Euler's formula: A cool proof

numbers and it satisfies i^2=-1. This famous formula has been around for hundreds of
years and it was first proved by Euler in the 1740s using power series. You write cosine as a
power series, sin as a power series, the exponential as a power series and then you equate real
and imaginary parts. In this presentation I am going to do something a little bit different I
am going to use an approach that is related to differential equations. And it is based on the fact
that solutions to certain initial value problems involving ordinary differential equations the
solutions are unique. Euler did the
power series method and there is a Khan academy video about that as well there is also
a limit way to prove it which you will find on Wikipedia and there is also a way that Gil Strang
does it using calculus and assuming that this can be written in the r cis θ form. Well let me show you and one of the
nice things about the method that I am going to show you is that you can prove all sorts of other
identities with the method just by using some basic differential equations. The proof for this result I am just going to
motivate it by a mathematician called Richard Bellman so certainly the proof I am going to
show you would be known but you just do not see it very many places. I am going
to let a function be this e^i*x. I am going to differentiate this function and see what
differential equation f satisfies. So if I differentiate just by treating i as a constant
bring the i to the front now if I differentiate again okay so i would come to the front and we know
that i^2=-1 so I will get the following. Notice that f'' is the negative of f. Now I
have got a differential equation 00:00-04:20.
If I plug in say x=0 I will get e to the 0 which is e^0 which is 1 and if I go up here and
plug in x=0 I will get i*e^0 which is just i. Now we have an initial value problem okay in
particular this is a linear second order problem with constant coefficients and we have some
initial conditions associated with it. Okay this IVP or initial value problem has a unique solution
and that is you would learn that say in a first course in ordinary differential equations. Now
because the coefficients are constant in this problem it is actually easy to solve this problem for
f.
You can look at the characteristic equation the characteristic equation would be
something like r^2+1 =0 and so the roots will be complex and you can write the general
solution to this problem as a linear combination of cosine and sines where the a and b, because
of the initial conditions, could be complex numbers. If you put in x=0 instead x=1, differentiate said
x=0 and equate it with i then you will get these two values for a and b okay. So If I go up to here
and plug in a=1, b=i I will get the following. 04:20-07:53.
We have shown that e^(i*x) satisfies a certain initial value problem and then we
have taken that initial value problem and we solved it. Now because the initial value problem
under consideration is a special initial value problem it has special properties because of the
uniqueness there is one and only one solution to that problem. The two functions that we have
started with or derived here, the f and the g, must be equal. Okay so by uniqueness of solutions to our IVP or our
initial value problem this and this must be the same. So f must be identically equal to g, that
what these sort of 3 horizontal lines mean. That is e^(i*x) must equal cos(x) +i*sin(x). 07:53-09:39.
Alright so there are some advantages and disadvantages of this proof. You need to
know a little bit about differential equations to really appreciate the style of proof that I have
mentioned. One of the positives as far as I can see for the method of proof that I have just
shown you is that it gives you a nice way of applying the idea of uniqueness of solutions to
differential equations and initial value problems. You can use the theory of differential
equations to prove all sorts of nice formulae or identities. Now is it as simple as
say Euler's method using power series well no okay no. But i still think it is really nice to talk
about this proof because I have not seen it anywhere I do not think it is on youtube. I have seen
a first order type approach to prove this identity but I have never seen second order approach
okay so I think I am sort of adding something new there alright. Now if you get a chance see
what other identities you can prove involving trig functions using differential equations. There is a
whole bunch of them you can prove and it provides a nice alternative than say a geometric
approach.

[grantmoore]

Isn't this a bit circular? You solve the differential equation using the characteristic equation, but the characteristic equation is derived by assuming the solution to the differential equation is exponential. When the roots don't exist, the solution is imaginary, which assume's Euler's Identity can provide the solution as the linear combination of sine and cosine.

