Author Topic: TT's Maths Thread (Read 157294 times) Tweet Share

kamil9876 · « **Reply #750 on:** January 15, 2010, 01:51:12 am »

you can do it geometrically

The red line is |sina-sinb| while |a-b| is that arc length right next to it (obviously longer)

kamil9876 · « **Reply #751 on:** January 15, 2010, 02:01:14 am »

But if you must... bore yourself away with calculus:

$(sin(x))'=cosx$

Therefore there exists a $c$ such that:

$<br />sin(a)-sin(b)=cos(c)(a-b)$

$cos(c)=\frac{sin(a)-sin(b)}{(a-b)}$

but $-1\leq cos(c)\leq 1$

so:

$-1\leq \frac{sin(a)-sin(b)}{a-b}\leq 1$

result now easily follows.

TrueTears · « **Reply #752 on:** January 15, 2010, 02:04:35 am »

Nice, well I don't have a choice do I, Stewarts is a calculus book afterall

TrueTears · « **Reply #753 on:** January 15, 2010, 11:38:27 pm »

Show that $\sqrt{1+x} < 1+ \frac{1}{2}x \ \forall \ x>0$ using MVT.

I know this question seems pretty trivial (it is

) but how would one do it using MVT?

kamil9876 · « **Reply #754 on:** January 15, 2010, 11:53:06 pm »

$(\sqrt{1+x})'=\frac{1}{2\sqrt{x+1}}$

Therefore there exists a $c$ such that $0<c<x$ and:

$x* \frac{1}{2\sqrt{1+c}}=\sqrt{1+x}-1$

$\implies \frac{1}{2}>\frac{1}{2\sqrt{1+c}}=(\sqrt{1+x}-1)(\frac{1}{x})$

$\therefore \frac{1}{2}>(\sqrt{1+x}-1)(\frac{1}{x})$

From which the result follows.

TrueTears · « **Reply #755 on:** January 15, 2010, 11:59:41 pm »

Thanks that was pr0

TrueTears · « **Reply #756 on:** January 16, 2010, 08:13:56 pm »

$\lim_{x \to \infty} x^{\left[\frac{\log_e(2)}{1+\log_e(x)}\right]}$

How to find the limit to this indeterminate case?

Damo17 · « **Reply #757 on:** January 16, 2010, 08:42:55 pm »

Quote from: TrueTears on January 16, 2010, 08:13:56 pm

$\lim_{x \to \infty} x^{\left[\frac{\log_e(2)}{1+\log_e(x)}\right]}$

How to find the limit to this indeterminate case?

$y=x^{\frac{\ln(2)}{1+\ln(x)}}$

$ln(y)=\frac{ln(2)ln(x)}{1+ln(x)}$

$\therefore \lim_{x \to \infty} \frac{ln(2)ln(x)}{1+ln(x)} =\lim_{x \to \infty} \frac{ln(2) \frac{1}{x}}{\frac{1}{x}}=\lim_{x \to \infty} ln(2)=ln(2)$

$\therefore \lim_{x \to \infty} x^{\frac{\ln(2)}{1+\ln(x)}}=\lim_{x \to \infty} e^{ln(y)}=e^{ln(2)}=2$

TrueTears · « **Reply #758 on:** January 16, 2010, 08:45:38 pm »

Thanks, that is the way Stewarts does it and is pretty formulaic but yeah again I was wondering if there is a method besides L'hospital to bring out the aesthetic nature of this question.

humph · « **Reply #759 on:** January 17, 2010, 03:47:44 am »

You don't need l'hopital, just take logs, divide top and bottom by $\log x$ , and use the fact that $\log x \to \infty$ as $x \to \infty$ implies that $1/\log x \to 0$ as $x \to \infty$ .
That is, we use the properties of the exponential and logarithm functions:
$\begin{split}\lim_{x \to \infty} x^{\log 2/(1 + \log x)} & = \lim_{x \to \infty} \exp \left(\log x^{\log 2/(1 + \log x)}\right) \\ & = \lim_{x \to \infty} \exp \left(\frac{\log 2 \log x}{1 + \log x}\right) \\ & = \lim_{x \to \infty} \exp \left(\frac{\log 2}{\frac{1}{\log x} + 1}\right) \\ & = \exp \left(\lim_{x \to \infty} \frac{\log 2}{\frac{1}{\log x} + 1}\right) \\ & = \exp(\log 2) \\ & = 2\end{split}$
(you can take the limit inside the exp because the exponential function is continuous)

TrueTears · « **Reply #760 on:** January 17, 2010, 04:02:12 am »

Quote from: humph on January 17, 2010, 03:47:44 am

You don't need l'hopital, just take logs, divide top and bottom by $\log x$ , and use the fact that $\log x \to \infty$ as $x \to \infty$ implies that $1/\log x \to 0$ as $x \to \infty$ .
That is, we use the properties of the exponential and logarithm functions:
$\begin{split}\lim_{x \to \infty} x^{\log 2/(1 + \log x)} & = \lim_{x \to \infty} \exp \left(\log x^{\log 2/(1 + \log x)}\right) \\ & = \lim_{x \to \infty} \exp \left(\frac{\log 2 \log x}{1 + \log x}\right) \\ & = \lim_{x \to \infty} \exp \left(\frac{\log 2}{\frac{1}{\log x} + 1}\right) \\ & = \exp \left(\lim_{x \to \infty} \frac{\log 2}{\frac{1}{\log x} + 1}\right) \\ & = \exp(\log 2) \\ & = 2\end{split}$
(you can take the limit inside the exp because the exponential function is continuous)

Exactly what I was looking for!!!!

