Mean Value Theorem proofs (1 Viewer)

leehuan · Apr 15, 2016

So I got part a) out. Need only a small starter on part b). But it's mainly cause of the process.

Q: Use the MVT to prove that
a) ln(1+x)<x for x>0
b) -ln(1-x)<x/(1-x) for 0<x<1

So for part a) I defined f:[0,x], f(t)=ln(1+t) and used the MVT to get ln(1+x)/x=1/c (for some c in (0,x))
So cause c is positive ln(1+x)/x < 1 and x is positive so ln(1+x)<x. Nice and easy

But I'm only really concerned with defining the function. For x>0 I could just use [0,x].
How would I approach 0< x
< 1
?

Edit: This forum can't handle too many <'s it seems. Sorry

InteGrand · Apr 15, 2016

leehuan said:
So I got part a) out. Need only a small starter on part b). But it's mainly cause of the process.

Q: Use the MVT to prove that
a) ln(1+x)<x for x>0
b) -ln(1-x)<x/(1-x) for 0<x<1

So for part a) I defined f:[0,x], f(t)=ln(1+t) and used the MVT to get ln(1+x)/x=1/c (for some c in (0,x))
So cause c is positive ln(1+x)/x < 1 and x is positive so ln(1+x)<x. Nice and easy

But I'm only really concerned with defining the function. For x>0 I could just use [0,x].
How would I approach 0< x
< 1
?

For b), try defining f(t) = -ln(1-t) for 0 <= t<= x, where x is some arbitrary fixed number in (0,1).

kawaiipotato · Apr 16, 2016

Can also be done by defining f(t) = (x-1)ln(1-x) for t E (0,1)

leehuan · Apr 16, 2016

Oh yea I also used that method for some of the later questions lol

laters · Apr 16, 2016

Can anyone prove $\frac{4}{27}<3-23^{\frac{1}{3}}<\frac{3}{16}$ using the mean value theorem? I've been trying to come up with functions that I can apply the MVT to but I've gotten stuck.

Also for $\frac{8}{145}<\tan^{-1} \frac{9}{8}-\frac{\pi}{4}<\frac{1}{16}$

InteGrand · Apr 16, 2016

laters said:
Can anyone prove $\frac{4}{27}<3-23^{\frac{1}{3}}<\frac{3}{16}$ using the mean value theorem? I've been trying to come up with functions that I can apply the MVT to but I've gotten stuck.

Also for $\frac{8}{145}<\tan^{-1} \frac{9}{8}-\frac{\pi}{4}<\frac{1}{16}$

$\noindent Hint for the first one: we can use $f(t) = t^{\frac{1}{3}}$, defined for $t\in \left[a,b\right]$, where $a=23$ and $b=27$.$

$\noindent Hint for the second one: an arctan function would be used, with the domain having $1$ being involved of course, as well as $\frac{9}{8}$.$

leehuan · Apr 16, 2016

laters said:
Can anyone prove $\frac{4}{27}<3-23^{\frac{1}{3}}<\frac{3}{16}$ using the mean value theorem? I've been trying to come up with functions that I can apply the MVT to but I've gotten stuck.

Also for $\frac{8}{145}<\tan^{-1} \frac{9}{8}-\frac{\pi}{4}<\frac{1}{16}$

Basically what InteGrand said, but it actually takes nothing more than a tiny bit of intuition to figure out part d). It's just arctan over [1, 9/8] (Or you can do arctan [1,x] and then let x=9/8 but there's no need)

laters · Apr 16, 2016

Thanks leehuan and Integrand!! I managed to get the arctan one out.

However for the first one I asked, I just want some clarification. I know 23 < c < 27 so if c < 27 then I can obtain the lower bound. But obviously if I use c>23 then I would require a calculator to show that the 3-23^(1/3)< 0.16..< 3/16, which defeats the purpose of the question. c>512/27 would allow this, but I didn't apply the MVT in (512/27, 27). What did you do?

InteGrand · Apr 16, 2016

laters said:
Thanks leehuan and Integrand!! I managed to get the arctan one out.

