Turn on thread page Beta
    • Thread Starter
    Offline

    3
    ReputationRep:
    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

    ================================ ==========================
    Offline

    3
    ReputationRep:
    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)
    • Thread Starter
    Offline

    3
    ReputationRep:
    (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...
    Offline

    3
    ReputationRep:
    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.
    • Thread Starter
    Offline

    3
    ReputationRep:
    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...
    Offline

    3
    ReputationRep:
    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.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (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)
    Offline

    8
    ReputationRep:
    (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
    Offline

    0
    ReputationRep:
    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: )
    Offline

    2
    ReputationRep:
    Count me in too
    Offline

    15
    ReputationRep:
    (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!
    Offline

    0
    ReputationRep:
    what is this all about?

    i want to join...
    Offline

    0
    ReputationRep:
    (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
    Offline

    0
    ReputationRep:
    (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?
    Offline

    1
    ReputationRep:
    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
    Offline

    2
    ReputationRep:
    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
    Offline

    1
    ReputationRep:
    (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.
    Offline

    2
    ReputationRep:
    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
    Offline

    1
    ReputationRep:
    (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.
    Offline

    2
    ReputationRep:
    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.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: April 21, 2018

University open days

  1. University of Cambridge
    Christ's College Undergraduate
    Wed, 26 Sep '18
  2. Norwich University of the Arts
    Undergraduate Open Days Undergraduate
    Fri, 28 Sep '18
  3. Edge Hill University
    Faculty of Health and Social Care Undergraduate
    Sat, 29 Sep '18
Poll
Which accompaniment is best?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.