Mathematical Curiosities. (3 Viewers)

asianese · Dec 13, 2013

There is some theory called the 'p integrals' and explains that that for p>1, the integral $\int_1^\infty \dfrac{1}{x^p}\mathop{dx} $ converges.$$ The point is that the successive parts being added are getting smaller and smaller at a fast enough rate so that the total area is finite.

The length of a curve $y=f(x)$ is given by $\mathcal{L} = \int_a^b \sqrt{1+(f'(x))^2}\mathop{dx}.$ Which is derived from pythagoras' theorem essentially. In terms of velocity/accel it doesn't have any significance.

If there constants were different the world would still function just in a different way. This is more of a philosophical question.

anomalousdecay · Dec 13, 2013

When graphing the curve of:

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

You get the area beneath the curve, over the x-axis and between x=1 and x= -1 as pi/2.

This can be done by integration.

$\int_{-1}^{1}\frac{dx}{1+x^2} = \\ \\ \\ |tan^{-1}(x)|^{1}_{-1} = \\ \\ \\ \frac{\pi}{4} + \frac{\pi}{4} = \\ \\ \\ \frac{\pi}{2}$

My curiosity is:

How can an exact area bounded by a curve have an irrational value, where the curve is finite, by the area beneath is not exact.

asianese · Dec 13, 2013

You contradicted yourself in that sentence. Please clarify.

braintic · Dec 13, 2013

anomalousdecay said:
When graphing the curve of:

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

You get the area beneath the curve, over the x-axis and between x=1 and x= -1 as pi/2.

This can be done by integration.

$\int_{-1}^{1}\frac{dx}{1+x^2} = \\ \\ \\ |tan^{-1}(x)|^{1}_{-1} = \\ \\ \\ \frac{\pi}{4} + \frac{\pi}{4} = \\ \\ \\ \frac{\pi}{2}$

My curiosity is:

How can an exact area bounded by a curve have an irrational value, where the curve is finite, by the area beneath is not exact.

But pi/2 IS exact. Which part is not exact?

braintic · Dec 13, 2013

asianese said:
In terms of velocity/accel it doesn't have any significance.

It is a length - or a displacement along the curve. So its derivative with respect to time IS the velocity.

asianese said:
If there constants were different the world would still function just in a different way. This is more of a philosophical question.

Not sure I see a connection to the length of a curve.

asianese · Dec 13, 2013

I'm not quite understanding what he is saying...

So if you said that $v = \dfrac{dx}{dt}$ then $\mathcal{L} = \int_a^b \sqrt{1+\left(\dfrac{dv}{dt}\right)^2} \mathop{dt}$ is the ...???

RealiseNothing · Dec 13, 2013

braintic said:
Not sure I see a connection to the length of a curve.

He was answering 2 separate questions.

braintic · Dec 13, 2013

RealiseNothing said:
He was answering 2 separate questions.

Sorry ... as there was no quote I didn't realise he was answering someone's questions.

asianese · Dec 13, 2013

Also the stuff about mathematical constants...the symbols $\pi$ and $e$ are just SYMBOLIC representations of a concept which happen to be constant. For pi, it is the ratio between the circumference of a circle to its diameter (many will argue circumference to radius would be more appropriate, see tau.) and e is the base of the natural logarithm. It is the number 'a' for which $\log_{a}a = 1$ and the limit of $\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n$

RealiseNothing · Dec 13, 2013

asianese said:
It is the number 'a' for which $\log_{a}a = 1$

Are you sure about this?

nightweaver066 · Dec 14, 2013

1. So we have the triangle inequality... but where did it come from?

2. Why/how does the method of variables separable work?

nerdasdasd · Dec 14, 2013

I wonder how did they discover radians :3, or figure out the value of pi.

RealiseNothing · Dec 14, 2013

nightweaver066 said:
So we have the triangle inequality... but where did it come from?

Straight line distance is always the shortest route between any two points.

Another way is to consider the cosine rule and make the cosine the subject:

$cosA = \frac{b^2+c^2-a^2}{2bc}$

Now since the cosine function oscillates between -1 and 1 we can say that:

$b^2+c^2-a^2 < 2bc$

$a^2 > (b-c)^2$

$a > b-c$

$a+c > b$

RealiseNothing · Dec 14, 2013

nerdasdasd said:
I wonder how did they discover radians :3, or figure out the value of pi.

Consider some angle subtended at the centre of a circle of radius R, by an arc PQ, then:

$\theta = \frac{PQ}{R}$

Consider now that the angle is a revolution such that it covers the whole circle:

$\theta = \frac{2\pi R}{R} = 2\pi$

So $360=2\pi$

asianese · Dec 14, 2013

nightweaver066 said:
2. Why/how does the method of variables separable work?

