Turn on thread page Beta
    Offline

    0
    ReputationRep:
    (Original post by Gaz031)
    (4+x)^{\frac{1}{2}}=2(1+\frac{x}  {4})^{\frac{1}{2}}
    Clearly |\frac{x}{4 }|<1 ie |x|<4 for the series to converge.

    So basically if it was larger then it wudnt be converging, it would be diverging? ie each term would be getting larger and larger?
    Offline

    15
    ReputationRep:
    (Original post by melbourne)
    So basically if it was larger then it wudnt be converging, it would be diverging? ie each term would be getting larger and larger?
    That's it. If the y value in (1+y)^{n} is such that |y|>1, then y^{2},y^{3},y^{4} etc. progressively get larger and larger. Only in the case |y|<1 is the approximation sensible.
    Offline

    0
    ReputationRep:
    (Original post by Gaz031)
    That's it. If the y value in (1+y)^{n} is such that |y|>1, then y^{2},y^{3},y^{4} etc. progressively get larger and larger. Only in the case |y|<1 is the approximation sensible.
    ah that makes sense, it explains why in c2 we get those questions "now use your answer to make an approximation............." so the more terms you get the closer to the answer, whereas if it was diverging the answer would go beyond.
    Offline

    0
    ReputationRep:
    Integration By Parts

    I'm just doing the Integration by parts - and i am a lil stuck.

    This is what i have done so far:

    int X(x-1)^½

    V=x dv/dx=1

    du/dx= (x-1)^½ u=2/3(x-1)^3/2


    I whack them into the eqn. and get

    2/3x(x-1)^3/2 - 4/15(x-1)^5/2

    I think this is correct, im doing it from the P3 Heinemann book pg.119 q 8....Any help would be appreciated!

    :tsr:
    • Thread Starter
    Offline

    2
    ReputationRep:
    Looks right!
    Offline

    0
    ReputationRep:
    in the book they simplify it and i am unsure of how to do this?
    Offline

    2
    ReputationRep:
    (Original post by bavman)
    in the book they simplify it and i am unsure of how to do this?
    What do they simplify it to?

    [[ don't have copy of book ]]
    Offline

    0
    ReputationRep:
    lol sorry - they make it

    2/15(3x+2)(x-1)^3/2

    my algebra is awful
    Offline

    2
    ReputationRep:
    Right so you have got:

    2/3x(x-1)3/2 - 4/15(x-1)5/2 (1)

    and you want:

    2/15(3x+2)(x-1)3/2 (2)

    I'll just to a check to see if they are the same (I am aware this will not prove anything):

    Put x = 2: (1) gives: 16/15

    (2) gives: 16/15.

    Alright they could be equivalent.

    EDIT: poor maths removed.

    => Anyone else help me out?
    Offline

    0
    ReputationRep:
    2/3x(x-1)3/2 - 4/15(x-1)5/2
    2/3[x(x-1)3/2 - 2/5(x-1)5/2]
    2/3(x-1)3/2[x - 2/5(x-1)]
    2/3(x-1)3/2[3/5x+2/5]
    2/3(x-1)3/2 x 1/5 (3x+2)]
    = 2/15(3x+2)(x-1)3/2

    EDIT: would be much easier if you just take out 2/15(x-1)3/2 first.
    Offline

    0
    ReputationRep:
    thanks mate, your a legend!

    Nah, i understood better because you left it in. I just didnt realise that if youi take the indices out the 5/2 becomes to the power of 1 thus eliminatin gthe indicies on the inside of the brackets
    Offline

    0
    ReputationRep:
    (Original post by bavman)
    thanks mate, your a legend!
    np. just play around with it, and you'll soon find what you need!
    Offline

    0
    ReputationRep:
    (Original post by samdavyson)
    => (x-1)1/2[2/3x(x-1)3 - 4/15(x-1)5]

    => (x-1)3/2[2/3x(x-1) - 4/15(x-1)5/3]

    => (x-1)3/2[2/3 - 2/3x - 4/15(x-1)5/3]
    be careful with your powers.
    when you took out (x-1)3, you have to MINUS the powers of the rest inside the brackets (while you divided).
    Offline

    2
    ReputationRep:
    (Original post by Zuber)
    be careful with your powers.
    when you took out (x-1)3, you have to MINUS the powers of the rest inside the brackets (while you divided).
    Yeah sorry about that.

    [[ think sam think ]]
    • Thread Starter
    Offline

    2
    ReputationRep:
    Crapp i thought this'd be piss

    im basically implicitly differentiating:

    x³-2xy-4x+y³-51 =0

    when I differentiate -2xy using chain rule i get -2x dy/dx - 2y, but it should be -2x dy/dx + 2y. Whys that? Cheers
    Offline

    2
    ReputationRep:
    (Original post by TheWolf)
    Crapp i thought this'd be piss

    im basically implicitly differentiating:

    x³-2xy-4x+y³-51 =0

    when I differentiate -2xy using chain rule i get -2x dy/dx - 2y, but it should be -2x dy/dx + 2y. Whys that? Cheers
    It is a brackets problem I think.

    x³-2xy-4x+y³-51 =0

    => 3x² - (2x(dy/dx) + 2y) - 4 - 3y²(dy/dx) - 51 = 0

    => 3x² - 2x(dy/dx) - 2y - 4 - 3y²(dy/dx) - 51 = 0
    Offline

    16
    ReputationRep:
    (Original post by TheWolf)
    Crapp i thought this'd be piss

    im basically implicitly differentiating:

    x³-2xy-4x+y³-51 =0

    when I differentiate -2xy using chain rule i get -2x dy/dx - 2y, but it should be -2x dy/dx + 2y. Whys that? Cheers
    you should differentiate it via the product rule.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by samdavyson)
    It is a brackets problem I think.

    x³-2xy-4x+y³-51 =0

    => 3x² - (2x(dy/dx) + 2y) - 4 - 3y²(dy/dx) - 51 = 0

    => 3x² - 2x(dy/dx) - 2y - 4 - 3y²(dy/dx) - 51 = 0
    Yes I've got that, but the answer (q2 mock) says another thing? http://imaad.pwp.blueyonder.co.uk/mockanswer1.jpg

    naz i do use the product rule
    Offline

    2
    ReputationRep:
    (Original post by TheWolf)
    Yes I've got that, but the answer (q2 mock) says another thing? http://imaad.pwp.blueyonder.co.uk/mockanswer1.jpg

    naz i do use the product rule
    They have just missed out the brackets.

    If you look at the working for the rest it follows what I (& you) got.
    • Thread Starter
    Offline

    2
    ReputationRep:
    Ah coolio.. thanx sam =)
 
 
 
Turn on thread page Beta
Updated: June 28, 2005
The home of Results and Clearing

1,983

people online now

1,567,000

students helped last year

University open days

  1. Bournemouth University
    Clearing Open Day Undergraduate
    Fri, 17 Aug '18
  2. University of Bolton
    Undergraduate Open Day Undergraduate
    Fri, 17 Aug '18
  3. Bishop Grosseteste University
    All Courses Undergraduate
    Fri, 17 Aug '18
Poll
A-level students - how do you feel about your results?
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.