Question

Is this algebra trickery or math?

Answer 1

This looks like calculus well with the 0 to 1 interval

Answer 2

This isn't algebra sorry.
This is just understanding why a integral evaluates to this answer.

Answer 3

With the complex equation

Answer 4

Hmm. This is because e^(1-j2pi*k) evaluates to e^(1)*e^(-j2pi*k).
This e^(-j2pi*k) evaluates to 1, why?

Answer 5

Well I can say that you are plugging one in for some variable? I don't know which and you subtracting it from which you get 0 your finding the integral and using the evaluation theorem

Answer 6

t is variable there

Answer 7

Yeah your right

Answer 8

Sorry!

Answer 9

You plug in 1 into the t and everything else either gets simplified or stays the same like the jk2pi

Answer 10

The trick is this identity when k is an integer.

\[e^{jk2 \pi} =1\]

Why is this true? Well if you know this identity:

\[e^{jk2 \pi} = \cos(k2\pi) + j \sin(k2\pi)\]

Now you can see we're looking at evaluating some integer multiples of a period, so it's the same as if you didn't move at all from k=0:

\[e^{jk2 \pi} = \cos(k2\pi) + j \sin(k2\pi) = \cos(0) + j \sin (0) = 1\]

Answer 11

That's totally correct, can't believe I always neglect that identity! @Kainui !

Answer 12

Thanks for showing this @Kainui I learned something new!

Answer 13

Haha glad I could help! :P Yeah that identity is pretty awesome, you can use it to derive a ton of trig identities, like the angle sum identities because it turns trig functions into exponents, and exponent rules are much easier to play with I think.