Function

Hello

I just disovered some new functions: sinIntegral(x) (Si(x)) and cosIntegral(x) (Ci(x)) and just played around with them on GeoGebra.

Now, I found something kind of 'beautiful': the average area between those functions, limited by [math]a-1[/math] and [math]a[/math] (so just basically any 'area-block' with width 1), equals [math]\frac{\pi}{2}[/math] when the number of elements of which the average is taken reaches [math]\infty[/math]:

[math]\lim_{n\to\infty}{\left[\frac{\int_{1}^{n}{Si(x) dx}-\int_{1}^{n}{Ci(x) dx}}{n-1}\right]}=\frac{\pi}{2}[/math].

In order to prove it, I worked out the left side, but I don't really know how to do it further:

[math]\lim_{n\to\infty}{\left[\frac{n\cdot Si(n)+\cos{n}-Si(1)-\cos{1}-n\cdot Ci(n)+\sin{n}+Ci(1)-\sin{1}}{n-1}\right]}[/math]

According to L'Hôpital:

[math]=\lim_{n\to\infty}{\left(n\cdot\frac{\sin{n}}{n}+Si(n)+\sin{n}-n\cdot\frac{\cos{n}}{n}-Ci(n)-\cos{n}\right)}[/math]

[math]=\lim_{n\to\infty}{\left(Si(n)-Ci(n)+2\sin{n}-2\cos{n}\right)}[/math]

[math]=\lim_{n\to\infty}{\left(Si(n)-Ci(n)\right)}+2\lim_{n\to\infty}{\left(\sin{n}-\cos{n}\right)}[/math]

[math]=\frac{\pi}{2}+2\lim_{n\to\infty}{\left(\sin{n}-\cos{n}\right)}[/math]

Now, I know that the first term is [math]\frac{\pi}{2}[/math], so the second term should be [math]0[/math], but how can one prove this? As far as I know, [math]\lim_{n\to\infty}{\left(\sin{n}-\cos{n}\right)}[/math] is not specified, for the values of [math]\cos{n}[/math] and [math]\sin{n}[/math] alternate.

Can someone help me?

Thanks

Function

Hello everyone

For maths, we need to do a research of a mathematical subject. We chose the theorems of Fermat (both the big and small theorem).

We are to make a question about this subject, a question worth investigating. (One big question and perhaps some smaller questions about that primary question)

However, we can't seem to find one.

Can someone help us?

Thanks.

Function

Hello

For a research we have to do in class, I'd like to use mathematically correct notations.

There is, however, one expression from which I don't know how it's written: a false equation.

e.g. Fermat's big theorem:

[math]\nexists (x,y,z,n) \in \mathbb{N}_0 : x^n+y^n=z^n[/math]

[math]\forall(x,y,z,n)\in\mathbb{N}_0 : \left\{x^n+y^n=z^n\right\}=\emptyset[/math]

[math]\forall(x,y,z,n)\in\mathbb{N}_0 : \left\{x^n+y^n=z^n\right\}=0[/math]

Or just a very simple example:

[math]\left\{1=0\right\}=0[/math]

[math]\left\{1=0\right\}=\emptyset[/math]

Is any of these notations correct? If not, which one could I use?

Thanks.

Function

Battery's had it by the sound of it or at least it's regulation mechanism is faulty if not the cells themselves.

Ah crap.. Thanks. I'll make sure to get another one.

What does it do with no battery in and plugged in when you power on?

It then just boots normally.

 

You really are a glutton for the extreme.

 

 

I haven't ever had need to think about it but the usual thing in those circumstances would be to generalise the binomial expansion with the multinomial expansion.

 

http://en.wikipedia.org/wiki/Multinomial_theorem

I'm a glutton for the most general formulas

Thanks for the link.

Hello

I have a Dell Inspiron Mini 10, and since 2 days ago, the battery is acting very strange, and extremely annoying.

1. When the notebook is plugged in, it won't start. It will start when the cable isn't plugged in, and will continue its boot when I then plug in the cable.

