*3 hours later* kid in the back who didn"t listen: yooooo how can you have the sqrt of -1 ??? I just checked on my calculator and it says "undefined" worse even: Everyone in the class: fuck he's right, we think you made a mistake sir
I had a physics professor who would just cartoonishly wave his arms and say "as any fool can see!" and then write down the last statement of the derivation. It was his way of not-so-subtly hinting that the physics is more important than the n-th degree of the math- he would show us resources that contained the full derivations later. That might not work in a math class, though.
You forgot to mention that Euler’s identity also employs the four principal mathematical operators: addition, multiplication, exponentiation, and equality. ;)
If you're ever curious about where those series come from (or you don't feel like remembering them), just derive them yourself! Suppose cos(x) = a + bx + cx^2 + dx^4 + ... . Set x = 0, solve for a, then differentiate both sides with respect to x and solve for b. Differentiate yet again and keep going! Within four or five iterations, you should see a pattern emerge. You can do the same thing for any function, like sin(x) or e^x or ln(1+x). It's much easier than having to remember them.
@@catnip202xch. Deriving a Taylor series shouldn't be that hard. Do you know how to differentiate most functions? Can you set x = 0 and solve for coefficients one after another?
Taylor series just assumes every function can be expressed as a polynomial function of infinite order. Let f(x)= a0 + a1x + a2x²+.......+anxⁿ+....upto infinity f(0)=a0 f'(0)=a1 f"(0)=2a2 ....... And so on fⁿ(0)=n!an Then just multiply individual an values to xⁿ and thats the mauclaurin series. I should mention this is a specification of Taylor series. The original one's a bit more complicated to derive: f(a+h)=f(a) + hf'(a)+..... Remembering things is hard. Try to comprehend the underlying concept and you wont need memorization.
I agree that this is the most beautiful equation in math. I learned it in my fifth and final year at the university, taking a course i didnt even need. Needless to say, I am quite grateful I took that course.
Not only does it have the 5 most important numbers, it also has the 4 most important operations: equality, addition, multiplication and exponentiation.
I am watching your 100 infinite series little by little and the difference in hair style is so nuts compared to now. Also the mic is now a pokeball amazing
A technician takes X hours to visit the stores in a couple of streets in one shift, and when he finishes his tour, another technician revisits the same stores again, needing other X hours to finish the second shift, and so on. For example, if the first technician started his shift at 1:00pm, he finishes it at 7:00pm and the second technician starts his shift at 7:00pm and finishes it at 1:00am, and so on. The supervisors used to make a quick meeting with all technicians twice a day at 1:00 am and 1:00 pm, so they need to finish their tours exactly 1:00. Identify the mathematical notation for the number of shifts should be made by the technicians in order to achieve this, write the name of this mathematical value, and find it for two tours, one with X=7 and another with X=11 I need the solution please
Well, it is good and valid proof/derivation, but I think, that using infinite series in that case is like shooting at fly with a bazooka. Why not to use limit definition of e^x and geometric meaning of complex number operations?
Doesn't the geometric interpretation come from the fact that you can represent e^iz as cosz+isinz? I mean without it we could represent a complex number as a point in coordinate system, but we wouldn't know where e^iPi is.
@@DawidEstishort no, geometric interpretation can be derived from trigonometry. Any complex number is defined by real and imaginary parts or x and y coordinates, like vectors. But you can go to polar coordinates r and t. Then real part x = r*cos t , imaginary part y = r*sin t. So we have z = x + iy = r(cos t + i sin t). If you try to multiply two complex numbers with coordinates x1 y1 and x2 y2 (or r1 t1 and r2 t2) in cartesian you will get a mess, but in polar, after some trigonometric exercise, you will get a new number with r = r1*r2 and t= t1 + t2. (Lengths are multiplied and angles are added). With this as a starting point, you can use the limit definition. e^it=lim (1+it/n)^n So you need to multiply the number 1+it/n by itself n times and tend n to inf. Now recall, lengths are multiplied so the resulting length is lim sqrt(1+t^2/n^2)^n = 1, and angles are added so lim (t/n)*n =t. And so e^it=1*(cos t + i sin t). Hope, I've convinced you.
That makes sense. I didn't notice limits work nicely with lengths of vectors here. Although I think in this case series approach is more elagant. Using this approach he just had to use the definitions of functions as a series and everything else was shown in the video. I think if he wanted to use lim approach in the video he would need to spend too much time with trigonometric calculations to show where everything comes from (like showing why multiplicating complex numbers in polar coordinates works how it works). Cheers
@@DawidEstishort indeed, moreover, this video is targeted to calc 2 students. So, yes, I guess series approach is better when you study them. I just really like proofs with geometric origin, even if they are longer or not intuitive. It's like some kind of art or sport for me.
