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"surely you can see that (r')^n = r"
No, I can't. Why does (r')^n = r? I would have thought all we can say is that:
and .Why can we assume that when taking the nth root of a number, the modulus of the nth root is equal to the nth root of the original modulus?
Ahh okay, thanks
But my book says
Thanks.
The techniques used to find the nth roots of unity can be extended to find the nth roots of any complex number.
Suppose that
is an nth root of thenThe book then says that
and , but I don't see why this would necessarily be the case?Wouldn't you have to say that
and ?Thanks.
I've just been doing some investigtion and I've found something quite nice.
The number of distinct factors a number has is given by the product of one more than the maximum power of each of the prime factors.
i.e. If
then the number of distinct factors n has is given by .This is relatively easy to prove. We have
. There is a total number of α_1 + 1 numbers that can be made from taking powers of p_1 from 0 to α_1, and similarly a total number of α_2 + 1 numbers that can be made from taking powers of p_2 from 0 to α_2. Each number from the first α_1 + 1 numbers can be paired with each of the α_2 + 1 numbers, giving a total of (α_1 + 1)(α_2 + 1) numbers. These can then be paired with the α_3 + 1 numbers that can be made from powers of p_3, giving a total of (α_1 + 1)(α_2 + 1)(α_3 + 1) numbers. This process can then be continued to give the desired result.For example, if n = 40 = 2^3 x 5^1, the number of distinct factors is (3+1)(1+1) = 8. This can be verified using the principals involved in the proof.
1 1
2 5
2^2
2^3
The possible factors of 40 are therefore: (1*1, 1*5), (2*1, 2*5), (2^2*1, 2^2*5), (2^3*1, 2^3*5), a total of 8.
I'm assuming you mean
In which case, remember that
The area of a surface formed when arc AB is rotated completely about the x-axis is:
It would be easier if you parametrised the equation of the circle and used:
It would be helpful if we could multiply by a number n such that 13n is 1 more than a mutliple of 29. The smallest such number is 9. Mutliplying by 9 gives:
i.e.
Thank you Jane
Rather than me setting exercises for others to do, I'd like people to give me some challenging questions It's a thread where people can share nice problems, but I'd like the benefit of seeing if I can get to the result myself too.
By challenging questions I mean ones which require insight and clever tricks, or ones with nice results, rather than something that requires a lot of knowledge and can be done by appyling heavy theorems.
Content-wise they can contain anything that doesn't assume too much university-level knowledge, although I would like to learn some things along the way so don't hold back.
I'm mainly looking at people like Jane, mathsy and Ricky, but anyone else is of course welcome
Thank you.
Ahh, woops.
Thanks.
For the curve with equation y = arsinh(x+1), find
a) the coordinates of its point of inflexion P
b) an equation of the normal to the curve at P.
But this derivative is never 0???
Thanks.
I think pictures are a good idea. It's nice to see the face behind the words I'll get one up soon.
thanks.
I'm almost there with this one... I just can't make the final leap.
"Prove that if x,y,z are integers such that 9|(x^2+y^2+z^2), then 9|(x^2-y^2) or 9|(y^2-z^2) or 9|(z^2-x^2)."
The quadratic residues (mod 9) are 1, 4 and 7. Therefore the only ways that three squares can be summed to give 0 (mod 9) are (0,0,0); (1,1,7); (1,4,4); and (4,7,7) and any permutations of these triples. In all cases there is a repeat, but does this / why does this mean that 9|(x^2-y^2) or 9|(y^2-z^2) or 9|(z^2-x^2)?
I can sort of see it, but I can't explain it.
Thanks
The probability that you will not select 2 green apples is equal to (1 - the probability that you will select 2 green apples).
The probability that you will select two green apples is 3/8 x 2/7.
EDIT: I'm assuming this is without replacement.
The probability that the first one you pick is white is 8/16. The probability that the second one you pick is white, given that the first one was white, is 7/15. Multiply these two numbers together.
EDIT: I'm assuming this is without replacement.
For part d), I've drawn a graph. It just continues to spiral outwards:
Which gives:
EDIT: where [x] is x rounded down to the nearest integer. Obviously this could be written differently i.e. without (mod 2), but I think it's nicer this way.
You've made a mistake in the third line.
x^2 + (mx+c)^2 = a^2
⇒ x^2 + m^2x^2 + 2mxc + c^2 = a^2
⇒ x^2(1+m^2) + x(2mc) + (c^2 - a^2) = 0
Okay, I get the answer as 3/4.
Consider the unit circle.
Define the argument of A,B,C to be the angle made with the positive x-axis, where 0 ≤ θ ≤ 2pi.
Fix A at the point shown. We can confine B to the top half of the circle without any loss of generality i.e. 0 < arg(B) < pi
For the triangle to have no obtuse angles, A must not lie on the same semicircle's arc as B and C i.e. arg(A) < arg(B) + pi.
The probability that angle BAC is acute given the position of point B = the probability that arg(A) is smaller than arg(B) + pi = (arg(B) + pi)/(2pi).
Assume that arg(B) is uniformly distributed over [0,pi]. The expected value of arg(B) is therefore pi/2.
Thus the probabiltiy that BAC is acute = (pi/2 + pi)/(2pi) = 3/4
EDIT: Swap A and C.
What I meant was "are you sure the question shouldn't be
?"Maybe it should be 2008?
stick things around your house.
Ahh I see, thanks Jane. When I was investigating a^n (mod 10) it was cycling through 1, 3, 7 and 9. I can see why now.