Intermediate Value Theorem Question (1 Viewer)

VBN2470 · Apr 7, 2014

Can someone please answer the following

Thanks

Carrotsticks · Apr 7, 2014

Can you show us what you've done so far? So we can help you from there rather than just doing it for you.

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Sy123 · Apr 7, 2014

For the second part I am getting

$a \in \left[0 , \frac{n-1}{n} \right]$

VBN2470 · Apr 7, 2014

Sy123 said:
For the second part I am getting

$a \in \left[0 , \frac{n-1}{n} \right]$

Yep, sorry I've fixed it up for you now, what you said is right

Would you mind explaining the question to me so I can see where to start and learn how to approach this type of question?

By the way, the actual question has given a hint by letting $g(x)=f(x+\frac{1}{2})-f(x)$ . I just wanted to see how you would approach this question given a hint was not provided.

Thanks

Sy123 · Apr 7, 2014

I did not use the hint, I did this instead:

$\\ $a) Take some$ \ k \ $in$ \ [0,1] \ $since$ \ f \ $is continuous on$ \ [0,1] \ $then is continuous on$ \ [0,k] \ $and$ \ [k, 1] \\ $By the intermediate value theorem then$ \ f \ $takes every value in$ \ [f(0) , f(k)] \ $by the first interval and takes every value in$ \ [f(k) , f(1)] \ $by the second interval$ \ \\ $Since$ \ f(0) = f(1) \ $then it takes every value from$ \ [0,1] \ $twice$ \\ $Hence for some arbitrary$ \ m \ $in$ \ [0,1] \ $there is$ \ n \ $in$ \ [0,1] \ $so that$ \ \ f(m) = f(j) \\ $Since this holds for any$ \ m \ $in$ \ [0,1] \ $then pick$ \ a \ $such that$ \ j = a + \frac{1}{2} \\ $hence$ \ f(a) = f\left( a + \frac{1}{2} \right) \ $and since$ \ \frac{1}{2} \leq a + \frac{1}{2} \leq 1 \ $then$ \ a \ $must be in$ \ [0 , \frac{1}{2}]$

$\\ $b) Using the same method we pick some$ \ a \ $such that$ \ j = a + \frac{1}{n} \\ $and since$ \ j \leq 1 \ $then$ \ a \ $is in$ \ [0, \frac{n-1}{n} ]$

RealiseNothing · Apr 7, 2014

USYD, VBN2470? caps

Sy123 · Apr 7, 2014

Also some changes should be made slightly to the terminology, on line 5 it should say

"then it takes every value from [0,1] at least twice"

Sy123 · Apr 7, 2014

But using the hint:

$g(x) = f\left( x + \frac{1}{2} \right) - f(x)$

$\\g\left(\frac{1}{2} \right) + g(0) = f\left(1\right) - f \left(\frac{1}{2}\right) + f\left(\frac{1}{2} \right) - f(0) = 0 \\ $Hence$ \ g\left(\frac{1}{2} \right) = - g(0)$

$\\ g \ $is continuous over$ \ [0, 1/2] \ $and since$ \ g(0) < 0 < g(1/2) \ $or$ \ g(1/2) < 0 < g(0) \ $there must be some$ \ x \ $so that$ \ g(x) = 0 \\ $Hence, some$ \ x= a \ g(a) = 0 \Rightarrow \ f(a+1/2) = f(a)$

VBN2470 · Apr 7, 2014

Sy123 said:
I did not use the hint, I did this instead:

$\\ $a) Take some$ \ k \ $in$ \ [0,1] \ $since$ \ f \ $is continuous on$ \ [0,1] \ $then is continuous on$ \ [0,k] \ $and$ \ [k, 1] \\ $By the intermediate value theorem then$ \ f \ $takes every value in$ \ [f(0) , f(k)] \ $by the first interval and takes every value in$ \ [f(k) , f(1)] \ $by the second interval$ \ \\ $Since$ \ f(0) = f(1) \ $then it takes every value from$ \ [0,1] \ $twice$ \\ $Hence for some arbitrary$ \ m \ $in$ \ [0,1] \ $there is$ \ n \ $in$ \ [0,1] \ $so that$ \ \ f(m) = f(j) \\ $Since this holds for any$ \ m \ $in$ \ [0,1] \ $then pick$ \ a \ $such that$ \ j = a + \frac{1}{2} \\ $hence$ \ f(a) = f\left( a + \frac{1}{2} \right) \ $and since$ \ \frac{1}{2} \leq a + \frac{1}{2} \leq 1 \ $then$ \ a \ $must be in$ \ [0 , \frac{1}{2}]$

$\\ $b) Using the same method we pick some$ \ a \ $such that$ \ j = a + \frac{1}{n} \\ $and since$ \ j \leq 1 \ $then$ \ a \ $is in$ \ [0, \frac{n-1}{n} ]$

OK Thanks

Another question I was sort of stuck on:

How many real zeroes does $p(x)=x^3-12x^2+45x-51$ have?