I’m a 10th grade student in Taiwan, and I chose pre-calculus as my elective course (it’s really PRE-calculus, we won’t even mention Epsilon-Delta definition of a limit this semester!) Though I can’t fully understand most of your calculus videos, I still find them interesting! And I wish to have a teacher like you ：）
There was a problem once that I found but I could not understand the answer and why it works. The problem sounded like this. Given a board of M rows an N columns find the number of ways to tile an MxN rectangle with 1x2 (they can be rotated) dominoes. (something related to aztec treasure problem but on a rectangular board) and the formula that I found on a forum is like (I tested it with a program and it does indeed give the right answer for several M and N test cases). 2^(M*N/2) * ( cos^2(m*pi/(M+1)) + cos^2(n*pi/(N+1)) ) ^ (1/4) over all m,n (integers) in the range 0
The rabbit hole goes deeper. Euler's identity contains the five fundamental mathematical constants, three fundamental operations (addition, multiplication, exponentiation) and equality. Of course, this invariably brings up the tau debate and the fact that e^iτ=1.
I don't think most calc 2 students can show that the derivative of e^z is e^z given that the understanding of the derivative that we learn kinda relies on us using the reals (I think)
@@General12th you have to be very careful how you do that. In many cases yes, but you need complex analysis to actually show why it’s reasonable and the answers we get. You cannot just assume that things play nice like that, you first need to prove it before you use it.
@Max V you’re right, the definitions taught up to calc 2 and 3 are dependent on using real number inputs. You need complex analysis to extend the definitions into the complex world before you can start taking derivatives of functions with complex inputs.
What is this man? @blackpenredpen Should we consider pie 3.14159.... or 180 degree angle? Then e to the power i0 will be equal to 1. And put something else if you want a different result of your choice.
I have a genuine question for you, if e^(iθ)=cos(θ)+i*sin(θ) then wouldn't that mean that if you plug in 2π you would get +1, meaning that i×2π=ln(1) so therefore i=0?
Is this the only way to prove as we can use the analytical way of splitting the graph of e^ix into the real and imaginary part and obtain cos(x) as the real part and sin(x) as the imaginary part and then obtain this identity
No. I know of a proof via calculus. If you consider a function f(t)=e^(it)/(cos(t)+i sin(t)). f is a nice function, since cos(t) and sin(t) can not be both zero, for any t, so our function is devoid of poles and discontinouties. Taking a derivative f'(t)=(i e^(it)(cos(t)+i sin(t))-e^(it)(-sin(t)+i cos(t)))/(cos(t)+i sin(t))^2 ; f'(t)=(i e^(it)cos(t)-e^(it)sin(t))+e^(it)sin(t)-i e^(it)cos(t))/(cos(t)+i sin(t))^2 f'(t)=(i ( e^(it)cos(t)-e^(it)cos(t)) - (e^(it)sin(t)-e^(it)sin(t)))/(cos(t)+i sin(t))^2 f'(t)=(i ( 0 ) - ( 0 ))/(cos(t)+i sin(t))^2 f'(t)=(0)/(cos(t)+i sin(t))^2=0 This implies f is a constant for all t (f(t)≡A, A is some constant). But if f is constant for every t, we can evaluate it at any t, to see what A is. Since f(0)= e^(0 i)/(cos(0)+i sin(0)) and i 0=0 i=0,e^(0)=1, cos(0)=1, sin(0)=0 so f'(0)=1/(1+i 0)=1/1=1. So f is constantly 1, so e^(it)/(cos(t)+i sin(t))=1, or e^(it)=cos(t)+i sin(t). ▧
Is there any way possible to solve the equation x^^x=4? The answer is 2 obviously, but can we confirm thats the only one? Talking about tetrarion, not just regular exponentiation
Would it be a proper derivation to say the space in the circle with r=1 equals pi, so integral from 0-1 of sqrt(1-x^2) = pi/4. Then complex substitution gives pi/4 = i/2*ln(-i) which can be transformed into the euler formula e^ipi = -1
Yes, the recursive form of Euler's Identity should be considered the most beautiful and important of lone equations, but the solution to the Basel problem and its reciprocal to define the moments in time where black holes get crushed into dark matter as part of a Big Bounce event, well...
That's a nice validation. but how did we discovered that? Did someone just randomly tried to taylor expand e^ix? Probably not. How to invent math, that is the question.
