how big a change is yr 11 maths to yr 10 maths (2 Viewers)

Shadowdude · Jul 16, 2011

benji_10 said:
Isn't that the perpendicular distance formula?

$P(0,0)\ l:Ax+By+C$
$\therefore\ x_1=0,\ y_1=0,\ a=A,\ b=B,\ c=C$
$\begin{align*}d &=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\\ \therefore\ d &=\frac{|A(0)+B(0)+C|}{\sqrt{A^2+B^2}}\\ &=\ \frac{|C|}{\sqrt{A^2+B^2}}\end{align*}$

For some stupid reason there are stupid br/ tags in my post. Solution to that would be nice.

And no cheating shadow, it's moar fun this way =)

I think that's cheating using the perpendicular distance formula. I think you have to derive it somehow.

And hey, the Mean Value Theorem is perfectly fine Calculus. It's just first year uni maths.

benji_10 · Jul 16, 2011

Lol not my problem

. You said show, not prove through the use of part a) or diagram 1.2 or whatever. The formula is quotable because it work in all cases with a point and a straight line. I applied it (Y) Although I wasn't surprised you'd say that. The question could have been done by anyone in year 10 using the formula. Knew there was some sort of requirement or condition that would make it ridiculously harder than it should be

And what is the mean value theorem?
EDIT: I had a peek in wikipedia. I'm fine for the first couple of paragraphs or so, then I got lost

Shadowdude · Jul 16, 2011

Okay fine, you got me on a technicality.

The mean value theorem is basically saying that if a function is continuous on a closed interval [a,b] and differentiable on an open interval (a,b) then a point c, where a < c < b has the property:

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Essentially, the tangent at c has the same slope as the 'rise over run' straight line from f(b) to f(a).

You can use that to prove inequalities like the one above.

Is that... clear? Or would you like a diagram =P

benji_10 · Jul 16, 2011

Cheers shadow! I get it now (Y)
Just wondering if there were any applications to it, besides solving inequalities like that one. And could you post up a solution to the inequality using the theorem? I'm interested in uni maths, and its use in 3/4U. It would be great if you could. Thanks!

cutemouse · Jul 16, 2011

Shadowdude said:
Okay fine, you got me on a technicality.

The mean value theorem is basically saying that if a function is continuous on a closed interval [a,b] and differentiable on an open interval (a,b) then a point c, where a < c < b has the property:

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Essentially, the tangent at c has the same slope as the 'rise over run' straight line from f(b) to f(a).

You can use that to prove inequalities like the one above.

Is that... clear? Or would you like a diagram =P

Should really say that THERE EXISTS a point c such that ...

Shadowdude · Jul 16, 2011

benji_10 said:
Cheers shadow! I get it now (Y)
Just wondering if there were any applications to it, besides solving inequalities like that one. And could you post up a solution to the inequality using the theorem? I'm interested in uni maths, and its use in 3/4U. It would be great if you could. Thanks!

Well, one of the chapters in MATH1141 (Higher Maths 1A) is called: "The Mean Value Theorem and its applications". It's one of the very basic and powerful theorems in calculus theory. Applications include identifying where a function is increasing/decreasing, identifying types of stationary points (these first two you know as your basic 'if the first derivative is positive, the function is increasing' and 'if the second derivative is positive where the derivative is zero, the stationary point is a minimum' - but they can be proved via the mean value theorem), determining how many zeros a polynomial has, evaluating indeterminate limits (of form 0/0 or infinity/infinity), proving inequalities like the one you gave and estimating errors in approximations.