2. With the cable plugged in and the notebook booted, there's a pretty annoying phenomenon:

• For about 16 seconds, the battery indicator indicates that the battery is charging (3% available (plugged in, charging))
• After those 16 seconds, it suddenly jumps to 0% remaining (so not plugged in and not charging, according the the symbol) for about 7 seconds; this is accompanied by a verry annoying high-frequence sound. (The screen brightness is also much less then)
• After the 7 seconds of non-charging, it jumps back to the 16 seconds of charging (which just stays on 3% available), making the annoying sound stop
• And it continues to do this..

Yesterday, I managed to stop this annoying phenomenon, but don't ask me how... Can't remember.

So yesterday, in the evening, I shut down my notebook, which was to download updates, and now I experience the problem again.

Is my battery f*cked? How could I resolve this problem?

Thanks!

Function

UPDATE: just saw that the battery indicator now just holds on "3% available (plugged in, charging)", not doing the 'filling battery animation' (the symbol is static).

UPDATE 2 (UPD 1 +10 min): 4% charged

UPDATE 3 (UPD 1 +20 min): all the sudden: 100% charged

Indeed there are.

 

You have uncovered Leibnitz rules.

 

Well done.

 

This develops the nth derivative of a product of two functions u and v via the binomial coefficients of lower derivatives,

For example

 

[math]\frac{{{d^2}}}{{d{x^2}}}\left( {uv} \right) = u\frac{{{d^2}v}}{{d{x^2}}} + 2\frac{{du}}{{dx}}\frac{{dv}}{{dx}} + v\frac{{{d^2}u}}{{d{x^2}}}[/math]

 

[math]\frac{{{d^3}}}{{d{x^3}}}\left( {uv} \right) = u\frac{{{d^3}v}}{{d{x^3}}} + 3\frac{{du}}{{dx}}\frac{{{d^2}v}}{{d{x^2}}} + 3\frac{{{d^2}u}}{{d{x^2}}}\frac{{dv}}{{dx}} + v\frac{{{d^3}u}}{{d{x^3}}}[/math]

 

 

 

 

http://en.wikipedia.org/wiki/General_Leibniz_rule

Ah.. Those binomials are everywhere Thanks.

But how about m-numbers of functions? E.g. 3 functions instead of 2, or maybe 4, 5, ...?

Hello everyone

I was wondering if there was a general formula (sum, product, ...) to define the n-th derivative of the m-th power of a function f(x)

So far, I've found that for [math]n[/math] up to 3, the following should be right (if I made no mistakes):

[math]\frac{d^{n}\left[f(x)\right]^m}{dx^n}=\frac{d^{n}f}{dx^n}m\left[f(x)\right]^{m-1}+\left(\frac{df}{dx}\right)^n\left[f(x)\right]^{m-n}\cdot\prod_{i=0}^{m-1}{(m-i)}+n\left[f(x)\right]^{m-n+1}\cdot\prod_{i=1}^{m-1}{\frac{d^{i}f}{dx^i}}\cdot\prod_{i=0}^{m-2}{(m-i)}[/math]

Can anyone tell me if it's even possible to make such formula, and if yes, what that formula is exactly?

Thanks.

Function

Hello

Now that I get the basics of a chemical equilibrium, I wanted to remake an excercise from my book, which we solved in class:

In a closed space with a volume of 2 litres, 18,4 g N2O4 is added. As a consequence of heating this up to 50°C, half of the molecules is being transferred into NO2.

a) Calculate the equilibrium constant for this analytic reaction.

b) What are the concentrations when the volume is now changed to 1 litre?

a) K = 0,22 (this is not the question I'm having problems with)

b) Volume is being decreased [math]\Rightarrow[/math] equilibrium shifts to the side with the fewest amount of molecules [math]\Rightarrow[/math] shift to N2O4.

According to me:

Start with 0,22 moles of N2O4 (just as in (a)) and 0 moles of 2NO2.

As the equilibrium shifts to N2O4, to get the equilibrium concentration of N2O4, I would add [math[x[/math] to the number of N2O4-particles, resulting in a total of N2O4[math]+x[/math] moles of N2O4.