[UzunKamis]

If you consider the respective solutions with the initial values c(0)=1, c'(0)=0 and s(0)=0, s'(0)=1, this also leads to the following important trigonometric properties:
• c'(x)=-s(x) and s'(x)=c(x)
• c²(x)+s²(x)=1
• c(-x)=c(x) and s(-x)=-s(x)
• c(x±y)=c(x)c(y)∓s(x)s(y) and s(x±y)=c(x)s(y)±s(x)c(y)

[abdelrahmangamalmahdy]

in addition, you can proof that by solving [the second order differential equation with constant coefficients y''=f(y)] using chain rule (y''=y'.dy'/dy), then you can integrate both sides, then you get a solution c1.sin(x+c2), then you can use a trig identities, you get (c1.sinx +c2.cosx ), if you substitute the initial values of the original function exp(ix) you get c1=i and c2=1, and you get eular's formula even if you don't know what is the function identically to exp(ix) ... there is imagination, yea

[MotorFlaps]

I have already seen this very proof in Brad Osgood´s book on Fourier Transforms, page 407 (bottom of the page). Congratulations Dr Tisdell as he agree to see this as the most elegant way of proving Euler´s formula. Thanks for your work

[kstahmer]

Thanks for the video. I’m guessing you’ve seen this previously...

Set f(x) = exp(-ix)(cos(x) + i sin(x)).

Then:

f'(x) = -i exp(-ix)(cos(x) + i sin(x)) + exp(-ix)(-sin(x) + i cos(x))
= exp(-ix)( -i cos(x) + sin(x) – sin(x) + i cos(x))
= exp(-ix)(0) = 0

Hence f(x) is constant. Since f(0) = 1, then for all x, f(x) = 1.

Consequentially:

exp(-ix)(cos(x) + i sin(x)) = 1
exp(ix) = cos(x) + i six(x)

[JustFamilyPlaytime]

Well this sure works, but given that you need also to know how to prove the uniqueness of the solution to this IVP, and that's harder than just going with the Taylor expansion or considering the function exp(-ix)(cos(x) + i sin(x)) and its derivative.

[abdelrahmangamalmahdy]

it is a fantastic proof ... that means this two function identically, and the same at every value for x .... i see this proof now for the first time, there is a good way to proof identities by the unique solutions of the differential equations  

[sinnathambyganeshalingham]

sir, you are assuming here a complex function is differentiable. how?

[abdulhalim]

Thank you, Dr. Could you please prove the addition formula?

[SynapticMachines]

Thanks for posting this! I am new to maths, filling the gaps in my scholar education for fun. So that video really clarified an issue I had with the Euler formula in terms of understanding where it comes from... I saw a bunch of proofs: series, 1st order diff eq and now second order. But all make me uncomfortable... They all operate via algebraic analogy and not a fundamental relation (I would dare say physical...). I would like to witness the proof of some kind of solid relation between exponential and trig functions, for example a geometrical one. Haven't found it yet. Any videos to suggest?

[alawodeadewale]

Sir help me with this question
Find the solution of difference equations
Y(n+4) - 4Y(n+3) +5Y(n+2)-4Y(n+1)+4Y(n)
= 4
With y(0)=5, y(1)=0, y(2)=-4, y(3)=-12

Solve y(n+1) -2(sin x) y(n)+y(n-1)=0
When y(0)=0 and y(1)=cos x
Find y(n) from the difference equations
Δ^2y(n+1) + 1/2 Δ^2y(n)=0 when y(0)=0, y(1)=1/2, y(2)=1/4

Solve Δ^2y(n) + 3 Δy (n) -4y(n)=n^2 with condition y(0)=0, y(2)=2

[TheLastXCloud]

thank you, i always use this formula but i didn't know it belong to euler, at first when i read the title, i thought it's euler equation (fluid dynamics)

[josephavant]

Best explanation I've seen - THANK YOU so very much for sharing!!!

[asparagii]

I found this helpful - thanks Professor

[palvindarchhokar]

What Chris was trying to say at the end was...




"First!"

[billyjean]

liked it, but the camera going in and out is kind of annoying... Thanks!!

[ukidding]

how can i apply this formula in everyday life situations?

[barryhughes]

Dr Chris Tisdell. I m not a mathematician, however there is I think a nicer way of doing it.. Let z=cosx+jsinx, so dz/dx=-sinx+jcosx, = both sides, dx/dz=I/jz and integrating with respect to z,   x=I/j times the integral of [I/z]dz= I/jlogz + c. When x=0, cos x=I and sinx = 0, i.e z=1 therefore c=0. [ log to any base of 1 =0] therefore x=I/logz, so jx=logz and so e^jx=z [ definition of logs ]And thus cosx+jsinx= e^jx.

[ryanpowell]

There's no intuition behind this really, would never be derived like this. It is a simple proof of you have taken diff Eq though.

[rgqwerty]

for goodness sake stop smacking your lips