Here's how I'd solve your question:

$\textbf{Prove\;that}\;ln(x)<x\;\textbf{for\;all\;real\;positive}\;x$

$\textup{Note\;that\;}x-1<\,x.\;\textup{Suppose\;}x>1\;\textup{and\;consider\;the\;closed\;interval\;[1,x]}.$
$\textup{Define\;a\;function}\;f:\;\textup{[1,x]}\,\rightarrow \;\mathbb{R}\;\textup{by}\;f(t)=ln(t). \;f\;\textup{is\;continuous\;on\;[1,x]\;and\;differentiable\;on\;(1,x)}.$
$\textup{We\;apply\;the\;mean\;value\;theorem\;on\;f\;on\;[1,x]}$
$f'(t)=\frac{1}{t}.\;\textup{So:}$
$\frac{f(x)-f(1)}{x-1}=\frac{1}{c}\;,\textup{for\;some}\;c\;\textup{in\;[1,x]\;by\;the\;mean\;value\;theorem}$
$\frac{ln(x)}{x-1}=\frac{1}{c},\;\textup{for\;some\;c\;between\;1\;and\;}x.$
$\textup{Since}\;c>1\;\textup{then}\;\frac{1}{c}<1$
$\textup{Hence:}\;\frac{ln(x)}{x-1}<1$
$\therefore\;ln(x)<\,x-1 \Rightarrow \, ln(x) < x$

A bit fudgy I think at the end... but, yeah. That's the case for x > 1, and then for the case x = 1 and lesser than 1, I'd probably go for a graphical approach or something else as the mean value theorem only works on closed intervals and we can't really use the interval [0, x] as ln(0) isn't defined. Though if I put it in my iPhone, it gives negative infinity, so maybe...

(I realise I didn't fully answer the question but that's the general gist of the mean value theorem)

benji_10 · Jul 17, 2011

Mmmmkay. A few questions:
$Prove\ that\ lnx<x\ for\ all\ real\ positive\ x.$
But we were trying to prove that:
$Prove\ that\ lnx<\sqrt{x}\ for\ all\ real\ positive\ x.$

Was your reasoning that since x belongs in the all real number field, then $\sqrt{x}<x$ , hence proving that $lnx<x$ would answer the question as well? Otherwise, I don't get/see why you were trying to prove $lnx<x$ .

Also, there is no use in using the mean value theorem on linear curves (y=x-a) right?
Say f(x) = 5x-3 ---->
$f(x)=5x-3$
$f'(x)=5$
$\frac{f(b)-f(a)}{b-a}=f'(c)=5$
And since $\frac{f(b)-f(a)}{b-a}$ is just the gradient of a straight line; doing the mean value theorem on straight lines would basically find the gradient in a demented way?

Lol thanks for your solution anyways. Uni maths looks fun

And you couldn't really cheat could you because you could only prove it for an interval, and not as x went to infinite. Which brings me to my next question: Can you, or why can't you take the closed interval as [1,inf.]? The curve is diff. and continuous between this interval.

Shadowdude · Jul 17, 2011

Ohh, well... it was 11pm at night - I probably got carried away with the Latex. But I'll leave the proper proof as an exercise to you (you'll see a lot of that at uni) - similar kind of style, might be a bit more difficult. But no, you can't say ln(x) < x and then jump to say that ln(x) < sqrt(x) because sqrt(x) < x as this: ln(x) > sqrt(x) can occur.

And yeah... the mean value theorem is basically rise over run and will give you the slope of a straight line.

Also I supposed that x > 1 in my shoddy attempt at a proof at 11pm at night - so x can range to infinity. Infinity is not a number, so you cannot 'close' the interval over infinity.

benji_10 · Jul 17, 2011

If this can occur: lnx>sqrtx, then why would they ask me to prove a false result?
The question said prove lnx < sqrtx for ALL POSITIVE values of x. If lnx>sqrtx for some large value of x, then they wouldn't ask me to prove it true for all pos. values of x. Rather, they would ask me: "Is this true for all real pos values of x: lnx<sqrtx"

And assuming that my tutor got the question wrong and the statement was false: where would lnx>sqrtx?

deterministic · Jul 17, 2011

^^ dw, Shadowdude's "method" is needlessly complicated for a question that could be more easily done by high school calculus.

benji_10 · Jul 17, 2011

Lol. I thought the mean value theorem would be easier because I he said that he could "cheat" with it. Meh.

