# Official TSR Mathematical Society

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#1
Please feel free to discuss interesting mathematical problems here.

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0
15 years ago
#2
Surely (1) is not correct. I.e. the equation x^2-4x+4=0 only has 1 repeated root.

(2) Follows from the definition of the exponent.

a^n = a*a*a*a...*a

so a^x*a^y = a*a*a...(x times)*a*a*a...(y times) = a^(x+y)
0
#3
(1) We count the repeated roots as two...
for your information it is correct as its called "the fundamental theorem of algebra"

(2) your argument does not prove e^x.e^y = e^(x+y) as u dont know what e is...is it it an arbitrary number? ur argument merely shows the result but doesnt prove it using first principles...
0
15 years ago
#4
Righto, I and III are equivalent then aren't they. As they are both formulations of the FTA.

i.e. if the polynomial has n roots then you can always express it as (x-a1)(x-a2)...(x-an), and as the complex roots come in complex conjugate pairs you can multiply those together to get a quadratic with real coefficients.

But for II, I can't see your point.

e can be any number you want and e^x*e^y = e^(x+y), if you take it to be THE e then it is just as valid as if you take e to be 1.
0
#5
First you are certainly right that (3) is formulated from (1)....and the proof of (3) depends on the proof of (1), so if u can prove (1) thenas u said that multipying conjugate pairs of complex roots will give a quadratic with real coefficients thus it can be reduced in linear and quadratic factors with real coefficients (however u might have to prove that complex roots occur in conjugate pairs)

For (2) i mean ur argument SHOWS the desired result but cannot be taken as a valid PROOF...

surely a^n = a.a.a....a (n times)

but u cannot say that exp(x) is the same as e^x we know that it is as exp(2) does equal [exp(1)]^2 but we need a definition of exp(x) and then we can start the proof...so a hint is find how exp(x) is defined and then embark on a proof...
0
15 years ago
#6
I am not sure about whether it is neccesary to prove what you are asking. We can define e^x as an infinite series, but e is still a real number and as such e^x has the same properties as any other exponential function.
0
#7
(Original post by AntiMagicMan)
I am not sure about whether it is neccesary to prove what you are asking. We can define e^x as an infinite series, but e is still a real number and as such e^x has the same properties as any other exponential function.
infinite series is a good definition, use thsi to prove (2)
0
15 years ago
#8
(Original post by IntegralNeo)
infinite series is a good definition, use thsi to prove (2)
one way to show question 1 is to use liouville's theorem (from complex analysis)
the details are gone from my brain but basically you assume f(z)=0 has no solutions and define g=1/f so g is defined and continuous since f(z) does not equal 0 for any z. using some complex analysis it is shown g is bounded
and using liouville's theorem g is constant.hence only polynomials with no roots are the constant roots.
let a be a root of f(z)=0
then f(z)=(z-a)h(z) for some poly h of degree <f
then apply above to h to show h is constant or has root
so if b is a root
f(z)=(z-a)(z-b)j(z) etc
0
15 years ago
#9
take the infinite series for exp(x), defined as 1+x+(x^2)/2!+(x^3)/3!...., multiply that by exp(y), collect terms, it comes out to exp(x+y)....

(meeheehee )
0
15 years ago
#10
Count me in too
0
15 years ago
#11
(Original post by IntegralNeo)
================================ ==========================

Problems: (green is solved, red is unsolved)

(1) Prove that every polynomial of degree n with real coefficients has n roots in complex field.

(2) Prove that exp(x).exp(y) = exp(x+y) , where exp(x) is exponential function

(3) Prove that every polynomial of degree n with real coefficients can be reduced to the product of linear and/or quadratic factors.

these three were all mentioned in the first week of our maths course! is that why you posted them neo?
1st i'll get back to you.

2nd. we define exp (x) = 1 + x + x^2/2! + x^3/3! + ....
so exp (x) * exp (y) =

(1 + x + x^2/2! + x^3/3! + ....)(1 + y + y^2/2! + x^3/3! + ....) =
1 + x + x^2/2! + x^3/3! + .....
+ y + xy + yx^2/2! + yx^3/3! + ....
+ y^2/2! + xy^2/2! +x^2 y^2/2!2! + ....
+ y^3/3! + .....

etc. looking at the terms diagonally, you have:
1 + (x+y) + 1/2! (x+y)^2 + 1/3! (x+y)^3 + ...
= exp (x+y)

QED.