My friends, next time your significant other asks how beautiful you think they are, just tell them they’re as beautiful as Euler’s identity. And then link this video to explain to them what it is
I think what's totally mind boggling about this is that e and pi are supposedly two completely unrelated numbers *found in nature*. e occurs naturally in anything that has to do with exponential growth, and pi occurs naturally in all circles. We did not invent or define those numbers. We observed them. And yet, we've proven that those two numbers are intrinsically linked through the imaginary world.
Hello Sir , Facing a lottt of difficulty in solving Definite Integration of Greatest Integer Function questions , Plz If possible then , Make a separate video for it 🔥🔥🔥🔥🔥🔥🔥🔥🔥
We know that e^(iø) = cos ø + i. sin ø and when we substitue ø = 2π, we get e^(i.2π) = 1 Say, take "ln" to both side, we get ln e^(i. 2π) = ln 1 = 0 i. 2π = 0 How could it be? Can you explain this case, why i. 2π can be 0?
Taking the natural logarithm (\(\ln\)) of both sides in \(e^{i2\pi} = 1\) yields \(i2\pi = 0\), which is not true. The issue here is that the natural logarithm is a multivalued function for complex numbers, and in this case, it leads to an incorrect result. Thats what chatgpt says
It also can be solved as following 🙃😅 e^i@ =cos@+isin@ (euler's identity) So, @=π (here) e^iπ= cosπ+isinπ , (sinπ=0 & cosπ=-1) Hence, e^iπ= -1 Now, e^iπ+1= -1+1=0 hence proved.
I want to purchase brilliant premium but i don't have even any card to pay and also i don't have money.You might be surprised but it is normal in my country like Bangladesh.But i hope ,when i would earn, i will purchase brilliant premium version
Notice 2^3 < 3^2 but 3^4 > 4^3. Since both functions are continuous between x=2 and x=3 they must intersect in that range. Therefore, there is (at least) one solution.
Am I wrong but wasn't there something that needed to be considered when infinate sum's are rearanged like that? Just curious why it is possible in this case while you can also get pure nonsense out of it othertimes.
The faceted dimensional transition operator annihilates itself during a transcendental time stretch to remain within the singularity, and create the count, and 1st ordinal..
The count exists before the 1st ordinal, but has no presence until being recognized by the 1st ordinal, which then: a. bootstraps the count, b. then defines the 5 aspects of the presence of the singularity, c. conducts the euler matriculation of presence to exhaust the count.
@blackpenredpen Hello, after watching the video I made a little google search about the topic and people seem to say that the identity is "Euler's proof of God" and they are saying it because the equation is made from all the basic constants from every field of mathematics, could you elborate on that or is it just because the beauty in that math Euler said that in order for this to be true God must be real?
"Everybody knows..." is highly context-dependent. You pretty much have to actively choose to study a bunch of optional math courses in high school to even so much as get introduced to imaginary numbers.
I don't get why Euler's identity is so significant. Like sure, it has five cool features, but there are other irrational numbers. Wouldn't it be more impressive if phi was also part of the equation?
@@robertmcknightmusic If you think about d/dx b^(x) where b is real and b > 1 using the difference equation: lim h->0 [ b^(x + h) - b^(x) ]/h you find this simplifies to: b^(x) * lim h->0 [ b^(h) - 1 ]/h and that: lim h->0 [ b^(h) - 1 ]/h == 1 if and only if b == e This means using the Taylor (Maclaurin) Series to evaluate b^(x) is only tractable when b == e. It can easily be shown that lim n->∞ ( 1 + x/n)^(n) yields the Taylor series for e^(x).
Here's a simpler proof that requires no knowledge of Taylor series. Let complex numbers z and w be on the unit circle. Using trig identities and simple geometry, one can show that Arg(zw) = Arg z + Arg w. This theorem looks surprisingly like the theorem for logarithms, ln(ab) = ln a + ln b. So let's hypothesize that ln z = k * Arg z = k * Theta, for some constant k. (We'll write T = theta for short). Taking e to both sides, we get that z = e^(k T). What could k be? k can clearly not be real, because z is complex and if k is real, then e^(k T) is real, too. So if our hypothesis has any chance of being right, then k will have to be complex. Let k = a + bi. Let's deduce what a and b would have to be. We know that any complex number z on the unit circle can be written as cos T + i sin T by simple trig. So our hypothesis can be written as cos T + i sin T = e^(k T) = e^(a T + bi T) = e^(a T) e^(bi T). Clearly a = 0, because z is on the unit circle and has magnitude 1. Thus the equation reduces to cos T + i sin T = e^(biT). To deduce the value of b, differentiate both sides w.r.t. T. Then the equation becomes -sin T + i cos T = bi e^(biT). But e^(bi T) is just cos T + i sin T by our hypothesis. So we get -sin T + i cos T = bi (cos T + i sin T) = bi cos T - b sin T. Matching real and imaginary parts, we see that b = 1! Thus, k = i satisfyies the hypothesis! Thus, cos T + i sin T = e^(i T).