Anyways:
$Use\ mathematical\ induction\ to\ prove\ that\ 1^3+2^3+3^3+...+n^3>\frac{n^4}{4}$

There's a part 1 to this question but imo you don't need it. If you've worked with binomials, complex numbers or polynomials, then you should see it very easily.

b33g_boss · Jul 17, 2011

bleakarcher said:
lol prairiewood high school deep within the south-western suburbz

yea western suburbs

Shadowdude · Jul 17, 2011

benji_10 said:
If this can occur: lnx>sqrtx, then why would they ask me to prove a false result?
The question said prove lnx < sqrtx for ALL POSITIVE values of x. If lnx>sqrtx for some large value of x, then they wouldn't ask me to prove it true for all pos. values of x. Rather, they would ask me: "Is this true for all real pos values of x: lnx<sqrtx"

And assuming that my tutor got the question wrong and the statement was false: where would lnx>sqrtx?

I'm saying that after proving: ln(x) < x, we cannot go on to logically say ln(x) < sqrt(x) on the next line - even though it's true, it has to be proved separately.

The result ln(x) < sqrt(x) is true - it's mathematically proved, so it is definitely true. I'm not doubting that, I'm just saying that we can't go from ln(x) < x to ln(x) < sqrt(x) directly.

Like say, if I had this working:

Solution:
1 < 3
therefore: 1 < a

We can't do that, just in case a > 3. Similarly, we can't say ln(x) < sqrt(x) from ln(x) < x, just in case ln(x) > sqrt(x).

benji_10 · Jul 17, 2011

Shadowdude said:
Solution:
1 < 3
therefore: 1 < a

We can't do that, just in case a > 3. Similarly, we can't say ln(x) < sqrt(x) from ln(x) < x, just in case ln(x) > sqrt(x).

Ahhhhh I see where I went wrong. I foolishly assumed that sqrt(x) < x for all x, but this is only true when x>1. For 0 < x < 1, sqrt(x)>x. So my jumping from lnx < x to lnx < sqrt(x) is wrong.

And I think you meant to say that instead of "just in case a>3", it should be "a<1", because if a>3, 1 < a would still hold true.

Cheers shadowdude

bleakarcher · Jul 18, 2011

what happened lol? y is this thread now flooded wif university maths?

freestyler1 · Jul 18, 2011

kill yourselfs

freestyler1 · Jul 18, 2011

Shadowdude said:
I'm saying that after proving: ln(x) < x, we cannot go on to logically say ln(x) < sqrt(x) on the next line - even though it's true, it has to be proved separately.

The result ln(x) < sqrt(x) is true - it's mathematically proved, so it is definitely true. I'm not doubting that, I'm just saying that we can't go from ln(x) < x to ln(x) < sqrt(x) directly.

Like say, if I had this working:

Solution:
1 < 3
therefore: 1 < a

We can't do that, just in case a > 3. Similarly, we can't say ln(x) < sqrt(x) from ln(x) < x, just in case ln(x) > sqrt(x).

cant even construct a logical proof, and you do maths

"im saying",

"it's mathematically proved, so it is definitely true" - wtf

This is the level of intelligence required to get into advanced maths,

see its not hard , its piss easy

benji_10 · Jul 18, 2011

bleak have you done induction before? If you haven't, its pretty fun and you should self-learn it.

Shadowdude · Jul 18, 2011

benji_10 said:
Ahhhhh I see where I went wrong. I foolishly assumed that sqrt(x) < x for all x, but this is only true when x>1. For 0 < x < 1, sqrt(x)>x. So my jumping from lnx < x to lnx < sqrt(x) is wrong.

And I think you meant to say that instead of "just in case a>3", it should be "a<1", because if a>3, 1 < a would still hold true.

Cheers shadowdude

Yeah you get the gist... whatever =P

As long as you get it.

freestyler1 said:
cant even construct a logical proof, and you do maths

"im saying",
"it's mathematically proved, so it is definitely true" - wtf

This is the level of intelligence required to get into advanced maths,

see its not hard , its piss easy

I know... I know, but it's okay, I'm starting MATH1081 and that'll teach me how to do proper proofs. I even bought a book on it. =)

bleakarcher · Jul 18, 2011

yeh i have learnt mathematical induction

how big a change is yr 11 maths to yr 10 maths (2 Viewers)

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