just so you know i am the OP and this retarded spammer is not me. i wish i could delete this thread

Anonymous

Anonymous Tue 15 Mar 2022 03:17:12 No.14304608 Report

Quoted By: >>14304628 >>14304718

>>14304309
It might be best to think of this starting from the problem "if you have a quantity that grows at a rate equal to its current value, what function describes that quantity?" The answer depends on the starting value. But once you have a starting value, since you know the rate of growth, you have enough information to predict the quantity at any later point in time.

Let's call the function that describes growth from a starting value of 1 exp(t). Per the specification, exp(0) = 1, and its derivative must equal itself: exp'(t) = exp(t).

What if we start from a different value? Let's call the starting value C. Notice that the derivative of C*exp(t) is C*exp(t), and C*exp(0) = C*1 = C. So C*exp(t) is our solution. All we had to do to was multiply our original solution by the starting value.

Let's think a bit more about the case starting from 1 to learn a bit about exp(x). At 0 seconds (or whatever unit of time we're using), the quantity is 1, and at 1 second, the quantity is exp(1). What will the quantity be at 2 seconds? Well, exp(2), but there's another way to answer the question. The situation at t=1 second is exactly the same as it was at t=0 seconds except we have a different value. We can think of exp(1) as our starting value and use the multiplication trick to work out what it will be 1 second later at t=2. It will be exp(1) * exp(1) = exp(1)2. We can continue this reasoning: At t=3, the quantity will be exp(1) * exp(1) * exp(1) = exp(1)3, and so on. See a pattern? This is an exponential function: exp(t) = exp(1)^t. Or abbreviating exp(1) as e, we have exp(t) = e^t.

By the chain rule, the derivative of C*exp(k*t) is k*C*exp(k*t), or k times the original function. This can be used to solve a lot of problems where the rate of growth of a quantity is proportional to its current value. For example population growth, interest, radioactive decay. The function exp(t) = e^t shows up a lot. It's not really the number e that's important, but this function.

Anonymous

Anonymous Tue 15 Mar 2022 03:25:55 No.14304628 Report

Quoted By: >>14304664

>>14304608
>exp(1) * exp(1) = exp(1)2
>exp(1) * exp(1) * exp(1) = exp(1)3
Ah, 4chan brain damage. Hopefully it's clear that exp(1)^2 and exp(1)^3 is meant.

>>14304434
That explains why the questions seem to be from someone with less understanding. Well, I'm still happy to answer questions.

Anonymous

Anonymous Tue 15 Mar 2022 03:51:57 No.14304664 Report

Quoted By:

>>14304628
That's great anon, you are godsend, I just wasn't sure if he is actually asking questions in good faith or deliberately spamming with bait questions >>14304400
to make fun of this thread and discourage you from helping others.

Anonymous

Anonymous Tue 15 Mar 2022 04:25:05 No.14304718 Report

Quoted By:

>>14304608
So 2.7 whatever whatever

Has nothing truly to do with, rotating 360 degrees. And calling that equal to 0, because you are where you started?

If ? in this context means, turn 180. Is the equation equivelent to ?^2 = 0
Or ? +1 =0

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