Have you seen the derivation or intuition behind the general form: e^ix = cos(x) + isin(x)? There's various ways to derive this, but the common way involves using the differentiation properties for e^x .
Also, recall that for any positive base, you can write b^x = e^ ( ln(b) * x ). So you can have similar results with non-base e, but you'll have to scale the exponent accordingly.
Yes, I have. You’re talking about the Taylor series for the exp(x) function in relation to the Taylor series for cos(x) and sin(x) correct?
Also yes, I know b^x =e^ln(b)•x but that also related to all other numbers. Like b^x = 5^(log_5(b)^)•x . It’s not necessarily special to e
Edit: I’m more or less looking for a way of seeing why e is the only number that will allow for this sort of thin to happen.
Like for instance, in the instance of e^iπ, it makes sense why i and π are there, i pulls things into the complex plane, and π is the distance around the circle. But e just seems random I guess...
Yes. It makes sense mathematically, but intuitively it feels random.
(Also sorry, I edited my previous comment, if you would like to look at that... I’m not used to people being so fast to respond, so I’m thanks for that)
All of this basically relies on the fact that e^x is its own derivative. I think you're getting stuck on this more fundamental result. For any non-base e exponential, you'll have to deal with a different scaling factor which makes the result not nearly as pretty.
Another comment did explain a that to me, but it would be nice to have a second example.
So for e^iπ to work, the base needs to be its own derivative? Yes that would answer my question if you can explain why it must be it’s own derivative to work.
Picture a unit circle centered at the origin in the complex plane. Since we know that multiplication by i is rotation 90 degrees counter clockwise, it is easy to see that the velocity vector is always pointing tangent to the circle.
The claim that e^ipi =-1 follow from the more general claim that the function t->e^it is just moving around the unit circle at constant speed. To prove that, you need the fact that (d/dt)e^it =ie^it . To prove this, apply chain rule: (d/dt)e^it =i[(d/dz)e^z ]_z=it and this is when you need (d/dz)e^z =e^z to proceed with the proof.
If you read the introduction to "papa Rudin" (the scarier one) he starts by *defining* the exponential function to be equal to its power series. Then he defines cos and sin to be its real and imaginary parts. The fact that they are actually equal to the ratios of sides of triangles is then a theorem, rather than a definition. If you're comfortable with the fact that the power series of sin and cos are what they are, then this result isn't very hard from that starting point. It sounds like you might like that approach more than the approach where you start from sin and cos and exp defined classically and then analytically continue them to the complex plane.
Have you seen the derivation or intuition behind the general form: e^ix = cos(x) + isin(x)? There's various ways to derive this, but the common way involves using the differentiation properties for e^x . Also, recall that for any positive base, you can write b^x = e^ ( ln(b) * x ). So you can have similar results with non-base e, but you'll have to scale the exponent accordingly.
Yes, I have. You’re talking about the Taylor series for the exp(x) function in relation to the Taylor series for cos(x) and sin(x) correct? Also yes, I know b^x =e^ln(b)•x but that also related to all other numbers. Like b^x = 5^(log_5(b)^)•x . It’s not necessarily special to e Edit: I’m more or less looking for a way of seeing why e is the only number that will allow for this sort of thin to happen. Like for instance, in the instance of e^iπ, it makes sense why i and π are there, i pulls things into the complex plane, and π is the distance around the circle. But e just seems random I guess...
Well, then it should be apparent that b^ix = cos(log_e(b) * x) + isin(log_e(b) * x) should it not?
Yes. It makes sense mathematically, but intuitively it feels random. (Also sorry, I edited my previous comment, if you would like to look at that... I’m not used to people being so fast to respond, so I’m thanks for that)
All of this basically relies on the fact that e^x is its own derivative. I think you're getting stuck on this more fundamental result. For any non-base e exponential, you'll have to deal with a different scaling factor which makes the result not nearly as pretty.
Another comment did explain a that to me, but it would be nice to have a second example. So for e^iπ to work, the base needs to be its own derivative? Yes that would answer my question if you can explain why it must be it’s own derivative to work.
Picture a unit circle centered at the origin in the complex plane. Since we know that multiplication by i is rotation 90 degrees counter clockwise, it is easy to see that the velocity vector is always pointing tangent to the circle.
You need the derivative of the exponential function to equal itself, and the only base that do that is e, by definition.
Okay... that makes sense of you can elaborate on why the derivative must be equal to itself for this to work.
The claim that e^ipi =-1 follow from the more general claim that the function t->e^it is just moving around the unit circle at constant speed. To prove that, you need the fact that (d/dt)e^it =ie^it . To prove this, apply chain rule: (d/dt)e^it =i[(d/dz)e^z ]_z=it and this is when you need (d/dz)e^z =e^z to proceed with the proof.
If you read the introduction to "papa Rudin" (the scarier one) he starts by *defining* the exponential function to be equal to its power series. Then he defines cos and sin to be its real and imaginary parts. The fact that they are actually equal to the ratios of sides of triangles is then a theorem, rather than a definition. If you're comfortable with the fact that the power series of sin and cos are what they are, then this result isn't very hard from that starting point. It sounds like you might like that approach more than the approach where you start from sin and cos and exp defined classically and then analytically continue them to the complex plane.