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
Eddie8704
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
burjet
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
eunit
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
GHOSH5
RocTheMic
calicoeannecash
chinqu
Adampolar
davidcy147
amberlivable
advice_guru
Khodu
nchen5
IBiot Ash '08
Mitchy0283
Ninja Keg
alexmahone
Ignoramus
SprinterJr
Tallon
rapha
GHOSH5
somepeople
GHOSH5
RocTheMic
calicoeannecash
chinqu
Adampolar
davidcy147
amberlivable
================================ ==========================

Euler
 Follow
 1 follower
 3 badges
 Send a private message to Euler
 Visit Euler's homepage!
 Thread Starter
Offline3ReputationRep: Follow
 1
 13102004 11:10
Last edited by LadySmythe; 20112010 at 13:32. 
Revision help in partnership with Birmingham City University

 Follow
 2
 13102004 11:42
Surely (1) is not correct. I.e. the equation x^24x+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) 
Euler
 Follow
 1 follower
 3 badges
 Send a private message to Euler
 Visit Euler's homepage!
 Thread Starter
Offline3ReputationRep: Follow
 3
 13102004 12:30
(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... 
 Follow
 4
 13102004 12:37
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 (xa1)(xa2)...(xan), 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. 
Euler
 Follow
 1 follower
 3 badges
 Send a private message to Euler
 Visit Euler's homepage!
 Thread Starter
Offline3ReputationRep: Follow
 5
 13102004 12:46
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... 
 Follow
 6
 13102004 13:08
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.

Euler
 Follow
 1 follower
 3 badges
 Send a private message to Euler
 Visit Euler's homepage!
 Thread Starter
Offline3ReputationRep: Follow
 7
 13102004 14:52
(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. 
 Follow
 8
 13102004 15:23
(Original post by IntegralNeo)
infinite series is a good definition, use thsi to prove (2)
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)=(za)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)=(za)(zb)j(z) etc 
 Follow
 9
 15102004 13:42
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 ) 
 Follow
 10
 25102004 00:59
Count me in too

 Follow
 11
 26102004 01:16
(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.
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! 
ConfidentGenius
 Follow
 0 followers
 0 badges
 Send a private message to ConfidentGenius
Offline0ReputationRep: Follow
 12
 25112004 11:12
what is this all about?
i want to join... 
 Follow
 13
 27112004 23:40
(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 
Hags
 Follow
 0 followers
 0 badges
 Send a private message to Hags
 Visit Hags's homepage!
Offline0ReputationRep: Follow
 14
 28112004 01:32
(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 (a²2) 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? 
 Follow
 15
 28112004 02:17
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 
 Follow
 16
 18122004 13:29
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 3q1, 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 
 Follow
 17
 18122004 20:45
(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. 
 Follow
 18
 18122004 22:39
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 
 Follow
 19
 19122004 05:49
(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
So this is exactly what you have to prove. 
 Follow
 20
 19122004 13:09
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 (n3) + 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.
Related discussions
Related university courses

Biblical Studies and Mathematics
University of St Andrews

Chemistry with Mathematics
UCL

German and Mathematics
University of Leeds

Economics and Mathematics
University of Essex

Financial Mathematics
Cardiff University

Computer Science and Mathematics
Edge Hill University

Sport Science and Mathematics
Nottingham Trent University

Mathematics (with Study Year Abroad)
University of Bath

Mathematics and Physics
University of St Andrews

Mathematics/Business and Management
University of Glasgow
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
 Notnek
 charco
 Mr M
 Changing Skies
 F1's Finest
 RDKGames
 davros
 Gingerbread101
 Kvothe the Arcane
 TeeEff
 The Empire Odyssey
 Protostar
 TheConfusedMedic
 nisha.sri
 claireestelle
 Doonesbury
 furryface12
 Amefish
 Lemur14
 brainzistheword
 Sonechka
 TheAnxiousSloth
 EstelOfTheEyrie
 CoffeeAndPolitics
 Labrador99
 EmilySarah00
 thekidwhogames
 entertainmyfaith
 Eimmanuel
 Toastiekid
 CinnamonSmol
 Qer