As the equilibrium shifts away from 2NO2, [math]2x[/math] needs to be substracted from 0 to get the number of 2NO2-particles.

Here is the first problem: one cannot have a negative amount of particles.

Second problem is how we resolved this in class: we substracted x from N2O4 and added 2x to 2NO2. Why? If the equilibrium shifts to the left, more particles are formed at the left, no?

Thanks.

Function

Yes, it does. Thanks. Now, if an equilibrium moves to the left, for example, what does that exactly mean? My teacher has failed to explain it in a way that is clear to me.

From what I know, it's that the concentrations to the left will be bigger, but I'm afraid that isn't true.

K is a reactiondependant constant, not a concentrationdependant constant.

That something that simple is going to save my sorry *ss tomorrow. Thanks. +1

So temperature, volume & concentration change the 'position' of the equilibrium, but not K? (Apart from temperature)

Hello everyone

I'm afraid I don't really get some things in 'chemical equilibrium':

My book first states that if the equilibrium constant K is big, there's a big conversion from left side to right side (the equilibrium is right).

It also states that if a concentration [A] is being increased, the equilibrium is being shifted towards the side of A.

Now, it states at the equilibrium reaction [math]H_2+I_2\leftrightarrow 2 HI[/math] that a doubling of [H2] will shift the equilibrium to the right.

However: [math]K=\frac{[HI]^2}{[H_2][i_2]}[/math], making K smaller when [H2] is being enlarged, shifting the equilibrium to the left (K is smaller, so the reaction is less to the right and now more to the left, thus making it being shift to the left).

Can someone explain my misunderstanding?

Thanks.

Function

EDIT: is it maybe because K doesn't have to do much with the shifting of the equilibrium?

[math]\cos(\alpha + \beta) = \cos\left(\frac{\pi}{2}\right) = 0[/math], so the final line in the original post should read [math]0 = 0[/math] and not [math]1 = -1[/math].

Ah yes.. You see it was a stupid topicquestion

How could I've ever stated that if 0*a = 0*b, which is always true, a = b

What a most interesting subject.

iNow, thanks for mentioning 'solipsism', a new term to me. Learned something new there

Yet I'm afraid no-one will ever be able to give you a plausible, acceptable consensus on this subject. Is solipsism the case here? Are we all just parasites, bacteria, microbes, molecules, atoms, quarks of the earth, an organ of the universe, the organism of the universe of all universa, whereas what we see as molecules and atoms are, in comparison, even undiscovered things? Perhaps they are a universe, a planet on their own, with their own living beings, just like earth is to the universe? Perhaps our very universe is but an atom from another molecule, the brick of something that's very small to something/someone which/who lives outside this universe? Following this theorem, perhaps the Big Bang with the universe it created is just like the apparition of so-called "virtual particles" and is time as we know it actually very slow (perhaps trillions of years (lifetime of the universe) in comparison to about 10^(-22) seconds (average lifetime of a virtual particle)) and incomprehensible!

You see where I'm going? Don't sweat it. Don't f*ck up your mind like I did, thinking about such things. Let it be as it is and enjoy your time.

Hello everyone

Here in Belgium, there's a tv show in which in some round, 3 players have to bid on how many answers on a list they can give (e.g. the water brands sold in some super market), the one with the highest bid can try to name their answers.

Something remarkable: if a player picks a long list (let's say the bid 'n' answers, so they have to name 'n' answers), he/she will most likely name 'n-1' answers, leaving 1 open spot, failing the 'task'.

I have also remarked this in class: when students have to learn a list of e.g. consequences of an event, most can name all but one.

If they then take a look at the list, they remember the one they'd forgotten, but then forget another one.

Should they look again at the list, they remember the one they had now forgotten again, but forget the one they initially forgot again.

Is this just some coincidence and a result of lack of learning, or perhaps, learning too intensively, too much, too literally?

Thanks.

Function

 

Consider alpha = 20 degrees and beta = 50 degrees.

 

Are you really offering that cos (20+50) = - cos (70) ?