It's valid for imaginary numbers as well. You are probably simply taught that it is only valid for real numbers, to keep it simple, when first learning it.
![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
Shop an Euler's Identity t-shirt: amzn.to/427Seae
As a teacher, the ability to speed up like that would be amazing.
*3 hours later*
kid in the back who didn"t listen: yooooo how can you have the sqrt of -1 ??? I just checked on my calculator and it says "undefined"
worse even:
Everyone in the class: fuck he's right, we think you made a mistake sir
@@themibo899 lol
ruclips.net/video/3XjcbRYYyyQ/видео.html
proof of polynomial (x+a)^n
I had a physics professor who would just cartoonishly wave his arms and say "as any fool can see!" and then write down the last statement of the derivation. It was his way of not-so-subtly hinting that the physics is more important than the n-th degree of the math- he would show us resources that contained the full derivations later. That might not work in a math class, though.
You forgot to mention that Euler’s identity also employs the four principal mathematical operators: addition, multiplication, exponentiation, and equality. ;)
Oh yea! Thanks.
Whenever he says 'Ladies and gentlemen' I know magic is about to happen.
😇
more like maθ
@@plislegalineu3005 weird way to spell mathgic
i have been looking for this equation’s derivation for ages, thank you for this! this really is the most beautiful equation ever.
😃
If you're ever curious about where those series come from (or you don't feel like remembering them), just derive them yourself! Suppose cos(x) = a + bx + cx^2 + dx^4 + ... . Set x = 0, solve for a, then differentiate both sides with respect to x and solve for b. Differentiate yet again and keep going! Within four or five iterations, you should see a pattern emerge.
You can do the same thing for any function, like sin(x) or e^x or ln(1+x). It's much easier than having to remember them.
Yeah and it’s the Taylor series and I fucking hate it so bad it’s the bane of my existence lol
@@catnip202xch. Deriving a Taylor series shouldn't be that hard. Do you know how to differentiate most functions? Can you set x = 0 and solve for coefficients one after another?
Taylor series just assumes every function can be expressed as a polynomial function of infinite order.
Let f(x)= a0 + a1x + a2x²+.......+anxⁿ+....upto infinity
f(0)=a0
f'(0)=a1
f"(0)=2a2
....... And so on fⁿ(0)=n!an
Then just multiply individual an values to xⁿ and thats the mauclaurin series. I should mention this is a specification of Taylor series. The original one's a bit more complicated to derive:
f(a+h)=f(a) + hf'(a)+.....
Remembering things is hard. Try to comprehend the underlying concept and you wont need memorization.
Brilliant 👍
Bold of you to believe I can do this for any function :3
Amazing proof, thank you so much. I was struggling to wrap my head around Eulers identity for months.
I agree that this is the most beautiful equation in math. I learned it in my fifth and final year at the university, taking a course i didnt even need. Needless to say, I am quite grateful I took that course.
The second you wrote down "1-(1/2!)t^2" i immediately knew what was up, and it's beautiful.
like no joke i actually shouted "no way!" out loud the second i saw it
Not only does it have the 5 most important numbers, it also has the 4 most important operations: equality, addition, multiplication and exponentiation.
I am watching your 100 infinite series little by little and the difference in hair style is so nuts compared to now. Also the mic is now a pokeball amazing
The beauty of math in one episode
Are you proud to see mathematics making your alphabet famous?
Amazing !!! I don’t know that development of e^z using Taylor Series.
Thank you !!!!
A technician takes X hours to visit the stores in a couple of streets in one shift, and when he finishes his tour, another technician revisits the same stores again, needing other X hours to finish the second shift, and so on. For example, if the first technician started his shift at 1:00pm, he finishes it at 7:00pm and the second technician starts his shift at 7:00pm and finishes it at 1:00am, and so on. The supervisors used to make a quick meeting with all technicians twice a day at 1:00 am and 1:00 pm, so they need to finish their tours exactly 1:00. Identify the mathematical notation for the number of shifts should be made by the technicians in order to achieve this, write the name of this mathematical value, and find it for two tours, one with X=7 and another with X=11 I need the solution please
Excellent presentation of the topics in a beautiful manner by an experienced teacher . Vow !
