Euler
Badges: 3
Rep:
?
#1
Report Thread starter 15 years ago
#1
Please feel free to discuss interesting mathematical problems here.

If you want to join, please see this page.

================================ ==========================

Current Members:

beauford
Ralfskini
shiny
It'sPhil...
lou p lou
chrisbphd
lgs98jonee
Hoofbeat
silent p...?
kikzen
Babygal
Leeroy
FaerieLand
Daveo
Chud
Eddie87-04
Mysticmin
chandoug
orange_soap
Bosslady
IntegralAnomaly
ruth_lou
Katie Heskins
Zapsta
mikesgt2
ZJewelH
4Ed
Euler
zombie
JamesF
me!
capslock
jimmy_c
samdavyson
meepmeep
KOH
jyuk
Drederick Tatum
happysunshine
burje-t
musicboy
Kaiel
®eAl ©uTe Eye$
Mathemagician
David Frank
SiAnY
TheWolf
tammypotato
kamsingh
hornblower
IZZY!
jinsisi
Hawk
r perry
Undryl
shushimeng
mrjoe
sumi2000
Rustyk1
Kupo nut
ogs
jamierwilliams
Chinesegirl
Mr_Homosexual
freemyice
amo1
Widowmaker
Wherewathuwen?
James.R
D.S.
RobbieC
Ben.S
Jaq
BCHL85
distortedgav
choco_frog_chic
dvs
Eugene_Newton
eurasianfeline
bodomx
Newton
Invisible
jumpunderaboat
MRLX69
Simba
RDoh
riks2004
mandoman
Feria
franks
Amohan
ConfidentGenius
JFN
Hags
helpless
Nick Sutcliffe
Euclid
aiman
staylor
hajira
cktlee1
e-unit
AhmadMujtaba
JohnC
LHMarsh10
thescorpio
coldfusion
tokay
Sparkling_Jules
everybodylaughs
ashy2005
frixis
Phil23
Arvinda
saudia khan
ErastF
Madprof
Spenceman_
geekypoo
masterasg
Roger Kirk
darth_vader05
Master Gee
Zuber
KAISER_MOLE
goku999
rpotter
yazan_l
Theron
Scottus_Mus
Gexko
einsci
mizfissy815
SketchyCanvas
sayed_samed
flick
Metal_Mistress
HTale
alexsmithson
winged_one
Enfalas
ali p
kishan11
ljfrugn
Starsailor
joyous_zhang
wickedways99
Piggy
Hitonagashi
frixis
Hay_Hay
Abu
StudentTazzy
xx007
anterminator
Clark Black
sarz777
i_luv_maths
groseille
joe8232
Girdag
JamesEpsilon
saasmi
drumking1088
Aphrodite
pkiy
geekypoo
Esquire
Sentooran
vector771
rawkingpunkster
celeritas
little.princess.1991
zrancis
Yasser Hayatli
westhamfan
MatchDancer
davmarksman
foxygreg
mspen
adel_kadrah
n0b0dy
wackymathgal
DeathAwaitsU
Harryok
RdotR
Sarokrae
Dadeyemi
HBKss
nish81
soulsister_16
sweetfloss
erode87
thedemon13666
andrewlee89
katieinnit
fmathsgirly
MatchDancer
andco
shazzylai
GHOSH-5
RocTheMic
calicoeannecash
chinqu
Adampolar
davidcy147
amberlivable
advice_guru
Khodu
nchen5
IBiot Ash '08
Mitchy0283
Ninja Keg
alexmahone
Ignoramus
SprinterJr
Tallon
rapha
GHOSH-5
somepeople
GHOSH-5
RocTheMic
calicoeannecash
chinqu
Adampolar
davidcy147
amberlivable

================================ ==========================
0
reply
jpowell
Badges: 15
Rep:
?
#2
Report 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
reply
Euler
Badges: 3
Rep:
?
#3
Report Thread starter 15 years ago
#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
reply
jpowell
Badges: 15
Rep:
?
#4
Report 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
reply
Euler
Badges: 3
Rep:
?
#5
Report Thread starter 15 years ago
#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
reply
jpowell
Badges: 15
Rep:
?
#6
Report 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
reply
Euler
Badges: 3
Rep:
?
#7
Report Thread starter 15 years ago
#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
reply
mathz
Badges: 8
Rep:
?
#8
Report 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
reply
rafiees
Badges: 0
#9
Report 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 :rolleyes: )
0
reply
Jump
Badges: 16
Rep:
?
#10
Report 15 years ago
#10
Count me in too
0
reply
4Ed
Badges: 15
Rep:
?
#11
Report 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
reply
ConfidentGenius
Badges: 0
Rep:
?
#12
Report 15 years ago
#12
what is this all about?

i want to join...
0
reply
joe1212
Badges: 0
#13
Report 15 years ago
#13
(3) Prove/Disprove that square of an integer a is either 0 or 1 mod 4

Quadratic residues?

a mod 4 | a^2 mod 4
0 | 0
1 | 1
2 | 4 = 0 mod 4
3 | 9 = 1 mod 4
0
reply
Hags
Badges: 0
#14
Report 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.

Also, please could I join?
0
reply
JFN
Badges: 1
Rep:
?
#15
Report 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.

And please add me to the list
0
reply
Aitch
Badges: 2
Rep:
?
#16
Report 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.

Please add my name to your list.

Aitch
0
reply
JFN
Badges: 1
Rep:
?
#17
Report 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
reply
Aitch
Badges: 2
Rep:
?
#18
Report 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
reply
JFN
Badges: 1
Rep:
?
#19
Report 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
reply
Aitch
Badges: 2
Rep:
?
#20
Report 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.
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

If you do not get the A-level grades you want this summer, what is your likely next step?

Take autumn exams (240)
47.24%
Take exams next summer (68)
13.39%
Change uni choice through clearing (110)
21.65%
Apply to uni next year instead (53)
10.43%
I'm not applying to university (37)
7.28%

Watched Threads

View All