(-i)^13?
i (the imaginary unit) behaves much like any variable, it just has the added condition that \[\Large \color{green}i^2 = -1\]
So, you could readily say that \[\Large (-\color{green}i)^{13}=(-1)^{13}\cdot \color{green}i^{13}\] as the laws of exponents would dictate :)
So it is -i?
yeah
Yeah, that works :D \[\Large (-1)^{13}=-1\] \[\Large \color{green}i^{13}=\color{green}{i}^{12}\cdot \color{green}i = \color{green}i\]
IS there a trick to remembering these i's?
Sure there is :) If you're raising i to some exponent, you might want to divide that exponent by 4, and get the remainder.... for instance, what if we have \[\Large \color{green}i^{241}\] If you divide 241 by 4, what remains?
60.25
No, I'm asking for the remainder... but you could use that...
If you're going to do it like that, then use this guide... if the quotient ends with a .25 (like this one) then the answer is i If it ends with a .5 (or .50) then it's -1 If it ends with a .75, then it's -i And finally, if you get an integer (no decimal part) then it's just 1.
You rock
:) To avoid confusion, let me just stress that that guide was for i... your question involved a -i so just adjust accordingly (like how I separated the -1^13 first)
@terenzreignz , can you also help here? \[A _{n+1}=2A _{n}+1 \] for n= 0, 1, 2,
hmm... we need a base step... maybe A0 is already given?
I need to find A4
A0 is 1
Okay, good :) It REALLY helps to translate the maths terms into English :D Let's see how to do this... \[\Large A_{n+1}=2A_n+1\]
A_n simply means the nth term, right? So what this is telling you, is that every term is twice the previous term, plus 1. So... A0 = 1 What's A1?
3?
Very good :) So now you know that A1 is 3 So what's A2 ?
10?
no... remember... every term is twice the previous term... plus 1...
Duh. Sorry. 7
okay Much better :D
The answer is 31. Thanks.
So reading in english does help!
31? what...
I need to find A4. So A2 was 7 A3 is 15 A5 is 31
*A4
ahh okay :D Then that was correct :D See? You pretty much did it yourself... the key was just understanding the maths problem, and maybe translating it into plain English :P
Thanks
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