Intuitive Understanding Of Euler's Formula

There may be a mistake in the above derivation, which seems to say that ln(e^ipi/2)
is the same as e^ln(i
pi/2). Is it?

I found a nice derivation of i^i on Yahoo Answers:

i^i=e^ln(i^i)
=e^iln(i)
=e^i
ln(o + i)
=e^iln(cos(pi/2 + i sin(pi/2)
=e^i
ln(e^ipi/2)
=e^i
i*pi/2
=e^-pi/2

Hi,

I love this site! Thank you!

Quick question: if e^i*pi means going pi radians around a circle (in this case 1/2 circle), does that mean I can always think of the coefficient of i as radians? Is it correct to think about it this way?

Thanks again for a great site.

There is another approach for finding i to the i. e to the ipi is -1. The square root of -1 is i. So therefore i is equal to the square root of e to the ipi or in terms of exponents it is equal to e to the ipi to the one half power. From there you simply multiply the exponents by i giving e to the iipi/2 which equals e to -pi/2 .

I do like your using i as a rotator ( precessor). Seldom is it presented as such. In fact thatā€™s exactly what it is- a quaternion. Eulerā€™s identity is simply (actually very profoundly) the compact form of Hamiltonā€™s quaternions. That is: e to the ipi equals ii=jj=kk=ijk=-1

Since he did it for Khan Academy, Bill Gates should give you five million dollars so you can do this all the time.

Another idea: as e^x is the fastest way and ideal way to compound infinetisimal growth( 100%/unit of time), e^ix would be the ideal way to add compounded infinetisimal 90 deg pull which can be interprated as perpendiculary small vectors, I see it as a geometric vector addition:

=(1+i/10)*1+ (1+i/10)*i/10 +oo. hope I m not too wrong with thatā€¦ anyway ur article gave me plenty of brain storming moment and thanks for that! Cheers.

Hi Khalid.

Love what you do, just a remark ( correct me if I m wrong and sorry fro my poor english, native French speaker): I believe you should have wrote:

e^i =lim (n->+00) ( 1+ 100%*i/n)^n instead of e, I see it as the number growing from 1 => e^i at imaginary compounded growth.

obviously for n=1 , we are far from it as e^i is very different from (1+i), same as e^1 is very different from 2.

I believe it could be fun and useful to understand this formula by running it for n=let s say 10:

e^i ~ (1+i/10)*(1+i/10) ā€¦ ntimes

by developpin : (1+i/10)*1 + (1+i/10) *i/10+ etcā€¦

we can feel the 90 deg pull with the multiplication by i/10 ( 90 degree pull of a 1/20 of a full circle). as n tends to infinate and this pull applied quasi-instantly we are litterally and very preciselly running the cirlce circumference

Correct me if I m wrong and wish u the best!

Farid

I am an electrical engineer, 74 years old, and am most impressed by your approach to such mathematical problems. I always felt to live on solid ground, the tools were useful, but now I feel to have walked on an thin ice-layer of a deep lake. And I am most fascinated by what I see under the ice.

@Helmut: Thank you for the kind words and beautiful analogy! Learning is a constant quest to venture deeper :).

I think I should have wrote * (1+i/10) enables you to rotate the equivalent of 1/40 of a circleā€¦ should be 9 deg . and as you said it, if you go for large numbers (1+i* 1/100000) the hypotenus remains equal ~1 at the limit ( 1/100000*1/100000 is second order)ā€¦ and compounded growth rotates us in a very accurate way . Hope I m not too wrong with that insight. exponential and complex are a tricky but magic mix ā€¦ It could be cool to show an application of this in electronics as it s widely usedā€¦

Now let me see if Iā€™ve got this straight. We can look at e^i as 1e^1i. (This little trick of yours was really helpful.) The 1e means that weā€™re starting at 1, which in this case is 0 radians, our original amount, and then growing by e^1i, where 1 is 1 radian and i is a continuous change in direction. All this is equal to cos(1) + i sin(1), a point on a circle in the imaginary plane.

I think Iā€™ve got it now. Thanks for letting me think out loud.

@Tim: Thanks for the support ā€“ I wouldnā€™t sneeze at a few million :).

Happy to let you think it out: thatā€™s pretty much how I see it. e^x is 1 * e^x, which means ā€œLetā€™s grow continuously by xā€. Oh wait! It turns out x = 1 * i [grow continuously and perpendicularly for 1 second]. This will rotate you ā€œ1 unitā€ along the circle. Turning that rotational (polar) coordinate into a linear one means you are at cos(1) + i sin(1).

