Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    12
    ReputationRep:
    Hi there. did this paper and got my teacher to mark it. scored pretty high but a couple of queries.
    http://www.mathshelper.co.uk/notmine... Test 1999.pdf

    1(b) The inequality  2^n > n^2 is true for:

    i) no integers  n \geq 0
    ii) all integers  n \geq 0
    iii) all integers  n > 4
    iv) all intgers  n \geq 4
    my teacher marked it wrong, but im pretty sure the answer is iii):

    i) is not possible as it is true for n=0
    ii) is not possible as it not true for n=2
    iv) is not possible as it is not true for n=4
    therefore we must conclude that it is true for iii)
    is this correct?
    Spoiler:
    Show
    also, out of curiousity, im not sure how to just generally prove that on its own that it works for n > 4. would it be merely induction? or would it simply be a matter of considering the graphs, and that the growth rate of 2^n is greater than the growth rate of n^2 so it must follow that it will always be greater?


    4(d) The answers to a)i) and a)ii) are related in a particular way. Explain how the relationship can be seen without working out any integrals
    I wasn't sure when doing the paper so just waffled something about symmetry in the graphs. My teacher wasn't sure how to answer it either :confused: anyone got an idea?

    thanks for your help in advance.
    Offline

    16
    ReputationRep:
    (Original post by GHOSH-5)
    Hi there. did this paper and got my teacher to mark it. scored pretty high but a couple of queries.
    http://www.mathshelper.co.uk/notmine... Test 1999.pdf



    my teacher marked it wrong, but im pretty sure the answer is iii):

    i) is not possible as it is true for n=0
    ii) is not possible as it not true for n=2
    iv) is not possible as it is not true for n=4
    therefore we must conclude that it is true for iii)
    is this correct?
    Spoiler:
    Show
    also, out of curiousity, im not sure how to just generally prove that on its own that it works for n > 4. would it be merely induction? or would it simply be a matter of considering the graphs, and that the growth rate of 2^n is greater than the growth rate of n^2 so it must follow that it will always be greater?




    I wasn't sure when doing the paper so just waffled something about symmetry in the graphs. My teacher wasn't sure how to answer it either :confused: anyone got an idea?

    thanks for your help in advance.
    You are correct for the first one.

    There are lots of ways to prove it. Induction comes to mind, but are other ways.
    • Thread Starter
    Offline

    12
    ReputationRep:
    (Original post by SimonM)
    You are correct for the first one.

    There are lots of ways to prove it. Induction comes to mind, but are other ways.
    ok thanks for that.
    • Thread Starter
    Offline

    12
    ReputationRep:
    anyone got an idea for 4d)?
    Offline

    15
    ReputationRep:
    x^3 + x^2 - 2x = x^3 - x + x^2 - x.

    Does this help? If you're planning on applying to Oxford by the way, I recommend leaving the papers so you have something left.

    Further hint: look at the limits.
    Offline

    14
    ReputationRep:
    The integrand in the second one is x+2 times the first.
    I'm not sure if this helps or not:p:
    • Thread Starter
    Offline

    12
    ReputationRep:
    (Original post by Swayum)
    x^3 + x^2 - 2x = x^3 - x + x^2 - x.

    Does this help? If you're planning on applying to Oxford by the way, I recommend leaving the papers so you have something left.

    Further hint: look at the limits.
    im not too sure, but are you saying that (x^3 + x^2 + 2x) - (x^2 - x) = x^3 - x, which is symmetrical, so the difference between the areas between two x-coordinates -a and a would be zero, like -1 and 1, just as the difference between the areas of the two functions in the questions is 0?
    • Thread Starter
    Offline

    12
    ReputationRep:
    (Original post by Slumpy)
    The integrand in the second one is x+2 times the first.
    I'm not sure if this helps or not:p:
    you mean to say that:
     (\frac{x^3}{3} - \frac{x^2}{2}) \times (x+2) = \frac{x^4}{4} + \frac{x^3}{3} - x^2
    ?

    im not sure that it does...
    Offline

    15
    ReputationRep:
    What you've written there is hard to interpret :p:. What I mean is

    \int^1_{-1} x^3 + x^2 - 2x\ \mathrm{d}x = \int^1_{-1} x^2 - 2x\ \mathrm{d}x + \int^1_{-1} x^3 - x\ \mathrm{d}x

    Spoiler:
    Show
    Sketch the graph of x^3 and x, what do you think happens when you integrate between a and -a? (I think you've said the answer above). The property holds for all odd functions. We can interpret the latter integral as