Well, it is good and valid proof/derivation, but I think, that using infinite series in that case is like shooting at fly with a bazooka. Why not to use limit definition of e^x and geometric meaning of complex number operations?
It would be too boring
Doesn't the geometric interpretation come from the fact that you can represent e^iz as cosz+isinz? I mean without it we could represent a complex number as a point in coordinate system, but we wouldn't know where e^iPi is.
@@DawidEstishort no, geometric interpretation can be derived from trigonometry. Any complex number is defined by real and imaginary parts or x and y coordinates, like vectors. But you can go to polar coordinates r and t. Then real part x = r*cos t , imaginary part y = r*sin t. So we have z = x + iy = r(cos t + i sin t). If you try to multiply two complex numbers with coordinates x1 y1 and x2 y2 (or r1 t1 and r2 t2) in cartesian you will get a mess, but in polar, after some trigonometric exercise, you will get a new number with r = r1*r2 and t= t1 + t2. (Lengths are multiplied and angles are added). With this as a starting point, you can use the limit definition. e^it=lim (1+it/n)^n So you need to multiply the number 1+it/n by itself n times and tend n to inf. Now recall, lengths are multiplied so the resulting length is lim sqrt(1+t^2/n^2)^n = 1, and angles are added so lim (t/n)*n =t. And so e^it=1*(cos t + i sin t). Hope, I've convinced you.
That makes sense. I didn't notice limits work nicely with lengths of vectors here. Although I think in this case series approach is more elagant. Using this approach he just had to use the definitions of functions as a series and everything else was shown in the video. I think if he wanted to use lim approach in the video he would need to spend too much time with trigonometric calculations to show where everything comes from (like showing why multiplicating complex numbers in polar coordinates works how it works).
Cheers
@@DawidEstishort indeed, moreover, this video is targeted to calc 2 students. So, yes, I guess series approach is better when you study them. I just really like proofs with geometric origin, even if they are longer or not intuitive. It's like some kind of art or sport for me.
What I like best is that he is one of the few Americans to correctly pronounce Euler’s name is
I’m a 10th grade student in Taiwan, and I chose pre-calculus as my elective course (it’s really PRE-calculus, we won’t even mention Epsilon-Delta definition of a limit this semester!)
Though I can’t fully understand most of your calculus videos, I still find them interesting! And I wish to have a teacher like you ：）
Make sure you know unit circle very well haha 加油同学
This equation id really beautiful, like no words to describe this masterpiece
new inspiration on the formula,thank u
THIS EQUATION IS SO FUN!!!
It's not. It's wrong.
There was a problem once that I found but I could not understand the answer and why it works. The problem sounded like this. Given a board of M rows an N columns find the number of ways to tile an MxN rectangle with 1x2 (they can be rotated) dominoes. (something related to aztec treasure problem but on a rectangular board) and the formula that I found on a forum is like (I tested it with a program and it does indeed give the right answer for several M and N test cases).
2^(M*N/2) * ( cos^2(m*pi/(M+1)) + cos^2(n*pi/(N+1)) ) ^ (1/4)
over all m,n (integers) in the range 0
I really admire those people who really understand maths and its complexities. Wisdom is really a gift ❤️❤️
You might be confusing knowledge with wisdom.
@@eleSDSU No they are definitely wise.
Do you have a cal2 review? I need it your the best!
I got you. 100 calc 2 problems. ruclips.net/video/Kwyk_mteyNc/видео.html
@@blackpenredpen Thank you! 😊
Wow, it's really cool how easily I can show people that I'm done with the proof, thanks
Everybody knows they are watching RUclips,
But I know I am watching a Legend. 😁
If you use tau (equal to 2*pi) instead of pi then you get e^(i*tau) = 1, which to be honest looks even better!
It does not
@@PwerGuido I think you'll find that it does.
The rabbit hole goes deeper. Euler's identity contains the five fundamental mathematical constants, three fundamental operations (addition, multiplication, exponentiation) and equality. Of course, this invariably brings up the tau debate and the fact that e^iτ=1.
Does that “pi really equal 3,1415… or equal 180 degrees? What is the value of pi?
For example:
2 apples multiply 2 apricots = ?
OMG the sudden close up of your face for the sponsor message was so creepy lol 😅😂
आशीर्वाद from India ✋🏻
7:22 Ladies and gentlemen: Euler's Identity!
Very very nice presention. I highly appreciated you sir.
