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# Maths Syllabus watch

1. Hello, just interested to know if any of this stuff is in P1-6?

Matrices: Canonical Forms, Gaussian Elimination, Cayley-Hamilton Theorem, matrix decompositions.

Differential Equations: General solution methods for 2nd order Ordinary Differential Equations.

Mathematical Structure: Mathematical proof, set theory (i.e. Boolean algebra), group theory (i.e. proving Fermat's Little Theorem).

Numerical Methods: Solving differential equations using Runga-Kutta methods, integration with Simpson's Rule,

Complex Numbers: complex mappings (incl. conformal mapping)
2. Bloody hell, that looks complicated!

I think nearly all of that is Uni Level only; however:

"Differential Equations: General solution methods for 2nd order Ordinary Differential Equations." - This may appear in P6, I think I read something about it, where it appeared in P6. Maybe not though.
3. Second order DEs of the form a(d2y/dx2)+b(dy/dx)+cy = f(x) are in P4. Mathematical proof is covered in P1-6 but there is not a huge emphasis upon proof in the A-level. Matrices are covered in P6. However, most of the stuff you mention is above a-level.
4. (Original post by shiny)
Hello, just interested to know if any of this stuff is in P1-6?

Matrices: Canonical Forms, Gaussian Elimination, Cayley-Hamilton Theorem, matrix decompositions.

Differential Equations: General solution methods for 2nd order Ordinary Differential Equations.

Mathematical Structure: Mathematical proof, set theory (i.e. Boolean algebra), group theory (i.e. proving Fermat's Little Theorem).

Numerical Methods: Solving differential equations using Runga-Kutta methods, integration with Simpson's Rule,

Complex Numbers: complex mappings (incl. conformal mapping)
I think in MEI P6 you will find Gaussian Elimination and the cayley-hamilton theorem.

As mikesgt2 has said you learn 2nd order ODE solutions in p4 edexcel.

Proof, including by induction, is done in p6 edexcel, you should probably read up a bit on set theory any way, and group theory is in mei p6. How do you use group theory to prove fermats little theorem? I've always done it by a direct method.

I don't think you'll find Runga-Kutta in anything, you might find simpsons rule in mei or aqa. Complex mappings are dealt with a bit in edexcel p6. (transformations from one set of complex numbers to another)
5. (Original post by bono)
Bloody hell, that looks complicated!

I think nearly all of that is Uni Level only; however:

"Differential Equations: General solution methods for 2nd order Ordinary Differential Equations." - This may appear in P6, I think I read something about it, where it appeared in P6. Maybe not though.

No, differential equations is P3 mate. Only first order though.
6. (Original post by shiny)
, set theory (i.e. Boolean algebra), group theory (i.e. proving Fermat's Little Theorem).

LOL...the proof is over a hundred pages long.
7. (Original post by Ralfskini)
LOL...the proof is over a hundred pages long.
That's the last theorem
8. The proof of Fermat's Little Theorem comes from the fact that for a finite cyclic group G of order n, there is exactly one subgroup of order d for every d which is a divisor of n.
9. This is interesting since all the stuff I quoted was compulsory in the A-Level Further Maths syllabus I did!
10. (Original post by shiny)
The proof of Fermat's Little Theorem comes from the fact that for a finite cyclic group G of order n, there is exactly one subgroup of order d for every d which is a divisor of n.
Hmm, perhaps I still don't follow. How can conclude then that a^p-1 = 1 mod p where p is a prime that doesn't divide a? I've never really studied group theory properly so can't really see how this works...
11. (Original post by theone)
Hmm, perhaps I still don't follow. How can conclude then that a^p-1 = 1 mod p where p is a prime that doesn't divide a? I've never really studied group theory properly so can't really see how this works...
Let d be the order of a (where a is in the finite cyclic group Z which contains p-1 elements). From the theorem I just quoted, d divides p-1 or p-1 = qd ...

Hence,
a^(p-1) (mod p) = (a^d)^q (mod p) = 1^q (mod p) = 1 mod p
12. (Original post by shiny)
Let d be the order of a (where a is in the finite cyclic group Z which contains p-1 elements). From the theorem I just quoted, d divides p-1 or p-1 = qd ...

Hence,
a^(p-1) (mod p) = (a^d)^q (mod p) = 1^q (mod p) = 1 mod p
Ah, thanks for that (Far easier than my method of proving it )
13. (Original post by theone)
Ah, thanks for that (Far easier than my method of proving it )
What did you do? Induction?
14. (Original post by shiny)
What did you do? Induction?
Imagine the numbers a, 2a, 3a ... (p-1)a. where a is coprime to p

Now if any of these are congruent mod p, then we have the difference is 0 mod p. But the difference is going to be na where n < p. Now both of these are coprime to p, so the difference can not be 0 mod p. So none of the numbers are congruent mod p.

So the numbers are have different congruences mod p. Therefore they must, in some order, equate to 1, 2, 3, 4... p-1 mod p.

Taking the product. (p-1)! a^p-1 = (p-1)! mod p. Dividing through gives a^p-1 = 1 mod p.
15. (Original post by theone)
Imagine the numbers a, 2a, 3a ... (p-1)a. where a is coprime to p

Now if any of these are congruent mod p, then we have the difference is 0 mod p. But the difference is going to be na where n < p. Now both of these are coprime to p, so the difference can not be 0 mod p. So none of the numbers are congruent mod p.

So the numbers are have different congruences mod p. Therefore they must, in some order, equate to 1, 2, 3, 4... p-1 mod p.

Taking the product. (p-1)! a^p-1 = (p-1)! mod p. Dividing through gives a^p-1 = 1 mod p.
I don't think I have ever thought of it quite like that.

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