[ā€¦] we can show that Eulerā€™s formula (e^ix = cos(x) + i*sin(x)) is true because the Taylor series is the same on both sides. Accurate [ā€¦]

@Christian: Whoops, my mistake! I meant to write e^(-i * 5*pi/2) = e^(-i * pi/2) but that leads right back to your point!

You are correct, if you allow the base ā€œiā€ to be defined as any value e can be raised to (ipi/2, i5pi/2, ā€¦) then when you raise it to the ā€œiā€ power you could have e^(-pi/2), e^(-5*pi/2), etc. which clearly have different values.

@Dmitry: Thanks for the clarification! Google calculator, for example, will treat (e^(i * 5 * pi/2))^i the same as (e^(i*pi/2))^i. More interesting things to read up on!

@Alqazzafi: Funny how much we can study but not really learn, right? Anyway, happy if the article helped.

@Yonggook: Thatā€™s a really good question ā€“ check out this page which talks more about the limit http://www.cut-the-knot.org/arithmetic/algebra/Scott.shtml. Right now Iā€™m not sure what it means to have a complex number go to infinity.

Youre a GENIUS!!!

Ok, this is a very interesting and helpful reply, and it has definitely taken me at least one step closer to getting the underlying connection (which, by the way, I have been looking and looking for, and have not found anywhere but here).

I think what you are saying here - correct me if I am wrong - is that the sine curve and the exponential curve are just different manifestations of the same underlying function. If we plug in real numbers, we see an exponential curve because we are constantly pushing ā€œupā€, and if we plug in complex numbers, we see a circle because we are constantly pushing ā€œperpendicularā€. So if we do some sort of transformation from the real domain to the complex domain, an exponential curve maps to a circle, and vice versa.

Now, the next question that arises to me is, what is the deep relationship between the imaginary x and the 90 degree push. In other words, can we ask why it is that the one and only way to get 90 degree motion - which is the only motion that will give you a circle - in the complex plane is to use the same function that gives you continuous growth in the real domain? What FORCES that function to be the e^x function? Why was it impossible that an imaginary x give you a continuous 45 degree push, or an 88.63625 degree push? Why 90?

I think I might see the answer in this concept of continuity that you are using. Is it that, if you push continuously at any non-orthogonal angle, you canā€™t do it with continuity? If you push, say, at less than 90, then there is a component of your push that takes you away from the origin, and not only do you not make a circle, but you go spiraling off to some infinite imaginary destination. If you push at more than 90, then a component of your push is in toward the origin, and you spiral to zero. Therefore, the function that gives you maximum continuity of growth in the real plane HAS to trace a circle in the complex plane because no other ā€œshapeā€ could offer the same maximal continuity.

Did what I just said even make sense?

Hi Mike, I see i as a rotation.

For something like cos(x) + sin(x), note that you need a 2nd dimension otherwise the numbers run together. Try plugging in x=45 degrees: youā€™ll get cos(45) + sin(45) = .7 + .7 = 1.4. Which isnā€™t ā€œ45 degrees around the circleā€ like weā€™d expect. We need a way to track the vertical component separately from the horizontal one (i gives us an option here).

Kalid, I appreciate the response but I still donā€™t see it as an answer to the underlying question. You are just inviting me to rephrase the question, so I will: Why is a process of continuous growth related to a process of continuous circular movement? Itā€™s not enough to just observe that they are both continuous! You might as well say that e is related to c because they are both speed limits of a sort.

So the mystery reamins: why does ā€œthe fastest way to grow at 100% interestā€ just happen to be related neatly to the irrational number whose sign is 0?

mra: No problem, let me see if I can clarify.

e^x represents continuous, never-ending change. Usually, this ā€œchangeā€ is all in the same direction, and accumulates, and we call this exponential growth: 1, 2, 4, 8, 16, 32, etc.

Butā€¦ thatā€™s because the exponent (ā€œxā€) is assumed to be a real number. But if we let x be imaginary, something happens: our ā€œchangeā€ is NOT in the same direction. Each instant, we are changing by 90 degrees to our current direction. This is like swinging a rope over your headā€¦ every instant, the rope is moving in a direction perpendicular to where it is pointing [one of the characteristics of a circle: the radius, where youā€™re pointing, and the tangent, where you are going next, are perpendicular].

So, 100% continuous growth, when your interest is always in a 90-degree direction, will look like a circle. In other words, you can describe a circle as ā€œstart at 1, and always change 90-degrees to where you are. But as soon as you have changed a nanometer: stop, re-evaluate your direction, and grow 90 degrees to your current position.ā€ (Continuous means you are constantly changing, not ā€œmoving for a while and then deciding to changeā€).

Hope this helps!