2:07 that pause was funny, sorry)
Lol
I don't think most calc 2 students can show that the derivative of e^z is e^z given that the understanding of the derivative that we learn kinda relies on us using the reals (I think)
Doesn't the limit definition of the derivative apply in the complex world too? I don't see why not.
@@General12th you have to be very careful how you do that. In many cases yes, but you need complex analysis to actually show why it’s reasonable and the answers we get. You cannot just assume that things play nice like that, you first need to prove it before you use it.
@Max V you’re right, the definitions taught up to calc 2 and 3 are dependent on using real number inputs. You need complex analysis to extend the definitions into the complex world before you can start taking derivatives of functions with complex inputs.
Then, e^ipie must be =-1 right.
Thank you ❤❤❤
Esto prueba que e^x es una función sin ceros en todo C.
What is this man? @blackpenredpen Should we consider pie 3.14159.... or 180 degree angle? Then e to the power i0 will be equal to 1. And put something else if you want a different result of your choice.
See how we can integration e^(ax)*sin(bx) with Euler's formula and no integration by parts:
👉 ruclips.net/video/E_ecLWj_rcM/видео.html
I have a genuine question for you, if e^(iθ)=cos(θ)+i*sin(θ) then wouldn't that mean that if you plug in 2π you would get +1, meaning that i×2π=ln(1) so therefore i=0?
O km momok
Can you calculate integral of exp(x) over t^2 +1
another one I liked, in latex:
f(x) = e^{-ix}(\cos x + i \sin x)
\\
f^{\prime}(x) = e^{-i x}(i \cos x - \sin x) - i e^{-i x}(\cos x + i \sin x)
\\
f^{\prime}(x) = e^{-i x}(i \cos x - \sin x) - e^{-i x}(i \cos x + i^2 \sin x) \equiv 0
\\
f^{\prime}(x) = 0 \;\;\; \forall \; x \in \mathbb{R}\Rightarrow f(x) \text{ is a constant}
\\
f(0) = e^{0}(\cos 0 + i \sin 0) = 1 \cdot(1+0) = 1 \Rightarrow f(x) = 1 \;\;\; \forall \; x \in \mathbb{R}
\\ \\
1 = e^{-ix}(\cos x + i \sin x) \Rightarrow e^{ix}=\cos x + i \sin x \;\;\; \forall \; x \in \mathbb{R}
@@rosaorwinter910 No, because if you were to divide both sides by i aka multiply by -i you get 2π=0 and understand that ln(x) has infinite solutions
Why radians and not degrees
Is this the only way to prove as we can use the analytical way of splitting the graph of e^ix into the real and imaginary part and obtain cos(x) as the real part and sin(x) as the imaginary part and then obtain this identity
No. I know of a proof via calculus. If you consider a function f(t)=e^(it)/(cos(t)+i sin(t)). f is a nice function, since cos(t) and sin(t) can not be both zero, for any t, so our function is devoid of poles and discontinouties. Taking a derivative
f'(t)=(i e^(it)(cos(t)+i sin(t))-e^(it)(-sin(t)+i cos(t)))/(cos(t)+i sin(t))^2 ;
f'(t)=(i e^(it)cos(t)-e^(it)sin(t))+e^(it)sin(t)-i e^(it)cos(t))/(cos(t)+i sin(t))^2
f'(t)=(i ( e^(it)cos(t)-e^(it)cos(t)) - (e^(it)sin(t)-e^(it)sin(t)))/(cos(t)+i sin(t))^2
f'(t)=(i ( 0 ) - ( 0 ))/(cos(t)+i sin(t))^2
f'(t)=(0)/(cos(t)+i sin(t))^2=0
This implies f is a constant for all t (f(t)≡A, A is some constant). But if f is constant for every t, we can evaluate it at any t, to see what A is. Since f(0)= e^(0 i)/(cos(0)+i sin(0)) and
i 0=0 i=0,e^(0)=1, cos(0)=1, sin(0)=0 so f'(0)=1/(1+i 0)=1/1=1. So f is constantly 1, so
e^(it)/(cos(t)+i sin(t))=1, or e^(it)=cos(t)+i sin(t). ▧
E^(i*pi) + 1 = 0 literally reminded me of quadratic
It's beautiful ❤️
is e^z just taylor series proof for a=0 case? or if we proof a = 0 is true then it’s true for every a (in the case a= pi)?
Is there any way possible to solve the equation x^^x=4? The answer is 2 obviously, but can we confirm thats the only one? Talking about tetrarion, not just regular exponentiation
...and when you sketch e^(i pi) on the complex number plane with the unit circle drawn, e^(i pi) + 1 = 0 is just so obvious!
buuut you have to prove that r(cos x + isin x) = e^ix which comes from taylor series.