However for the first one I asked, I just want some clarification. I know 23<c<27 so if c<27 so I can obtain the lower bound. But obviously if I use c>23 then I would require a calculator to show that the 3-23^(1/3)<0.16..<3/16, which defeats the purpose of the question. c>512/27 would allow this, but I didn't apply the MVT in (512/27, 27). What did you do?

We don't need a calculator.

$\noindent Define $f:\left[a,b\right] \to \mathbb{R}$, where $a=23,b=27$, and $f(t) = t^{\frac{1}{3}}$ for $t\in \left[a,b\right]$. Then $f$ is continuous on the closed interval $\left[a,b\right]$ and differentiable on the corresponding open interval, so by the Mean Value Theorem, there exists a $c$ with $23<c<27$ satisfying$

$$\frac{f(b)-f(a)}{b-a}=f^\prime (c)\quad (*)$.$

$\noindent You should be able to get it out from here, if you note that $\frac{512}{27}<c<27$ $\Big{(}$because $23>\frac{512}{27}$. This part can be deduced without a calculator, e.g. by hand or mental arithmetic, $20\times 27 = 540 > 512 \Rightarrow \frac{512}{27}<20 < 23$.$\Big{)}$.$

leehuan · Apr 17, 2016

Please start me off on the processes in the inductive step.

$Use mathematical induction to show that the equation $\\ \sum_{k=0}^{n}{\frac{{x}^{k}}{k!}} = 0\\ $has: \\ - one real, negative solution if $n$ is odd\\ - no real solutions if $n$ is even$

I have no clue how to apply the hypothesis here...

InteGrand · Apr 17, 2016

leehuan said:
Please start me off on the processes in the inductive step.

$Use mathematical induction to show that the equation $\\ \sum_{k=0}^{n}{\frac{{x}^{k}}{k!}} = 0\\ $has: \\ - one real, negative solution if $n$ is odd\\ - no real solutions if $n$ is even$

I have no clue how to apply the hypothesis here...

$\noindent I think you proved the results for small $n$. Assume the result holds for some particular odd integer $(n-1)$ and even integer $n \geq 2$. We will show this implies the result for $n+1$ (odd) and $n+2$ (even).$

$\noindent Call the polynomial with degree $n$ $p_n$. Since $p_n$ has no real roots by inductive assumption, $p_{n+1}$ is an odd-degree polynomial whole derivative has no real roots (since $p_{n+1}^{\prime}=p_{n}$). So $p_{n+1}$ is monotone, since its derivative is of fixed sign. Since $p_{n+1}$ is a monotone odd-degree polynomial, it has precisely one real root (standard polynomials result). Also, the root has to be negative, because $p_{n+1}(x)$ is clearly positive for all $x\geq 0$. So half the induction step is done.$

$\noindent For the next part, we show that $p_{n+2}$ (even degree polynomial) has no real roots. This proof is similar to the one for the $p_{4}$ case in another thread. I.e. $p_{n+2}'' (x) \equiv p_{n}(x)>0$ for all real $x$ by the inductive hypothesis and the fact that $p_n(x)$ has positive leading coefficient. So the local min. value of $p_{n+2}(x)$ is in fact a global min. As before, we set the derivative to 0 to find where the global minimum is attained. It is attained at the point $a$ which is the solution to $p_{n+1}(x)=0$, since $p_{n+1}(x) \equiv p_{n+2}^{\prime}(x)$ (we know such a point exists and is unique as we proved it above). But $p_{n+2}(x) \equiv p_{n+1}(x) + \frac{x^{n+2}}{(n+2)!}$, so $p_{n+2} (a)=0 +\text{positive number}>0$. Hence the global minimum value of $p_{n+2} (x)$ is a positive number, so $p_{n+2} (x)>0$ for all real $x$. So there are no real solutions to the equation $p_{n+2} (x)=0$ and the inductive step is complete.$

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Mean Value Theorem proofs (1 Viewer)

leehuan

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