    \int^1_{-1} x^3 - x\ \mathrm{d}x = \int^1_{-1} x^3\ \mathrm{d}x - \int^1_{-1} x\ \mathrm{d}x
    Offline

    14
    ReputationRep:
    Integrand being bit inside the integral sign, no?
    Am i messing up my termage again?
    • Thread Starter
    Offline

    12
    ReputationRep:
    (Original post by Swayum)
    What you've written there is hard to interpret :p:. What I mean is

    \int^1_{-1} x^3 + x^2 - 2x\ \mathrm{d}x = \int^1_{-1} x^2 - 2x\ \mathrm{d}x + \int^1_{-1} x^3 - x\ \mathrm{d}x

    Spoiler:
    Show
    Sketch the graph of x^3 and x, what do you think happens when you integrate between a and -a? (I think you've said the answer above). The property holds for all odd functions. We can interpret the latter integral as

    \int^1_{-1} x^3 - x\ \mathrm{d}x = \int^1_{-1} x^3\ \mathrm{d}x - \int^1_{-1} x\ \mathrm{d}x
    yeh thats what i meant. as the integral between -a and a is '0' if we integrate directly between the two limits, and not separate the limits to calculate actual areas.

    and if we take the x^2 - x to the otherside, the difference between the two integrals should be zero, as it equals the x^3-x integral, which is what part a) demonstrates to us.

    is that it? thank you very much.

    sorry it was a little hard to intepret :p:
    Offline

    15
    ReputationRep:
    Well, the question isn't asking you to work out the difference between the integrals, I think they just want you to say "i and ii are equal because of [reason above]". And yeah, that's it. It'd hold for cos^3(x) and cos(x) as well since they're odd functions.

    And to clarify the other person's point, he meant

    (x^2 - x)(x + 2) = x^3 + x^2 - 2x, but I don't think it helps.
    • Thread Starter
    Offline

    12
    ReputationRep:
    (Original post by Swayum)
    Well, the question isn't asking you to work out the difference between the integrals, I think they just want you to say "i and ii are equal because of [reason above]". And yeah, that's it. It'd hold for cos^3(x) and cos(x) as well since they're odd functions.

    And to clarify the other person's point, he meant

    (x^2 - x)(x + 2) = x^3 + x^2 - 2x, but I don't think it helps.
    ok i see what you mean.

    although is cosx not an even function? sinx is an odd function...

    as i have it, odd functions are the same when rotated 180 about the origin, and even functions are symmetric across the y axis... unless im confusing something
    Offline

    14
    ReputationRep:
    (Original post by GHOSH-5)
    ok i see what you mean.

    although is cosx not an even function? sinx is an odd function...

    as i have it, odd functions are the same when rotated 180 about the origin, and even functions are symmetric across the y axis... unless im confusing something
    Yup, look at the graph of cosx, it's even.
    • Thread Starter
    Offline

    12
    ReputationRep:
    so we can say it works for sin(x) and sin^3(x).... but not with cosine...
    Offline

    15
    ReputationRep:
    (Original post by GHOSH-5)
    so we can say it works for sin(x) and sin^3(x).... but not with cosine...
    Yeah, sorry, I don't know what possessed me to say cosx. The way to show a function is even is to check

    f(x) = f(-x)

    For odd

    f(x) = -f(-x)
    • Thread Starter
    Offline

    12
    ReputationRep:
    (Original post by Swayum)
    Yeah, sorry, I don't know what possessed me to say cosx. The way to show a function is even is to check

    f(x) = f(-x)

    For odd

    f(x) = -f(-x)
    yeh thanks. knew about how to check for even functions, but wasnt sure about how to check for odd functions.

    with the odd functions, what you wrote, is it the same as writing:

    f(-x) = -f(x)

    id find it more useful in the way you wrote it, but just wanted to check that
    Offline

    16
    ReputationRep:
    (Original post by GHOSH-5)
    f(-x) = -f(x)
    Of course, multiply both sides of

    f(x)=-f(-x) by -1
    • Thread Starter
    Offline

    12
    ReputationRep:
    (Original post by SimonM)
    Of course, multiply both sides of

    f(x)=-f(-x) by -1
    yeh sorry, that was a bit obvious...>.<
 
 
 
Turn on thread page Beta
Updated: November 15, 2008
Poll
Do you agree with the proposed ban on plastic straws and cotton buds?
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.