Thanks humph!!

TrueTears · « **Reply #761 on:** January 17, 2010, 05:37:22 pm »

$\int_0^3 \left(x^3-6x\right)dx$ using Riemann Sum.

My attempt, not sure if it's right. I'm not too worried about the answer but rather the setting out.

$\int_0^3 \left(x^3-6x\right)dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x$

$\Delta x = \frac{3-0}{n} \implies x_i = i\left(\frac{3}{n}\right)$

$\lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x = \lim_{n \to \infty} \sum_{i=1}^n f\left(i\left(\frac{3}{n}\right)\right)\left(\frac{3}{n}\right)$

$\implies \lim_{n \to \infty} \sum_{i=1}^n \left[\left(\frac{3i}{n}\right)^3-6\left(\frac{3i}{n}\right)\right]\left(\frac{3}{n}\right)$

$\implies \lim_{n \to \infty} \left(\frac{3}{n}\right) \sum_{i=1}^n \left[\left(\frac{27i^3}{n^3}\right)-\left(\frac{18i}{n}\right)\right]$

$\implies \lim_{n \to \infty} \left(\frac{3}{n}\right) \left[\sum_{i=1}^n \frac{27}{n^3}i^3 - \sum_{i=1}^n \frac{18}{n}i\right]$

First remember the result $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$

Now to find $\sum_{k=1}^n k^3$ we need to find some polynomial containing $k^3$ .

To telescope: let $u_{k+1} - u_k = \mbox{some polynomial with} \ k^3$

Now let $u_k = k^4$

Thus $u_{k+1} - u_k = (k+1)^4 - k^4$

Now $\sum_{k=1}^n u_{k+1} - u_k = u_2 - u_1 + u_3 - u_2 + ... + u_{n+1} - u_n = u_{n+1} - u_1 = (n+1)^4 - 1$

But $\sum_{k=1}^n (k+1)^4 - k^4 = \sum_{k=1}^n (4k^3 + 6k^2 + 4k+1)$

$\implies \sum_{k=1}^n 4k^3 + 6k^2 + 4k+1 = (n+1)^4 - 1$

$\implies 4\sum_{k=1}^n k^3 = (n+1)^4 - 1 - 6\sum_{k=1}^n k^2 - \sum_{k=1}^n (4k+1)$

$\implies 4\sum_{k=1}^n k^3 = (n+1)^4 - 1 - n(n+1)(2n+1) - \frac{n}{2}(5+4n+1)$

$\therefore \sum_{k=1}^n k^3 = \frac{n^2}{4}\left(n^2+2n+1\right)$

$\implies \lim_{n \to \infty} \left(\frac{3}{n}\right) \left[\frac{27}{n^3}\left(\frac{n^2}{4}\left(n^2+2n+1\right)\right) - \frac{18}{n}\left(\frac{n(n+1)}{2}\right)\right]$

$\implies \lim_{n \to \infty} \left(\frac{3}{n}\right) \left[\frac{27}{n}\left(\frac{1}{4}\left(n+1\right)^2\right) - 18\left(\frac{(n+1)}{2}\right)\right]$

$\implies \lim_{n \to \infty} \frac{3(n+1)}{n}\left(\frac{27}{4}-9+\frac{27}{4n}\right)$

$\implies \lim_{n \to \infty} \left(3+\frac{3}{n}\right)\left(\frac{27}{4n}-\frac{9}{4}\right) = -\frac{27}{4}$

brightsky · « **Reply #762 on:** January 17, 2010, 06:07:49 pm »

TrueTears · « **Reply #763 on:** January 17, 2010, 06:12:12 pm »

I just found out both Chopin and Riemann died at the age of 39 and of the same cause of death: tuberculosis

fukn tuberculosis killed my fave composer and a great mathematician.

kamil9876 · « **Reply #764 on:** January 17, 2010, 07:10:50 pm »

I remember once proving for the general case of $\int x^n dx$ . Motivation for this came when i was once playing with $\sum_{i=0}^k i^n$ and realised it's an n+1 degree polynomial of k with where $k^{n+1}$ coefficient is $\frac{1}{n+1}$

Search the forums now!

We have moved!

Author Topic: TT's Maths Thread (Read 157294 times) Tweet Share

kamil9876

Re: TT's Maths Thread

kamil9876

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

kamil9876

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

Damo17

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

humph

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

brightsky

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

kamil9876

Re: TT's Maths Thread

Recent Posts

User:
Password: