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

    11
    ReputationRep:
    (Original post by Zacken)
    That's a nice analytic approach. I was looking into getting the inequality bounds by looking at the Laurent series for the function and deducing then from there but I don't have quite enough experience to know whether that's viable or not.
    I only use the term "Laurent series" for bad functions with singularities i.e. with 1/z^n terms in their expansion. I use "Taylor series" if it's a nice function. But anyway...

    I think the argument would go like this:

    Given f(x) =\cos x - \sin x = \sqrt{2}\cos(x+\frac{\pi}{4}) \Rightarrow f'(x) = -\sqrt{2}\sin(x+\frac{\pi}{4}) so that over the range [0, \pi/4] we have -1 \le f'(x) \le -\sqrt{2}.

    So f'(x) is more negative than the gradient of the tangent over the whole range, so we must have f(x) < 1-x as required.

    Also, f'(\pi/4) = -\sqrt{2} \approx -1.414 < -1.\dot{3} = -4/3 < -4/\pi so for some x_0 \in (0,\pi/4) we have f'(x_0) = -4/\pi.

    But in (0,x_0), -1 < f(x) < -4/\pi so we must have f(x_0) > 1-4x_0/\pi. But suppose that f(x_1) < 1-4x_1/\pi for some x_1 \in (x_0, \pi/4).

    Then we would have f(\pi/4) < 0 = 1-4(\pi/4)/4 since f'(x) < -4/\pi in (x_0, \pi/4) which is a contradiction.

    I'm hoping that 1) this makes sense and 2) there's an easier argument that doesn't waffle so much.
    • Thread Starter
    Offline

    12
    ReputationRep:
    Argument makes perfect sense up to the last two lines, granted I don't have much experience with Taylor expansions so that's no great suprise, it looks 'nicer' than the way I used though


    Posted from TSR Mobile
    Offline

    22
    ReputationRep:
    (Original post by drandy76)
    Argument makes perfect sense up to the last two lines, granted I don't have much experience with Taylor expansions so that's no great suprise, it looks 'nicer' than the way I used though
    I don't think he used Taylor expansions anywhere.
    • Thread Starter
    Offline

    12
    ReputationRep:
    (Original post by Zacken)
    I don't think he used Taylor expansions anywhere.
    I just assumed the bit where I lost it was that tbh :/


    Posted from TSR Mobile
    Offline

    17
    ReputationRep:
    https://en.wikipedia.org/wiki/Taylor_series Do some reading guys.
    • Thread Starter
    Offline

    12
    ReputationRep:
    (Original post by zetamcfc)
    https://en.wikipedia.org/wiki/Taylor_series Do some reading guys.
    Instructions not clear enough, ended up with 10 different tabs open


    Posted from TSR Mobile
    Offline

    22
    ReputationRep:
    (Original post by drandy76)
    Instructions not clear enough, ended up with 10 different tabs open


    Posted from TSR Mobile
    BTW: Look at Q3 here.
    Offline

    11
    ReputationRep:
    (Original post by drandy76)
    Argument makes perfect sense up to the last two lines, granted I don't have much experience with Taylor expansions so that's no great suprise, it looks 'nicer' than the way I used though


    Posted from TSR Mobile
    I don't use a Taylor expansion at all. The argument is as follows:

    1) For the upper line, i.e. the tangent, f'(x) is more negative over the whole range, so f(x) must always lie below that line as it decreases more quickly. So that part of the inequality is done.

    2) We need to show that f(x) always lies above the other line, the chord. I split this into two parts.

    First, I show that, since f'(x) eventually exceeds the gradient of the chord, then there is some point x_0 where f'(x) must be equal to that gradient - so in the interval between 0 and x_0, the average gradient of f(x) is less than than of the chord, so it decreases more slowly in this region, so is always above the chord in (0,x_0)

    Then we have to show that it is above the chord even in the region where f(x) is decreasing faster than then chord. I do this by contradiction: essentially I say that if at some point f(x) was below the chord in this region, then it would stay below it thereafter, since f(x) is decreasing faster than the chord - the chord could never "catch it up" again.

    But we know that the chord and f(x) meet at \pi/4 i.e. we know that f(x) is *not* below the chord, at the end of the region in question. Hence, there can be no point in the region where it was below the chord i.e. the chord is always below f(x). QED.
    • Thread Starter
    Offline

    12
    ReputationRep:
    (Original post by Zacken)
    BTW: Look at Q3 here.
    I'll give it a go now


    Posted from TSR Mobile
    • Thread Starter
    Offline

    12
    ReputationRep:
    (Original post by atsruser)
    I don't use a Taylor expansion at all. The argument is as follows:

    1) For the upper line, i.e. the tangent, f'(x) is more negative over the whole range, so f(x) must always lie below that line as it decreases more quickly. So that part of the inequality is done.

    2) We need to show that f(x) always lies above the other line, the chord. I split this into two parts.

    First, I show that, since f'(x) eventually exceeds the gradient of the chord, then there is some point x_0 where f'(x) must be equal to that gradient - so in the interval between 0 and x_0, the average gradient of f(x) is less than than of the chord, so it decreases more slowly in this region, so is always above the chord in (0,x_0)

    Then we have to show that it is above the chord even in the region where f(x) is decreasing faster than then chord. I do this by contradiction: essentially I say that if at some point f(x) was below the chord in this region, then it would stay below it thereafter, since f(x) is decreasing faster than the chord - the chord could never "catch it up" again.

    But we know that the chord and f(x) meet at \pi/4 i.e. we know that f(x) is *not* below the chord, at the end of the region in question. Hence, there can be no point in the region where it was below the chord i.e. the chord is always below f(x). QED.
    Thanks for elaborating for me, I maintain that your solution is nicer though, I think the confusing stemmed from the Xo and your comment at the start


    Posted from TSR Mobile
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 5, 2016
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.