 

Some of these statements depend upon which quadrant your sum ends up in, have you heard of CAST?

Let's say both alpha and beta are in the first quadrant; I haven't heard of CAST yet. Feel free to inform me, should you feel like doing so.

Hello everyone

I was, with another topic in the back of my head (the one about the integration constant), making the following reasoning:

In a right triangle, let's say that the two acute angles are [math]\alpha[/math] and [math]\beta[/math].

Then is true that [math]\alpha+\beta=\frac{\pi}{2}[/math]

[math]\Leftrightarrow \alpha=\frac{\pi}{2}-\beta[/math]

[math]\Leftrightarrow \tan{\alpha}=\tan{\left(\frac{\pi}{2}-\beta\right)}[/math]

[math]\Leftrightarrow \frac{\sin{\alpha}}{\cos{\alpha}}=\frac{\sin{\left(\frac{\pi}{2}-\beta\right)}}{\cos{\left(\frac{\pi}{2}-\beta\right)}}[/math]

[math]\Leftrightarrow \frac{\sin{\alpha}}{\cos{\alpha}}=\frac{\cos{\beta}}{\sin{\beta}}[/math]

[math]\Leftrightarrow \cos{\alpha}\cos{\beta}=\sin{\alpha}\sin{\beta}[/math]

According to Wolfram Alpha, left part and right part together would be (probably one of Simspon's formulas):

[math]\Leftrightarrow \frac{1}{2}\left(\cos{\left(\alpha-\beta\right)}+\cos{\left(\alpha+\beta\right)}\right)=\frac{1}{2}\left(\cos{\left(\alpha-\beta\right)}-\cos{\left(\alpha+\beta\right)}\right)[/math]

[math]\Leftrightarrow \cos{\left(\alpha+\beta\right)}=-\cos{\left(\alpha+\beta\right)}[/math]

[math]\Leftrightarrow 1=-1[/math]

Now, I will most certainly have made a mistake, and seen the result of this 'equality', a serious mistake, resulting in a pretty stupid topicquestion. Could someone help me? Perhaps I made a mistake in Wolfram? (I entered: cos(x)*cos(y) and sin(x)*sin(y) to get the equality after "According to ..."

Thanks.

Function

[math]I = \int {\frac{{du}}{{1 - {u^2}}}} [/math]

[math] = \frac{1}{2}\ln \frac{{1 + u}}{{1 - u}}[/math]

 

[math] = \frac{1}{2}\ln \frac{{1 + \sin x}}{{1 - \sin x}}[/math]

 

[math] = \ln |\tan \left( {\frac{\pi }{4} + \frac{x}{2}} \right)|[/math]

 

[math] = \ln |\sec x + \tan x|[/math]

 

 

(+C, of course.) Thank you.

Because you have made d(1/cosx) in your line 1 equal to sinx in your line 2 rather than sinx/cos²x

 

So work your integration by parts out again.

Ah yes, how could I've forgotten to divide by cos²x...

Continuing your method, the answer would be [math]\arctan{u}+C=\arctan{\sin{x}}+C[/math] or [math]\frac{\ln{\left|u\right|}}{2u}+C=\frac{\ln{\left|\sin{x}\right|}}{2\sin{x}}+C[/math]

Correct?

Hello

In Math class, we were to sole an integration. While my teacher was explaining and resolving the exercise, I did it myself, finding [math]\int{\cdots}=-\arctan{e^{-x}}=f_1(x)[/math]

Later, my teacher found [math]\int{\cdots}=\arctan{e^{x}}=f_2(x)[/math].

As she didn't find any mistake in my work, we agreed that the 'problem' MUST be that those two primitives are different from each other, from a constant.

I now find that [math]f_2(x)-f_1(x)=\arctan{e^{x}}+\arctan{e^{-x}}=\frac{\pi}{2}[/math]

And that this is not only true for [math]e[/math], but for every real number.

Can this be proven?

Thanks.