VBN2470 · Apr 7, 2014

RealiseNothing said:
USYD, VBN2470? caps

UNSW

Sy123 · Apr 7, 2014

Similarly:

$g(x) = f(x + 1/n) - f(x)$

$g(0) = f(1/n) - f(0)$

$g(1/n) = f(2/n) - f(1/n)$

$g(2/n) = f(3/n) - f(2/n)$

.
.
.

$g((n-1)/n) = f(1) - f(0)$

Add them side by side:

$(1): \ \ \ \sum_{k=0}^{n-1} g\left(\frac{k}{n} \right) = 0 \ \ \ $(telescoping and cancellation of$ \ f(1) \ $and$ \ f(0) $)$$

$\\ $Assume that$ \ g(x) \neq 0 \ $then$ \ g\left(\frac{k}{n} \right) > 0 \ $or$ \ g\left(\frac{k}{n} \right) < 0 \ \ $for all$ \k \\ $however this contradicts$ \ (1) \ $Therefore a contradiction arises meaning there is at least one$ \ k \ $so that$ \ g\left(\frac{k}{n} \right) < 0 \\ $Since$ \ g \ $is continuous, by IVT it follows that there must be a root to$ \ g \ $(similar logic to first part$ \\ $Therefore there is at least some$ \ a \ $so that$ \ g(a) = 0 \Rightarrow \ f(a + 1 /n) = f(a)$

RealiseNothing · Apr 7, 2014

VBN2470 said:
OK Thanks

Another question I was sort of stuck on:

How many real zeroes does $p(x)=x^3-12x^2+45x-51$ have?

$P'(x) = 3(x-3)(x-5)$

Check $P(3)$ and $P(5)$

Use IVT to justify the amount of real roots.

Sy123 · Apr 7, 2014

VBN2470 said:
OK Thanks

Another question I was sort of stuck on:

How many real zeroes does $p(x)=x^3-12x^2+45x-51$ have?

Try differentiating and looking at the stationary points, they should tell you something about the way the curve moves.

VBN2470 · Apr 7, 2014

Sy123 said:
Try differentiating and looking at the stationary points, they should tell you something about the way the curve moves.

Never mind I got it, as $P(3)$ and $P(5)$ yield opposite signs. Is there another way of doing this question, using IVT or Extreme Value Theorem or some other method? For example the polynomial $P(x)=x^3-x+2$ has only one real root, how would you be certain in finding the number of real zeroes existing?

Thanks

Sy123 · Apr 7, 2014

VBN2470 said:
Never mind I got it, as $P(3)$ and $P(5)$ yield opposite signs. Is there another way of doing this question, using IVT or Extreme Value Theorem or some other method?

Thanks

That is essentially IVT, since P(-n) and P(3) yield opposite signs for sufficiently large n (i.e. when n goes to inifinty), then there is some root in (-n, 3), similarly, P(3) and P(5) yield opposite signs, and P(5) and P(n), for sufficiently large n is yield opposite signs.

VBN2470 · Apr 7, 2014

Sy123 said:
That is essentially IVT, since P(-n) and P(3) yield opposite signs for sufficiently large n (i.e. when n goes to inifinty), then there is some root in (-n, 3), similarly, P(3) and P(5) yield opposite signs, and P(5) and P(n), for sufficiently large n is yield opposite signs.

OK Thanks

Carrotsticks · Apr 10, 2014

VBN2470 said:
Can someone please answer the following View attachment 30240

Thanks

In case anybody is curious, this is known as the 'Universal Chord Theorem'.

seanieg89 · Apr 10, 2014

Carrotsticks said:
In case anybody is curious, this is known as the 'Universal Chord Theorem'.

I'm pretty sure that this isn't the universal chord theorem. The universal chord theorem is that the numbers of the form 1/n are the ONLY real numbers such that the above theorem holds.

This is a little more subtle and difficult than the question that was asked.

Carrotsticks · Apr 10, 2014

seanieg89 said:
I'm pretty sure that this isn't the universal chord theorem. The universal chord theorem is that the numbers of the form 1/n are the ONLY real numbers such that the above theorem holds.

This is a little more subtle and difficult than the question that was asked.

My fault!

When I saw the f(a)=f(a+1/n), UCT was the first thing that came to mind and I had thought they were required to prove it.

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