Apply Lagrange's Theorem to demonstrate inequality

**Utherr** · June 21st 2010, 10:39 AM

Hi,

This is my first post. I have a slight problem. I know how to solve this exercise but there's something different here :/.

Ok, so i need to demonstrate this inequality by applying Lagrange's theorem on [k, k+1] interval, i have the solution in my book but the theorem i know is f ' (c) = [f(k+1) - f(k)]/(k + 1 - k), instead the book says its the other way around f(k) - f(k+1) ... etc

So if some1 can solve this problem for me from scratch ... i'd be grateful.

SOLVED

**HallsofIvy** · June 21st 2010, 10:49 AM

Originally Posted by Utherr

Hi,

This is my first post. I have a slight problem. I know how to solve this exercise but there's something different here :/.

Ok, so i need to demonstrate this inequality by applying Lagrange's theorem on [k, k+1] interval, i have the solution in my book but the theorem i know is f ' (c) = [f(k+1) - f(k)]/(k + 1 - k), instead the book says its the other way around f(k) - f(k+1) ... etc

So if some1 can solve this problem for me from scratch ... i'd be grateful.

No, no one will do your homework for you- but we are willing to help. You do understand that f(k)- f(k+1) is just the negative of f(k+1)- f(k) don't you? Now, what was the rest of that "etc."?

**maddas** · June 21st 2010, 10:51 AM

Isn't Lagrange's theorem is something about the order of groups and subgroups?

Anyway, there is a $k < \xi < k+1$ s.t. $f'(\xi) = f(k+1) -f(k)$ by the mean value theorem. Since $f'(x) = {-1\over 2 x\sqrt{x}}$ , f' is monotonically increasing. Therefore ${-1\over 2 k\sqrt{k}} < {1\over\sqrt{k+1}} - {1\over\sqrt{k}} < {-1\over 2 (k+1)\sqrt{k+1}}$ . Negate this.

**Utherr** · June 21st 2010, 10:53 AM

Originally Posted by HallsofIvy

No, no one will do your homework for you- but we are willing to help. You do understand that f(k)- f(k+1) is just the negative of f(k+1)- f(k) don't you? Now, what was the rest of that "etc."?

I don't need some1 to do my homeworking, i can do it myself, i just needed some help.

Thanks, for the poster above me, i know where i made my mistake, forgot the minus sign from the derivative of f(x) :shy:

Here's how i solved it:

I'm on my last year of high-school ... i have my exams in a week, im not from the US or UK and i don't know how the theorems sound like in english (or the maths terminology in english)

**Utherr** · June 23rd 2010, 05:11 AM

Hey, i decided to post my math problems here instead of making a bunch of "little" topics, that'd be messy, i hope you don't mind this.

And here i am with a new problem:

I know that f(x) is monotonically increasing for x < 0, and monotonically decreasing for x > 0, f(0) = 1, f(1) = f(-1) = 1/4

Please don't think i registered on this forum just so you can make my homework, it's not my homework, i'm trying to learn for the exam (SAT equiv. in the US), i have the book with all the variants, 100 in number, the exam is 7 days, and i still have some "minor" exercises from all those 100 variants.

**roninpro** · June 23rd 2010, 05:44 AM

Did you try taking the integral first? There is a fairly clean closed-form antiderivative.

**Utherr** · June 23rd 2010, 06:17 AM

this is all i could to it

:-s

**Utherr** · June 24th 2010, 12:05 PM

Another problem :-s

I have to prove that the integral $\to\infty$ so i can apply L'Hospital ... but how to prove this i have no idea, can some1 enlighten me?

**Ackbeet** · June 24th 2010, 12:11 PM

First off: the rule in this forum is, no more than two problems per thread. You should move this problem. It's for your benefit as much as anyone else's. People are less likely to help with a problem if it has lots of replies.

I would go with a $u=\ln(t)$ substitution first. [EDIT] Whoops, that doesn't work. Let me think about that a bit more.

**Utherr** · June 24th 2010, 12:13 PM

ok, sorry. I didn't want to "overload" this board with my topics

**Ackbeet** · June 24th 2010, 12:13 PM

Use the Fundamental Theorem of the Calculus to take the derivative upstairs. I don't think you'll have to evaluate that integral.

**Ackbeet** · June 24th 2010, 12:14 PM

This is the Calculus forum. There are over 35,000 threads. Don't worry about overloading!

**Also sprach Zarathustra** · June 24th 2010, 12:18 PM

$arctgx<\frac{\pi}{2}$

so, $arctg(lnx)<\frac{\pi}{2}$

**Ackbeet** · June 24th 2010, 12:22 PM

You have $\lim_{x\to\infty}\tan^{-1}(x)=\frac{\pi}{2}$ . The derivative of the arctan function is always positive, hence you have a monotone increasing function. Moreover, $\tan^{-1}(1)=\frac{\pi}{4}$ . From those pieces of information, and the fact that the logarithm function is monotone increasing, I think you can infer that the integral in the numerator is infinite.

**Utherr** · June 24th 2010, 12:29 PM

The book gives me a solution but it's ambiguous for me, i don't understand much from it

I swear, this is all they give me for the solution why that integral -> inf

Maybe this will help you, because it didn't helped me at all.

Damn, not too familiar with mathematics in english O.o (i don't know the terminology in English)

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