Function

[EDIT]

I found - pure by luck - a plausible proof:

[math]\arctan{a^x}+\arctan{a^{-x}}=\frac{\pi}{2}[/math]

[math]\Leftrightarrow \arctan{a^{x}}=\frac{\pi}{2}-\arctan{a^{-x}}[/math]

[math]\Leftrightarrow a^x= \tan{\left(\frac{\pi}{2}-\arctan{a^{-x}}\right)}[/math]

[math]\Leftrightarrow a^x=\frac{\sin{\left(\frac{\pi}{2}-\arctan{a^{-x}}\right)}}{\cos{\left(\frac{\pi}{2}-\arctan{a^{-x}}\right)}}[/math]

[math]\Leftrightarrow a^x=\frac{\cos{\arctan{a^{-x}}}}{\sin{\arctan{a^{-x}}}}[/math]

[math]\Leftrightarrow a^x=\frac{1}{\tan{\left(\arctan{a^{-x}}\right)}}[/math]

[math]\Leftrightarrow \frac{1}{a^x}=\tan{\arctan{\frac{1}{a^x}}}[/math]

[math]\Leftrightarrow \frac{1}{a^x}=\frac{1}{a^x}[/math]

True

[math]\Leftrightarrow \arctan{a^x}+\arctan{a^{-x}}=\frac{\pi}{2}[/math] Q.E.D.

You have the right idea with your 1/1 but with the wrong part of the integral

[math]I = \int {\frac{1}{{\cos x}}dx} [/math]

 

 

[math] = \int {\frac{{\cos x}}{{{{\cos }^2}x}}dx} [/math]

 

[math] = \int {\frac{{\cos x}}{{1 - {{\sin }^2}x}}dx} [/math]

 

If we make the substitution sinx=u

 

[math] I = \int {\frac{{du/dx}}{{1 - {u^2}}}} dx = \int {\frac{1}{{1 - {u^2}}}} du[/math]

 

This has three different (logarithmic and trigonometric) forms.

 

Does this help?

Hmm.. I understand what you're doing, but I still don't see why my reasoning wasn't all correctly:

[math]\int{u\cdot v' dx}=\int{u\cdot dv} = uv-\int{v du}[/math]

So why can't [math]u=\frac{1}{\cos{x}}[/math] and [math]v=x[/math]

Hello everyone

One of the example questions of the acceptance exam for Med school, is the following:

(http://www.ond.vlaanderen.be/toelatingsexamen/nl/modelvragen/files/2013/modelvragen-wiskunde-2013.pdf "Vraag 7")

Given:

[math]\int{\frac{dx}{x}}=\ln{x}[/math] and [math]\int{\cos{x}dx}=\sin{x}[/math]

Which one of the following statements is true?

[math]\int{\frac{dx}{\cos{x}}}=\ln{\sin{x}}+C[/math]

[math]\int{\frac{dx}{\cos{x}}}=\sin{\ln{x}}+C[/math]

[math]\int{\frac{dx}{\cos{x}}}=\ln{x}+\frac{1}{\sin{x}}+C[/math]

[math]\int{\frac{dx}{\cos{x}}}[/math] cannot be calculated on base of the given.

The last option should be the correct one.

But what about this:

[math]\int{\left[1\cdot \frac{1}{\cos{x}}dx\right]}=\frac{x}{\cos{x}}-\int{\left[x d\left(\frac{1}{\cos{x}}\right)\right]}[/math]

[math]\cdots = \frac{x}{\cos{x}}-\int{\left[x\cdot\sin{x}\right]}=\frac{x}{\cos{x}}-\int{\left[x\cdot d\left(-\cos{x}\right)\right]}[/math]

[math]\cdots = \frac{x}{\cos{x}}+x\cdot\cos{x}+\int{\cos{x}dx}=\frac{x}{\cos{x}}+x\cdot\cos{x}+\sin{x}+C[/math]

It's confusing to say the least. I wonder what the penguins think?

It's as easy as (-(-(+1)) = +1

Do birds even think? O.O

 

Antiantarctic penguins?

Aren't Antarctic penguins already 'anti-Arctic' penguins?

So.. anti-anti-Arctic penguins would be.. Arctic penguins?