3. I'll type up the proof tomorrow... it's about 1am atm!
0
15 years ago
#12

i want to join...
0
15 years ago
#13
(3) Prove/Disprove that square of an integer a is either 0 or 1 mod 4

a mod 4 | a^2 mod 4
0 | 0
1 | 1
2 | 4 = 0 mod 4
3 | 9 = 1 mod 4
0
15 years ago
#14
(3) Prove/Disprove that square of an integer a is either 0 or 1 mod 4

It seems joe1212 has beaten me to it and I don't even know what quadratic residues are. I don't know much about modular arithmetic so I hope I'm not missing something but here's an attempt:

suppose a² = 2mod4

a² = 4b + 2 (where b is an integer)

a² is even => a is even => a² is divisible by 4

(a² - 2)/4 = b

since a² is divisible by 4 (a²-2) is not => b is not an integer, which is a contradiction, so a² is never 2mod4

similarly, suppose a² = 3mod4

a² = 4c + 3 (where c is an integer)

a² is odd, so a is odd. Let a = (E+1) where E is even

(E+1)² = 4c + 3
E² + 2E + 1 = 4c +3

(E² + 2E - 2)/4 = c

since E² and 2E are both divisible by 4, (E² + 2E - 2) is not => c is not an integer => contradiction => a² is never 3mod4.

a² is either 0 or 1 mod4.

0
15 years ago
#15
Proving that a^2 is 1mod4 or 0mod4 is pretty simple.

Any integer a can be generated by one of these equations:
a=4k
a=4k+1
a=4k+2
a=4k+3

This means that we can write a modular congruence as,
a=0mod4,
a=1mod4,
a=2mod4,
a=3mod4

Squaring each of the above equations, we get (respectively):
a^2=0mod4
a^2=1mod4
a^2=4mod4=0mod4
a^2=9mod4=1mod4

So all a^2 is either 0mod4 or 1mod4.

0
15 years ago
#16
Tour of perfect squares starting from 4 involves adding the next odd number, so add 5 to get the next perfect square, then 7 to get the next, then 9, and so on.

Since we are interested in proving that the next perfect square is NEVER 3q-1, convert all of the above into an expression relating the number to multiples of 3, and then discard the multiples of 3. 4 becomes +1 (multiple of 3 +1), 5 becomes –1, 7 becomes + 1, 9 becomes 0, 11 becomes –1, and so on.

The sequence of odd numbers will then become (starting with 5) –1, +1, 0, -1, +1, 0, -1, +1, 0, and so on. Provable if req.

This sequence is added, in turn, to the starting perfect square 4 (which is converted into +1) You can also start with 1(which is also +1)

Adding each converted odd number in turn gives 0, +1, +1, 0, +1, +1, 0, +1, +1….i.e. always 3q or 3q+1. Interesting.

Aitch
0
15 years ago
#17
(Original post by Aitch)
The sequence of odd numbers will then become (starting with 5) –1, +1, 0, -1, +1, 0, -1, +1, 0, and so on. Provable if req.
My impression is that this is exactly what you have to prove.
0
15 years ago
#18
I think it looks OK -

Dividing odd numbers from 5+ by 3 leaves remainders of

-1, +1, 0, -1, +1, 0... etc.

Does this need a formal proof? I suppose it might...

Aitch
0
15 years ago
#19
(Original post by Aitch)
I think it looks OK -

Dividing odd numbers from 5+ by 3 leaves remainders of

-1, +1, 0, -1, +1, 0... etc.

Does this need a formal proof? I suppose it might...

Aitch
Your -1, +1, 0, -1,+1,0... argument is precisely what you want to prove.. that argument is equivalent to proving that n=0mod3 or n=1mod3 (In other words, n=3q or n=3q+1). This argument is equivalent because 5 divided by 3 mod3 is 2, which is equivalent to your -1(in mod3), and 7 divided by 3 mod3 is 1, and 9 divided by 3 mod3 is 0, etc...

So this is exactly what you have to prove.
0
15 years ago
#20
Sequence 5,7,9,11,13,15,17,...

If we divide the list into blocks of 3 terms, then since for any term T(n) we know that T(n) = T (n-3) + 6, we can reduce the list to 3 general terms:

5+6b
7+6b
9+6b
where b is some + integer.

So, since adding 6b to any of these does not change the relationship of the expression to multiples of 3 (as 6b is a multiple of 3), it should be sufficient to prove that

5=(multiple of 3)-1
7=(multiple of 3)+1
9=(multiple of 3)+0

and since all following blocks of 3 terms are of the form

5+6b
7+6b
9+6b

this sequence will be repeated ad infinitum.
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