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I have this question on my assignment which I just cannot seem to wrap my brain around. I've been reluctant to post it to any forum because I don't like the answer handed to me, but I seriously do need some help with this:

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I'm not sure how I am meant to use the MVT with this, but I have been thinking about this question for awhile. So since I've made an assumption that the second derivative is positive, and through this I can make the statement that, since our approximation is at a (the start of the interval), the worst approximation you can get is when x is the furthest the way it can be on the interval, i.e. x=b. here is my working so far:

|t(x)-f(x)|

t(x)= f(a)+f'(a)(x-a)

MVT: f(b) = f(a)+f'(c)(b-a)

Worst approximation, which will give maximum error when x=b

|f(a) +f'(a)(b-a)-f(b)|

|f(a) +f'(a)(b-a)-f(a) +f'(c)(b-a)|

|f'(a)(b-a)-f'(c)(b-a)|

|(f'(a)-f'(c))(b-a)|

I am very lost, and am not sure how to apply the MVT twice? Have I even applied it once with my f(b)= ... ? I would appreciate any help, but would prefer a push in the right direction over being told how to do it straight up. Thanks in advance. :)

Also, I believe it can be solved using Taylor's Series, but seeing as we haven't learnt about it and that it specifically says MVT in the hint, I don't think I should be using it.

Skylineblue
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1 Answers1

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First, $f(x) - T(x) = f(x) - f(a) - f'(a)(x - a)$. By MVT, $f(x) - f(a) = f'(c)(x - a)$ for some $c\in (a,x)$. Then $f(x) - T(x) = [f'(c) - f'(a)](x - a)$. Apply the mean value theorem again to $f'(c) - f'(a)$ and try to complete the argument.

kobe
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  • Thanks for your help.

    Was the question worded wrongly? Why are you saying f(x) - t(x) instead of |t(x) - f(x)|?

    I've also been stuck at that point, how can I apply MVT to f'(c) - f'(a) ?

     – Skylineblue Oct 02 '16 at 06:45
  • @Skylineblue the question was worded correctly. Note that $f(x) - T(x)$ and $T(x) - f(x)$ have the same absolute value. It just may be easier (mentally speaking) to deal with $f(x) - T(x)$ than with $T(x) - f(x)$. To apply MVT to $f'(c) - f'(a)$, note that $f'$ is continuous on $(a,c)$ and differentiable on $[a,c]$; then MVT gives $f'(c) - f'(a) = f''(d)(c - a)$ for some $d\in (a,c)$. – kobe Oct 02 '16 at 06:48
  • So I get f''(d)(c-a)(x-a). This is more what I'm looking for, how can I transform it into the form I want? I can see I want to make c and x = b. And then the f''(d) to = k. I'm contemplating how I can do this. Would I be working under the assumption that the worst case (in order to get the maximum possible error) that both c and x would have to be b? How do I turn the f''(d) into k? – Skylineblue Oct 02 '16 at 07:24
  • @Skylineblue by hypothesis, $\lvert f''(x)\rvert \le \kappa$ for all $x\in (a,b)$. In particular, $\lvert f''(d)\rvert \le \kappa$. Therefore $$\lvert T(x) - f(x)\rvert = \lvert f(x) - T(x)\rvert = \lvert f''(d)\lvert (c - a)(x - a) \le \kappa (b - a)(b - a) = \kappa (b - a)^2$$ – kobe Oct 02 '16 at 07:30
  • Awesome. Thank you so much. – Skylineblue Oct 02 '16 at 07:31
  • Hey Kobe. I've gotten another assignment which carries on from this question, but I'm even more stuck than when I attempted this one the first time.

    https://gyazo.com/33b9fe31c4cb350d0d168addc62567c2

    Any thoughts on where I should start?

     – Skylineblue Nov 06 '16 at 03:33
  • @Skylineblue since $-\kappa \le f''(t) \le \kappa$ for all $t\in (a,b)$, then given $x\in (a,b)$, we have $\int_a^x -\kappa, dt \le \int_a^x f''(t), dt \le \int_a^x \kappa, dt$. Simplify these integrals, then integrate again from $a$ to $x$ and simplify. – kobe Nov 06 '16 at 03:48
  • Thanks Kobe, you're seriously a legend. I'll give this a try now. The integral of f''(t) is f'(t) right? – Skylineblue Nov 06 '16 at 03:52
  • I've done the first integration and have gotten: -k(x-a)≤f'(x)-f'(a)≤k(x-a) You've said to take the integral again from a to x. Is this with respect to t again? How can I integrate the functions in the middle? – Skylineblue Nov 06 '16 at 03:59
  • @Skylineblue what you have is correct. It's valid for all $x\in (a,b)$. So now you can write $-\kappa(t - a) \le f'(t) - f'(a) \le \kappa(t - a)$ and perform integrations from $t = a$ to $t = x$ like before. – kobe Nov 06 '16 at 04:05
  • I can see by integrating the outsides of the inequalities we get the (k(t-a)^2)/2 that we are looking for, and from this we can make the substitution for t=b since we are looking at the error. But I am struggling to see how I integrate the middle. Does it become f(t)-t*f'(a), then we use our from t=a to t=x? – Skylineblue Nov 06 '16 at 04:13
  • @Skylineblue you do $\int_a^x -\kappa(t - a), dt \le \int_a^x [f'(t) - f'(a)], dt \le \int_a^x \kappa(t - a), dt$. From this you should get $-\kappa(x - a)^2/2 \le f(x) - f(a) - f'(a)(x - a) \le \kappa(x - a)^2/2$. – kobe Nov 06 '16 at 04:17
  • Awesome. I've gotten to the point of writing |f(x)-t(x)|≤k(x-a)^2/2. How do I rationalise the substitution of using b instead of x? – Skylineblue Nov 06 '16 at 04:35
  • @Skylineblue since $a < x < b$, then $0 < x - a < b - a$. Therefore $(x - a)^2 < (b - a)^2$. Now you can complete the argument. – kobe Nov 06 '16 at 04:39
  • Thank you again, so much. You've been an incredible help twice now. – Skylineblue Nov 06 '16 at 04:41
  • @Skylineblue no problem, take care! – kobe Nov 06 '16 at 04:42
  • Why is it that for all t in (a, b) we have -k≤f''(t)≤k? – Skylineblue Nov 06 '16 at 06:21
  • @Skylineblue it's given that $f"$ is bounded by $\kappa$ on $(a,b)$. That means $|f"(t)| \le \kappa$ on $(a,b)$, or $-\kappa \le f"(t) \le \kappa$ for all $t\in (a,b)$. – kobe Nov 06 '16 at 06:25