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We know we are going to be able to bound
so the important thing is to bound x away from 0.If
then and soFor any
we have:Hence if
thenBladito wrote:Assume that the theorem is true for some value k.
One problem is right there, you can't assume that. You can't find any k where that is true to start the inductive process.
The statement is true when k=1 as Bladito has stated. You seem to agree.
Bladito, the problem is that A-1 and B-1 are not necessarily both positive integers.
Hi bobbym
My main complaint was that you were assuming that P(x) was linear.
There is no harm in assuming that the remainder is linear, since it must have degree less that (x+2)(x-1).
Hence you could start by writing P(x) = (x+2)(x-1)Q(x) + ax + b for some polynomial Q(x)
and then prove that a = 7 and b = -5 as you have done.
It probably is a little neater to do it that way than the way that I did it.
You really shouldn't assume that P(x) is a linear polynomial.
P(x) = x^2 + 8x - 7 is another polynomial with the required properties.
We know that when P(x) is divided by (x+2) the remainder in -19.
This means that P(x) = (x+2)Q(x) - 19 for some polynomial Q(x).
We also know that when P(x) is divided by (x-1) the remainder is 2.
As you observe this means that P(1) = 2.
So 2 = P(1) = (1+2)Q(1) - 19 so 3Q(1) = 21 and hence Q(1) = 7.
This means that Q(x) = (x-1)R(x) + 7 for some polynomial R(x).
Therefore P(x) = (x+2)[(x-1)R(x) + 7] - 19 = (x+2)(x-1)R(x) +7(x+2) - 19
so P(x) = (x+2)(x-1)R(x) + 7x - 5.
The remainder when P(x) is divided by (x+2)(x-1) is 7x - 5.
The trace map is a polynomial over the field F_{q^n}. How many roots can it have?
Sylow's theorems tell you that any two Sylow p-subgroups of a group G are conjugate in G so if you can find one it shouldn't be too hard to find them all. (I assume that "all" in quotation marks is acknowledging this.)
Sylow's theorems also tell you that any p-subgroup of G is contained in a Sylow p-subgroup of G, so in these small cases it shouldn't be too hard to find a p-subgroup and if it is too small find a bigger one containing it.
Can you determine the orders of these Sylow p-subgroups?
For instance a Sylow 5-subgroup of S6 must have order 5. Can you find a subgroup of order 5 in S6?
Finding a Sylow 3-subgroup of S6 isn't too much harder.
To find a Sylow 2-subgroup of S5 consider what its order has to be, and also what the order of a Sylow 2-subgroup of S4 must be.
We can naturally embed S4 in S5 so any Sylow 2-subgroup of S4 must be contained in a Sylow 2-subgroup of S5.
If you have found a Sylow 2-subgroup of S5, then you want to find a Sylow 2-subgroup of S6 containing it.
Consider what the order of such a subgroup should be and I don't think it is too hard to find.
The two answers are equivalent.
At any point on the curve defined by
Since you can differentiate a power series term by term within the radius of convergence the following are all valid when |x| < 1
and your result follows with x = 1/2
First prove that no element can be of order 9.
Z9 has an automorphism with order 3, so there is a semi-direct product of Z9 and Z3 that is non-abelian.
To answer the second part you should consider the subgroups of G/Z(G).
I don't think Sylow's theorems can be used to solve these problems.
Some results that I think are more useful are:
If G/Z(G) is cyclic then G is abelian.
If G is a finite p-group then Z(G) is non-trivial.
If G is a p-group and H is a subgroup of index p in G then H is a normal subgroup of G.
I assume that part 2 is supposed to be "How many subgroups of order 9 are there?".
If
is differentiable at and then on some neighbourhood of and soHence if
is differentiable at and then is also differentiable at and if and only ifIt follows that a critical point of
must be either a critical point of or a zero ofIf the maximum of
is attained at a zero of then we must have that is identically zero, and so every point is a critical point ofTherefore if
has a maximum value, it is attained at a critical point ofI like this kind of puzzle. I find that the best way to make any sense of sentences like these is to read them from right to left.
Since 4*6 mod 23 = 1 we have
Anyone have the faintest idea why this thread has over 35,000 views?
This thread is the first page that comes up when I search for its title (or even just 'work out average') on Google.
bobbym,
soroban's antiderivative is
Since
Since a and b are coprime we must have
and soSince a and b are positive it follows that
andNow
Assuming that "underoot" means "square root"
x^4 - 10x² + 1 = 0
you can see that the only rational roots are +- 1, contradiction
I would interpret this to mean that 1 and -1 are roots of this polynomial, which is clearly untrue.
It would be better to say something like, by the rational root theorem any rational root of this polynomial is either 1 or -1.
I will admit that my choice of 610, 987 was rather extreme. I certainly wouldn't expect this pair to be found by randomly trying pairs of integers, but even 1, 2 is a better starting pair than 1, 1.
The important thing is that the ratios of consecutive pairs of Fibonacci numbers are the best rational approximations to the golden ratio.
The convergence seems to work best if the two positive integers are the same, and worst if the second is zero
What do you mean by "work best"?
I can't see a measure of goodness of convergence such that both
i) Starting with 1, 1 works better than starting with 1, 0
ii) Starting with 1, 1 works better than starting with 610, 987
The trick doesn't quite work for any two starting numbers. Let
Since
the sequence we get isI got my computer to test if
is prime for and the only primes it found were 11 and 101. I don't have enough memory to make k any higher with my brute force algorithm.There is a conjecture that
is prime for only finitely many n, an extension of the conjecture that only finitely many Fermat numbers are prime.It is telling me that 1111111111111111111 is prime. However, the 1 in [ <1111111111111111111, 1> ] means that the prime factor 1111111111111111111 has multiplicity 1.
For instance, Factorization(20); will output [ <2, 2>, <5, 1> ].
I'm still unsure what RickIsAnIdiot means by "shows all the digits".