You should've done this in Calculus first year maths... (or are you probing the question?)

nightweaver066 · Dec 14, 2013

asianese said:
You should've done this in Calculus first year maths... (or are you probing the question?)

lol we've definitely used it, but I don't recall ever having it proven/justified, just explained very wish-washily by my tutor.

asianese · Dec 14, 2013

This is just from my course notes summarised:

$$Solve the ODE $ \dfrac{dy}{dx} = g(x)h(y) $ and write $ f = \dfrac{1}{h} $ so that $ \\ f(y) \dfrac{dy}{dx} = g(x). \\ $ Let $F_0(y) $ and $ G_0(x) $ be antiderivatives of $f(y) $ and $ g(x) $ respectively.$ \\ $They exist provided $ f(y) $ and $ g(x) $ are continuous.$ \\ $Then $ \\ F'_0(y) \dfrac{dy}{dx} = G'_0(x). \\ $ Now, consider $ \dfrac{d}{dx}\left(F_0(y(x) \right) = F'_0(y(x))y'(x) = F'_0(y)\dfrac{dy}{dx} $ by the chain rule. $\\$So thus $ \dfrac{d}{dx}\left( F_0(y(x)\right)= G'_0(x) $ and by FTC$ \dfrac{d}{dx}\left( F_0(y(x))-G_0(x)\right)=0.$

$$ This implies $F_0(y(x))-G_0(x)=C, $ C a constant. \\So then $ F_0(y) = G_0(x) + C$

$$Compare with $ \dfrac{dy}{dx} = \dfrac{g(x)}{f(y)} \Rightarrow f(y) \mathop{dy} = g(x)\mathop{dx} $ so then $\\ \int f(y) \mathop{dy} = \int g(x)\mathop{dx} \Rightarrow F_0(y)+C' = G_0(x)+C'' \Rightarrow F_0(y) = G_0(x) + C, C = C''-C'$ which is what we wanted. So the method is justified - but we skip a lot of theory!

nightweaver066 · Dec 14, 2013

Thanks a lot for that asianese!

The last part seems a bit confusing, so would it still be correct to see it this way?
$F_0(y) = G_0(x) + C$

$\implies F_0(y) + C' = G_0(x) + C'',\;\text{where } C = C''-C'$

$\implies \int f(y) dy = \int g(x) dx$

Comparing this to $\frac{dy}{dx} = \frac{g(x)}{f(y)}$ , we see that we can apply the 'trick' in rearranging the differentials to integrate.

anomalousdecay · Dec 14, 2013

braintic said:
But pi/2 IS exact. Which part is not exact?

Oops. What I mean is in terms of a decimal. How do you get an irrational area in terms of a curve which is supposed to be known as definite.

I don't know, maybe its a stupid question. The one thing I've never been good at is terminology in Maths.

nerdasdasd said:
I wonder how did they discover radians :3, or figure out the value of pi.

My reference to they below is Mathematicians.
So what they did was draw a circle. Then they drew a rectangle in the circle (concyclic) and found the area. Then, they added isosceles triangles to each side of the rectangle to touch the circles edge (concyclic) to get an octagon. Then they did this again to get a 16-sided polygon and so forth for concyclic polygons. Eventually you get a circular shape. Thus they calculated the area, and had an approximation pi. Then a couple hundred years later, calculus was invented and you can use graphing to get pi to a more accurate approximation (Just like my question from above you can get exactly pi/2).

asianese · Dec 14, 2013

nightweaver066 said:
Thanks a lot for that asianese!

The last part seems a bit confusing, so would it still be correct to see it this way?
$F_0(y) = G_0(x) + C$

$\implies F_0(y) + C' = G_0(x) + C'',\;\text{where } C = C''-C'$

$\implies \int f(y) dy = \int g(x) dx$

Comparing this to $\frac{dy}{dx} = \frac{g(x)}{f(y)}$ , we see that we can apply the 'trick' in rearranging the differentials to integrate.

Yeah so just whenever you integrate stick in a +C, its more natural to have 2 constants, but usually we only bother ourselves with one!

Mathematical Curiosities. (3 Viewers)

Σ

Premium Member

Σ

Well-Known Member

Well-Known Member

Σ

what is that?It is Cowpea

Well-Known Member

Σ

what is that?It is Cowpea

Well-Known Member

Dont.msg.me.about.english

what is that?It is Cowpea

what is that?It is Cowpea

Σ

Well-Known Member

Σ

Well-Known Member

Premium Member

Σ

Users Who Are Viewing This Thread (Users: 0, Guests: 3)