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Fourier Series

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Can someone clarify or explain what the Fourier Series is? From what I understand, its a way in which periodic funx can be expressed as a sine & cosine funxs together;

 

However, seeing as how my knowledge in Calculus is limited to Grade School Advanced Functions & Introductory Calculus, I'm just like :eek: when I see sites like http://www.nst.ing.tu-bs.de/schaukasten/fourier/en_idx.html#DIRI & http://mathworld.wolfram.com/FourierSeries.html

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That's really all there is to it.

 

Aside from that, there's just a lot of maths.

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Originally posted by MrL_JaKiri

Aside from that, there's just a lot of maths.

 

hm...I don't get those parts..hehe

 

At what point do you learn those trig. integral identities?

Or integration even?

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Originally posted by NSX

Can someone clarify or explain what the Fourier Series is? From what I understand, its a way in which periodic funx can be expressed as a sine & cosine funxs together;

 

Yes; the thing is that, in general, you need an infinite number of sines and cosines to represent the function. That is why you see the infinite sums.

 

However, seeing as how my knowledge in Calculus is limited to Grade School Advanced Functions & Introductory Calculus, I'm just like :eek: when I see sites like http://www.nst.ing.tu-bs.de/schaukasten/fourier/en_idx.html#DIRI & http://mathworld.wolfram.com/FourierSeries.html [/b]

 

There is really nothing on those websites that is beyond basic calculus and trigonometry, so with some instruction and study you should be able to get it. What specifically is giving you trouble?

 

At what point do you learn those trig. integral identities?

Or integration even?

 

You learn the first in trigonometry, the second in calculus.

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Originally posted by Tom

There is really nothing on those websites that is beyond basic calculus and trigonometry, so with some instruction and study you should be able to get it. What specifically is giving you trouble?

 

I haven't learned integration yet; care to teach me? :P

 

But I was just thinking, if you have an infinite number of sine & cosine funxs, wouldn't they just nullify each other causing 0 amplitude?

Or am I missing something here?

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Originally posted by NSX

But I was just thinking, if you have an infinite number of sine & cosine funxs, wouldn't they just nullify each other causing 0 amplitude?

Or am I missing something here?

 

Of different amplitudes and positions.

 

Such as sin(x) + sin(x+15).

 

As to the integration, basically

 

:int:ax^n dx = (ax^n+1) / (n+1) + c.

 

It gets a lot more complicated than that, but that's the basic idea.

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Originally posted by MrL_JaKiri

As to the integration, basically

 

:int:ax^n dx = (ax^n+1) / (n+1) + c.

 

It gets a lot more complicated than that, but that's the basic idea.

 

ooh...I gotta jot it down.

hehe

 

Originally posted by MrL_JaKiri

Of different amplitudes and positions.

 

Such as sin(x) + sin(x+15).

 

ooh...I see; cool thanks.

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Originally posted by MrL_JaKiri

As to the integration, basically

 

:int:ax^n dx = (ax^n+1) / (n+1) + c.

 

So ax^n is your original funx.

Is dx is the derivative of the original funx?

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No offense man, but if you are still in Highschool geometry right now(as i read in another of your posts) that stuff is going to probably be outside your comprehension for a while. I'm in calculus 3 and can't even remotely follow what they are talking about doing(a 50 step proof?!?). I think most people will never get to math like that, unless they are physics or math majors.

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Originally posted by Bsun

No offense man, but if you are still in Highschool geometry right now(as i read in another of your posts) that stuff is going to probably be outside your comprehension for a while. I'm in calculus 3 and can't even remotely follow what they are talking about doing(a 50 step proof?!?). I think most people will never get to math like that, unless they are physics or math majors.

 

Well, I'm going into that; and its just a interest right now; like, dave helped me to prove some of the other trig identities last time; but I'm still missing the link b/t Calculus & Geometry...

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Originally posted by NSX

So ax^n is your original funx.

Is dx is the derivative of the original funx?

 

No, in all cases 'dx' means 'with respect to x',

 

ie. in differenciation d(f(x))/dx means 'the differenciation of the function f(x) with respect to x', as opposed to dy or dz, say.

 

d(f(x))/dy = 0, because f(x) doesn't change as y changes.

 

It's just being precise what you mean.

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Originally posted by MrL_JaKiri

No, in all cases 'dx' means 'with respect to x',

 

ie. in differenciation d(f(x))/dx means 'the differenciation of the function f(x) with respect to x', as opposed to dy or dz, say.

 

d(f(x))/dy = 0, because f(x) doesn't change as y changes.

 

It's just being precise what you mean.

 

That's cool; lots of practice with implicit differentiation helps...lol

So with the original thingy; did you mean:

:lint: d(ax^n)/dx ?

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No.

 

[edit]

 

y = x^2

 

dy/dx = 2x

 

1 dy = 2x dx

 

:int: 1 dy = :int: 2x dx

 

y = x^2 + c

 

c = 0 in this case.

 

:int: d(f(x))/dx doesn't mean anything, as there's only one dx command, which goes with the differentiation.

 

:int: f'(x) has no meaning; how can you integrate the rate of change of something when you haven't got a variable that's changing?

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Originally posted by MrL_JaKiri

No.

 

[edit]

 

y = x^2

 

dy/dx = 2x

 

1 dy = 2x dx

 

:int: 1 dy = :int: 2x dx

 

ooh!!!

I never thought of that!

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Originally posted by NSX

ooh!!!

I never thought of that!

 

It's a bit of a cheat :confused:

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Originally posted by MrL_JaKiri

No.

 

[edit]

 

y = x^2

 

dy/dx = 2x

 

1 dy = 2x dx

 

:int: 1 dy = :int: 2x dx

||

\/

y = x^2 + c

 

c = 0 in this case.

 

How'd you go from that step to the other?

 

edit:

You'll have to excuse my cheesed arrow there.

=)

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Originally posted by MrL_JaKiri

:int:ax^n dx = (ax^n+1) / (n+1) + c.

 

Apply that to both sides (replacing x's with y's for the :int:dy ofc)

 

(remember that y^0 = 1)

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Are there second integrals as well?

 

I mean, dy^2 / d^2x is the second derivative of y=f(x)

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You can integrate things twice if you want to. Not much point most of the time unless you're doing second order differential equations, but those are mostly solved from specific guidelines already laid out.

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dave said in post #19 :

You can integrate things twice if you want to. Not much point most of the time unless you're doing second order differential equations, but those are mostly solved from specific guidelines already laid out.

 

What exactly are differential eq'ns[/i]. I've been meaning to ask my phys teacher, since that's where we've been applying it, but haven't got around to it since I wanted to look it up a bit first; but other courses have been...lagging my intuition

 

:P

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A differential equation is an equation which has a differential in it.

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Well, that kindof sums it up bluntly, but yes :P

 

e.g.

 

x dy/dx = y

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:eek:

 

:)

 

Cool, thanks;

 

I was reading some stuff on Worlfram @ school earlier on in the day, and that stuff ahd me worried.

:P

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The trouble with wolfram is that it's all highly technical. I shouldn't worry too much about it if you don't have a solid foundation in mathematics.

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i havent done fourier series but i have read up a lil about it. if i understand correctly then. all it says is that

 

any periodic fucntion (cant remember what the period must be . 2pi or something or other) can be written as a linear combination of sums of sine and sums of cosines.

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