Author Topic: SUPER-FUN-HAPPY-MATHS-TIME (Read 45040 times) Tweet Share

evaporade · « **Reply #90 on:** July 06, 2009, 10:21:43 pm »

Another example:

Suppose x,y E R, y = 2x + d. Do you interpret the value of x and the value of y are given constants?

evaporade · « **Reply #91 on:** July 06, 2009, 10:24:11 pm »

Quote from: kamil9876 on July 06, 2009, 10:19:48 pm

Forget the word 'constant' then. All I mean is that the question can be stated as "prove that for every given x, y and r (such that 2r>|x-y|) there are infinitely many z such that...". This is what I meant by constant, just like in your diagram x and y are constant and fixed points because you are proving that 'for every given x and y...'

But the question was not stated like that. You chose to interpret it that way.

zzdfa · « **Reply #92 on:** July 06, 2009, 10:25:21 pm »

it's the convention

evaporade · « **Reply #93 on:** July 06, 2009, 10:27:31 pm »

~~Rubbish~~. What convention. Read the example.

zzdfa · « **Reply #94 on:** July 06, 2009, 10:32:46 pm »

your example is completely different from the question.

a better one would be the one i posted earlier.
$\text{For } r >0\text{ how many }m\text{ are there such that } r=m$
sure you could say there are an infinite number of cases where r=m
but almost always it means 'for any particular r, how many m where m=r'

and in response to your example, i would say, there is exactly one d such that y=2x+d.

evaporade · « **Reply #95 on:** July 06, 2009, 10:34:54 pm »

I followed the wordings of your question. Your example is completely different from the original wording.

evaporade · « **Reply #96 on:** July 06, 2009, 10:35:47 pm »

Also, what convention?

humph · « **Reply #97 on:** July 08, 2009, 02:54:35 pm »

Lolz. The question is asking you to prove that if you FIX some value $r >0$ , then there exist infinitely many (uncountably infinite, in this case) points $t \in \mathbb{R}^k$ satisfying $|z-x|=|z-y|=r$ . That is, the phrase "suppose that $r >0$ " implies that $r$ cannot vary at all, it is one fixed value.

brightsky · « **Reply #98 on:** January 02, 2010, 06:15:48 pm »

Might have already been solved, but loved this one!! ><

13. Prove that $0.9\dot{9} = 1$

Let $x = 0.9\dot{9}$ .

Then $10x = 9.9\dot{9}$ .

Hence, $10x - x = 9.9\dot{9} - 0.9\dot{9} = 9 \implies 9x = 9 \imples x = \frac{9}{9} = 1.$
From the evaluation above:

$x = 0.9\dot{9}$

and

$x = 1$

So, $0.9\dot{9} = 1$ .

brightsky · « **Reply #99 on:** January 02, 2010, 09:37:40 pm »

Are we allowed to use L'Hospital's rule for this one?

17. $\lim_{x\to0} \frac {e^{x^2} -e^{3x}} {\sin(\frac{x^2}{2}) - \sin(x)}$

This is an indeterminate form 0/0, as if you x = 0 is undefined when you sub it in. So using L'Hospital's rule, where $f(x) = e^x^2 -e^{3x}$ and $g(x) = \sin(\frac{x^2}{2}) - \sin(x)$ we have:

$\implies \lim_{x\to0} \frac {\frac{d}{dx} \left(-e^x^2 - e^{3x} \right)}{\frac{d}{dx} \left(\sin(\frac{x^2}{2}) - \sin(x)\right)}$

$\implies \lim_{x\to0} \frac {3e^{3x} - 2e^x^2 x}{\cos(x) - x\cos \left(\frac{x^2}{2}\right)}$

$\implies \frac {\lim_{x\to0} 3e^{3x} - 2e^x^2 x}{\lim_{x\to0}\cos(x) - x\cos \left(\frac{x^2}{2}\right)}$

$\implies \frac {3\left(\lim_{x\to0} e^{3x}\right) - 2 \left(\lim_{x\to0}e^x^2 x\right)}{\lim_{x\to0}\cos(x) - x\cos \left(\frac{x^2}{2}\right)}$

Write $\lim_{x\to0} e^{3x}$ as $e^{\lim_{x\to0}}^3x$ using the continuity of $e^3x$ at $x = 0$ .

