GoldyOrNugget
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Like I know what the answer should be, but I did it a different way and got something different. Trying to work out why that is.Consider the quadratic:
If one of the roots of the quadratic is where is not real.
Find the range of values for
I got my answer which I found interesting but I'm not sure if it's 100% correct. I want to see if anyone gets the same as me.
Consider the quadratic:
If one of the roots of the quadratic is where is not real.
Find the range of values for
I got my answer which I found interesting but I'm not sure if it's 100% correct. I want to see if anyone gets the same as me.
Great question there, here is my attempt:
Really just some random results, might have done a few parts before.
That was quick lol, very nice.Great question there, here is my attempt:
iii)
Two letters k, and j, can only be arranged in a set of n letters, either that k is BEFORE j, or k is AFTER j.
The number of arrangements of k after j and k before j are equal.
Hence the number of arrangements are total/2
iv)
We must arrange n letters into n slots.
If k and j have no spaces between them, there are n-1 ways they can be arranged ( i.e. _ _ _ k j _ _ .. _ _ )
If k and j hve 1 space between them, there are n-2 ways to arrange them (i.e. _ _ k _ j _ _ .. _ _ )
and so on.
So if we consider the number of ways we can arrange k and j, such that k is before j, when they are both together, 1 space apart, 2 spaces ... n-2 spaces apart the number of arrangements is 1 + 2 + 3 + ... + n-1 = T_(n-1). Arrange other letters in (n-2)! ways
Hence total arrangemetns is: T_(n-1) (n-2)!
v)
They both are the number of arrangements of same scenario hence they are equal:
vi)
vii) Using the result from part ii
viii) From part v
From part ii
Hence when we sub it in, we get recurrence.
iii) Let the two letters be a and b. In exactly half the permutations, a will come before b. There are n! total permutations, therefore a can come before b in n!/2 ways. (this seems very obvious to me, so I'm not sure how much more mathematically it must be phrased. If you imagine that a comes before b in more than half the permutations, then since a and b can refer to any pair of letters, we can swap their assignments so that a now refers to the b letter and vice versa, but now you'd have a coming before b in less than half the permutations, which is a contradiction. Similarly if you start off by assuming that a comes before b in less than half the permutations.)
Really just some random results, might have done a few parts before.
Here's what I did which ended up wrong (I think?)Oh right you said not real, yeah, then just take the complement.
Bolded what was wrong, when you add the two equations, you get:Here's what I did which ended up wrong (I think?)
Since one root is , the other must be
So we get:
Using:
and adding the two equation together we obtain:
2cos2\theta + 2Acos\theta + 1 = 0
Using the identity:
We get a quadratic in which when solved using the quadratic formula gives:
or
Now the first solution works for all , but the second solution doesn't as for certain values of A within the domain of
Hence solving for A:
Which is a different result to
(note I forgot to add the fact that )
What's wrong with this solution?
Woops that was a typo, hit 1 instead of 2. In my working I did have +2, and it worked out to be what I had.Bolded what was wrong, when you add the two equations, you get:
when you use that formula, you get:
Intuitive method though
Why isn't it true? Am I just being a massive retard or...
? Assuming theta is meant to be the argument of z, this is not true.
I'm not sure why its untrue either.Why isn't it true? Am I just being a massive retard or...
I stated that they were lol (well it was a late edit lol).How do we know the roots are of unit modulus?