Author Topic: TT's Maths Thread (Read 119591 times) Tweet Share

kamil9876 · « **Reply #645 on:** January 06, 2010, 02:50:04 pm »

yeah exactly, that's the first thing that comes to mind since it reminded me of that trick in the proof of:

a,2a,3a,4a...(p-1)a are all distinct modulo p. (notice that the numbers we have is the modulus less one, just like here).

TrueTears · « **Reply #646 on:** January 06, 2010, 06:33:50 pm »

Quote from: kamil9876 on January 06, 2010, 12:37:40 pm

Yeah it is. Though I know it also assumes in many instancse, like my proof, that the order of 2 modulo 101 is 100.

Yeah heaps of iff's....

And yeah, freaking awesome solution kamil.

TrueTears · « **Reply #647 on:** January 08, 2010, 05:10:18 pm »

Quote from: TrueTears on January 04, 2010, 08:48:34 pm

2. The number $d_1d_2 \cdots d_9$ has nine (not necessarily distinct) decimal digits. The number $e_1e_2 \cdots e_9$ is such that each of the nine $9$ -digit numbers formed by replacing just one of the digits $d_i$ is $d_1d_2 \cdots d_9$ by the corresponding digit $e_i \ (1 \le i \le 9)$ is divisible by $7$ . The number $f_1f_2 \cdots f_9$ is related to $e_1e_2 \cdots e_9$ in the same way: that is each of the nine numbers formed by replacing one of the $e_i$ by the corresponding $f_i$ is divisible by $7$ . Show that for each $i$ , $d_i - f_i$ is divisible by $7$ . (For example, if $d_1d_2 \cdots d_9 = 199501996$ then $e_6$ may be $2$ or $9$ , since $199502996$ and $199509996$ are multiples of $7$ ).

omfg fukn finally got this question.

Let $D = d_1d_2 \cdots d_9, \ E = e_1e_2 \cdots e_9, \ F = f_1f_2 \cdots f_9$

Now let's take a look at $D=199501996$ .

If we replace $d_6$ by $2$ or $9$ then we have created a number that is divisible by $7$ .

Which means $E = e_1e_2e_3e_4e_5(2)e_7e_8e_9$ or $E = e_1e_2e_3e_4e_5(9)e_7e_8e_9$

This means the other digits of $E$ can be anything but $e_6$ must be either $2$ or $9$ .

Let $D', E'$ be the numbers which are divisible by $7$ .

Using the same example and assume that we are replacing $d_6$ with $2$

$\implies D' = D + 1000 = D + (e_6-d_6)\left(10^{3}\right)$

$\implies D' = D + (e_i-d_i)\left(10^{9-i}\right)$

$\implies D + (e_i-d_i)\left(10^{9-i}\right) \equiv 0 \pmod {7} \cdots [1]$

Likewise we have $E' = E + (f_i-e_i)\left(10^{9-i}\right)$

$\implies E + (f_i-e_i)\left(10^{9-i}\right) \equiv 0 \pmod {7} \cdots [2]$

$[1]+[2] \implies D + (e_i-d_i)\left(10^{9-i}\right)+E + (f_i-e_i)\left(10^{9-i}\right) \equiv 0 \pmod {7}$

$\implies D+E + \left(10^{9-i}\right)(e_i) - \left(10^{9-i}\right)(d_i) + \left(10^{9-i}\right)(f_i)-\left(10^{9-i}\right)(e_i) \equiv 0 \pmod {7}$

$\implies D+E+\left(10^{9-i}\right)\left[f_i-d_i\right] \equiv 0 \pmod {7}$

But $D+E+\left(10^{9-i}\right)\left[f_i-d_i\right] \equiv \left(10^{9-i}\right)\left[f_i-d_i\right] \pmod {7}$

Thus $\left(10^{9-i}\right)\left[f_i-d_i\right] \equiv 0 \pmod {7}$

But $10^{9-i} \not\equiv 0 \pmod {7}$

$\therefore \left[f_i-d_i\right] \equiv 0 \pmod {7}$

$\text{Q.E.D}$

YAY FINISHED ART N CRAFT NOW I CAN FINALLY MOVE ON TO STEWARTS!!!!!

TrueTears · « **Reply #648 on:** January 09, 2010, 01:30:47 am »

$\lim_{x \to 7} \left(\frac{\sqrt{x+2}-3}{x-7}\right)$

I've rationalised the numerator and I got $\left(\frac{\sqrt{x+2}-3}{x-7}\right) = \frac{x-7}{3x+\sqrt{x+2}(x-7)-21}$

Now what?

kamil9876 · « **Reply #649 on:** January 09, 2010, 01:47:49 am »

let x=u+7

$\lim_{x \to 7} \left(\frac{\sqrt{x+2}-3}{x-7}\right)$

$= \lim_{u \to 0} \left(\frac{\sqrt{u+9}-3}{u}\right)$

Now you can rationalise to get something better

TrueTears · « **Reply #650 on:** January 09, 2010, 01:55:36 am »

I see you did some wishful thinking

TrueTears · « **Reply #651 on:** January 09, 2010, 02:33:21 am »

Another one:

$\lim_{x \to 4^-}\left(\frac{\sqrt{x^2+9}-5}{x+4}\right)$

When we are trying to find left/right limits, can we still apply the limit laws?

Ie, in this question, can we use the direct substitution method and just sub in $x=4$ ?

humph · « **Reply #652 on:** January 09, 2010, 03:11:31 am »

I'm guessing it's supposed to be $-4$ instead of $4^{-}$ , as otherwise the question's trivial (although it's pretty trivial anyway because it's the same as the previous question).

TrueTears · « **Reply #653 on:** January 09, 2010, 04:14:56 pm »

Quote from: humph on January 09, 2010, 03:11:31 am

I'm guessing it's supposed to be $-4$ instead of $4^{-}$ , as otherwise the question's trivial (although it's pretty trivial anyway because it's the same as the previous question).

Thanks humph

TrueTears · « **Reply #654 on:** January 09, 2010, 04:38:53 pm »

Use squeeze theorem to show that $\lim_{x \to 0} \left( \sqrt{x^3+x^2} \sin\left(\frac{\pi}{x}\right)\right)$

How do we know what inequality to pick?

kamil9876 · « **Reply #655 on:** January 09, 2010, 04:58:15 pm »

$-1 \leq sin\frac{\pi}{x} \leq 1$

Pretty common thing to do.

TrueTears · « **Reply #656 on:** January 09, 2010, 05:10:27 pm »

Quote from: kamil9876 on January 09, 2010, 04:58:15 pm

$-1 \leq sin\frac{\pi}{x} \leq 1$

Pretty common thing to do.

Oh yeah, so basically all squeeze theorems involving cos and sin are the same lol

TrueTears · « **Reply #657 on:** January 09, 2010, 06:00:28 pm »

$\lim_{x \to 0.5^-}\left(\frac{2x-1}{|2x^3-x^2|}\right)$

If $2x^3-x^2 \le 0 \implies x \le \frac{1}{2}$

So $\lim_{x \to 0.5^-}\left(-(2x^3-x^2)\right) = 0$

So we get $\frac{\lim_{x \to 0.5^-} \left(2x-1\right)}{\lim_{x \to 0.5^-} \left(-(2x^3-x^2)\right)}$ which is indeterminant.

So how do I do this?

kamil9876 · « **Reply #658 on:** January 09, 2010, 06:03:29 pm »

factorise the denominator.
$x^2(2x-1)$

TrueTears · « **Reply #659 on:** January 09, 2010, 06:07:37 pm »

Thanks but then we have $\lim_{x \to 0.5^-} \frac{1}{-x^2}$

We treat left/right limits as just the actual limit so $\lim_{x \to 0.5} \frac{1}{-x^2} = \lim_{x \to 0.5^-} \frac{1}{-x^2}$

But how do we know $\lim_{x \to 0.5} \frac{1}{-x^2}$ exists (without sketching a graph ofcourse!)

Do I need to know the definition of a limit which is taught later on Stewarts?

So it's sufficient now to just do it graphically?

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