$\implies \frac {3e^{{\lim_{x\to0}}^3x} - 2\left(\lim_{x\to0} e^{x^2} x \right)}{\lim_{x\to0}\cos(x) - x\cos \left(\frac{x^2}{2}\right)}$

$\implies \frac {3e^3\left({\lim_{x\to0}}^x \right) - 2 \left(\lim_{x\to0}e^{x^2} x} \right)}{\lim_{x\to0}\cos(x) - x\cos \left(\frac{x^2}{2}\right)}$

Sub x = 0 in:
$ \implies \frac {3 - 2 \left(\lim_{x \to0} e^x^2 x \right)}{\lim_{x\to0}\cos(x) - x\cos \left(\frac{x^2}{2}\right)}$

$\implies \frac {3 - 2 \left(\lim_{x\to0} e^x^2 \right) \left(\lim_{x\to0}x \right)}{\lim_{x\to0}\cos(x) - x\cos \left(\frac{x^2}{2}\right)}}$

Write $\lim_{x\to0} e^x^2$ as $e^{\lim_{x\to0}}^x^2$ by using the continuity of $e^x^2$ at $x = 0$ :

$ \implies \frac {3 - 2e^{\lim_{x\to0}}^x^2 \left(\lim_{x\to0} x\right)}{\lim_{x\to0}\cos(x) - x\cos \left(\frac{x^2}{2}\right)}} $

Write $\lim_{x\to0} x^2$ as $\left(\lim_{x\to0} x \right)^2$ using the power law:
$ \implies \frac {3 - 2e^{\left(\lim_{x\to0} x \right)^2} \left(\lim_{x\to0} x \right)}{\lim_{x\to0}\cos(x) - x\cos \left(\frac{x^2}{2}\right)}} $
Sub x = 0 in:
$ \implies \frac {3}{\lim_{x\to0}\cos(x) - x\cos \left(\frac{x^2}{2}\right)}} $

$\implies \frac{3}{\lim_{x\to0}\cos(x) - \lim_{x\to0}xcos\left(\frac{x^2}{2}\right)} $
The limit of $\cos(x)$ as $x\to0 = 1$ . Sub that in:
$ \implies \frac {3}{1 - \lim_{x\to0} x \cos \left(\frac{x^2}{2}\right)} $

$\implies \frac {3}{1 - \left(\lim_{x\to0} x \right)(\lim_{x\to0} \cos \left(\frac{x^2}{2}\right)} $

Sub x = 0 in:
$ \implies 3$

TrueTears · « **Reply #100 on:** January 02, 2010, 09:40:03 pm »

I already posted using L'hopital's rule but dcc wants a less cheap way

Maybe try sandwich it, play around.

brightsky · « **Reply #101 on:** January 02, 2010, 09:48:10 pm »

Ahh ok. Took me forever to type up! :p

TrueTears · « **Reply #102 on:** January 02, 2010, 09:50:49 pm »

Quote from: TrueTears on April 16, 2009, 12:23:03 am

Find $\lim_{x\rightarrow0}\frac {e^{x^2} - e^{3x}}{\sin(\frac {x^2}{2}) - \sin x}$

17) Using l'hopital's theorem again

let $f(x) = e^{x^2} - e^{3x}$

$g(x) = sin(\frac{x^2}{2}) - sin(x)$

so $f'(x) = 2xe^{x^2} - 3e^{3x}$

$g'(x) = xcos(\frac{x^2}{2}) - cos(x)$

so $\frac{f'(x)}{g'(x)} = \frac{2xe^{x^2} - 3e^{3x}}{xcos(\frac{x^2}{2}) - cos(x)}$

limit $x \rightarrow 0$ yields $\boxed{\frac{f'(x)}{g'(x)} = 3}$

brightsky · « **Reply #103 on:** January 02, 2010, 09:52:46 pm »

Wow! That was fast!

TrueTears · « **Reply #104 on:** January 02, 2010, 09:55:26 pm »

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Author Topic: SUPER-FUN-HAPPY-MATHS-TIME (Read 45040 times) Tweet Share

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