Would it be a proper derivation to say the space in the circle with r=1 equals pi, so integral from 0-1 of sqrt(1-x^2) = pi/4. Then complex substitution gives pi/4 = i/2*ln(-i) which can be transformed into the euler formula e^ipi = -1
would there be a time when u do more induction
Yes, the recursive form of Euler's Identity should be considered the most beautiful and important of lone equations, but the solution to the Basel problem and its reciprocal to define the moments in time where black holes get crushed into dark matter as part of a Big Bounce event, well...
That's a nice validation. but how did we discovered that? Did someone just randomly tried to taylor expand e^ix? Probably not. How to invent math, that is the question.
isn't this maclern's series
My friends, next time your significant other asks how beautiful you think they are, just tell them they’re as beautiful as Euler’s identity. And then link this video to explain to them what it is
BPRP, I have been following u for o long time, can you describe Catalan’s constant that my teacher ask me to study it.😄
I think what's totally mind boggling about this is that e and pi are supposedly two completely unrelated numbers *found in nature*. e occurs naturally in anything that has to do with exponential growth, and pi occurs naturally in all circles. We did not invent or define those numbers. We observed them. And yet, we've proven that those two numbers are intrinsically linked through the imaginary world.
Sir make a group in which you solve our query
thanks so much
I'm going for a math minor in college and I can say I'm definetly using this video to impress my family and friends
Hello Sir ,
Facing a lottt of difficulty in solving
Definite Integration of Greatest Integer Function questions ,
Plz If possible then ,
Make a separate video for it
🔥🔥🔥🔥🔥🔥🔥🔥🔥
Why is this vid unlisted?
how did you find this video 2 days early?
lol
@@officiallyaninja it was linked on his home youtube page before it was posted lmao
We know that e^(iø) = cos ø + i. sin ø
and when we substitue ø = 2π, we get
e^(i.2π) = 1
Say, take "ln" to both side, we get
ln e^(i. 2π) = ln 1 = 0
i. 2π = 0
How could it be? Can you explain this case, why i. 2π can be 0?
Taking the natural logarithm (\(\ln\)) of both sides in \(e^{i2\pi} = 1\) yields \(i2\pi = 0\), which is not true. The issue here is that the natural logarithm is a multivalued function for complex numbers, and in this case, it leads to an incorrect result.
Thats what chatgpt says
It also can be solved as following 🙃😅
e^i@ =cos@+isin@
(euler's identity)
So, @=π (here)
e^iπ= cosπ+isinπ , (sinπ=0 & cosπ=-1)
Hence, e^iπ= -1
Now, e^iπ+1= -1+1=0 hence proved.
thats only for one value and thats in fact not a proof, by that logic x^2 = 2x if we only consider x=0 and x=2
Sir can u visit India once ? There are so many here to appreciate your work especially me.....
e^i*tau=i^4
* *runs away* *
Hey Steve I love your videos, I would like you do Reimann hypothesis. What's it and what is the problem there and what's so difficult to solve it 😂
I want to purchase brilliant premium but i don't have even any card to pay and also i don't have money.You might be surprised but it is normal in my country like Bangladesh.But i hope ,when i would earn, i will purchase brilliant premium version
I know this might be offtopic...but is there any solution for x^(x+1)=(x+1)^x ?
Notice 2^3 < 3^2 but 3^4 > 4^3. Since both functions are continuous between x=2 and x=3 they must intersect in that range. Therefore, there is (at least) one solution.
hi ive been wondering at what value A will A^x=logA(x) have a single solution
maybe make a video out of it? thanks
Is this log base A of x or log of Ax? Log base 10 or e if that’s the case?
I will get a tattoo of that in the future!
this man planning to become an assailant sage or something?DAM!!!
I have a music video of this equation too
“Everybody knows that i^2=1”. That moment when nobody in my grade knows what I is...
E^(i•tau)
what degree is θ
i = the square root of e to the i times pi power. This describes stable particles. Teach correctly, sir.
Am I wrong but wasn't there something that needed to be considered when infinate sum's are rearanged like that? Just curious why it is possible in this case while you can also get pure nonsense out of it othertimes.
not just irrational... _transcendental._
kayra kibar chill dude
edit: he deleted most of the comments
Ladies and gentlemen, we got em.
6:28
Yes senpai
Ruined because you didn't use tau :(
The faceted dimensional transition operator annihilates itself during a transcendental time stretch to remain within the singularity, and create the count, and 1st ordinal..
The count exists before the 1st ordinal, but has no presence until being recognized by the 1st ordinal, which then:
a. bootstraps the count,
b. then defines the 5 aspects of the presence of the singularity,
c. conducts the euler matriculation of presence to exhaust the count.
What are you talking about???
@@aymanad3437 the equation has depth antecedents b4 face value appreciation, and is not an identity.
@blackpenredpen
Hello, after watching the video I made a little google search about the topic and people seem to say that the identity is "Euler's proof of God" and they are saying it because the equation is made from all the basic constants from every field of mathematics, could you elborate on that or is it just because the beauty in that math Euler said that in order for this to be true God must be real?
Here is a queston for you
cos2x/1×3 + cos4x/3×5 + cos6x/5×7......infinity
I’m an algebra I student trying to understand as best I can... 😅
Not gonna lie, relating it to geometry is much easier. Just imagining it in the context of the complex plane makes it hilariously simple.
"Everybody knows..." is highly context-dependent.
You pretty much have to actively choose to study a bunch of optional math courses in high school to even so much as get introduced to imaginary numbers.
hold on what i learned that like 9th grade math
2:12 he paused I laughed
I don't get why Euler's identity is so significant. Like sure, it has five cool features, but there are other irrational numbers. Wouldn't it be more impressive if phi was also part of the equation?
The Euler identity is fundamental to the Fourier transformation that makes the Internet as we know it actually work.
@@ronaldjensen2948 ok interesting. Does it have to be E or could we have also made the internet using another irrational number?
@@robertmcknightmusic If you think about d/dx b^(x) where b is real and b > 1 using the difference equation:
lim h->0 [ b^(x + h) - b^(x) ]/h
you find this simplifies to:
b^(x) * lim h->0 [ b^(h) - 1 ]/h
and that:
lim h->0 [ b^(h) - 1 ]/h == 1 if and only if b == e
This means using the Taylor (Maclaurin) Series to evaluate b^(x) is only tractable when b == e. It can easily be shown that lim n->∞ ( 1 + x/n)^(n) yields the Taylor series for e^(x).
@@ronaldjensen2948 fair enough. that's over my head, but I believe you. Thanks for your response.
You may move -1 to the left side, I do not :p
1:10 is it just me who got jump scare?
Here's a simpler proof that requires no knowledge of Taylor series.
Let complex numbers z and w be on the unit circle. Using trig identities and simple geometry, one can show that Arg(zw) = Arg z + Arg w. This theorem looks surprisingly like the theorem for logarithms, ln(ab) = ln a + ln b. So let's hypothesize that ln z = k * Arg z = k * Theta, for some constant k. (We'll write T = theta for short). Taking e to both sides, we get that z = e^(k T). What could k be? k can clearly not be real, because z is complex and if k is real, then e^(k T) is real, too. So if our hypothesis has any chance of being right, then k will have to be complex. Let k = a + bi. Let's deduce what a and b would have to be. We know that any complex number z on the unit circle can be written as cos T + i sin T by simple trig. So our hypothesis can be written as cos T + i sin T = e^(k T) = e^(a T + bi T) = e^(a T) e^(bi T). Clearly a = 0, because z is on the unit circle and has magnitude 1. Thus the equation reduces to cos T + i sin T = e^(biT).
To deduce the value of b, differentiate both sides w.r.t. T. Then the equation becomes -sin T + i cos T = bi e^(biT). But e^(bi T) is just cos T + i sin T by our hypothesis. So we get -sin T + i cos T = bi (cos T + i sin T) = bi cos T - b sin T. Matching real and imaginary parts, we see that b = 1! Thus, k = i satisfyies the hypothesis! Thus, cos T + i sin T = e^(i T).
cos(θ)+isin(θ)=-1!!!
Did anybody else grab a pen and prove Euler's identity before watching the video ?
How does Taylor series apply for imaginary numbers when it is valid only in real plane but the result is true
It's valid for imaginary numbers as well. You are probably simply taught that it is only valid for real numbers, to keep it simple, when first learning it.
Now, e^2iπ=1 and e^0=1.
So doesn't that mean 2iπ=0 but here, all are constants.
No, you're going to need grouping symbols around that exponent for that exponent of e,
because of the Order of Operations.
Not at all, e^(i x) is no bijection
After 8 time of calculus 1 this is drug for me
cos(x)+!sin(x)=-1
e^-1=1/e !!!
why e number to x power changed to curve why e^-πi
+1=1
Love you sir
This one has captions let's goooo