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dumbish question: if cosine is simply sine but offset by pi/2 radians, why are we approximation these functions with an infinite sum of sines and cosines and no

dumbish question: if cosine is simply sine but offset by pi/2 radians, why are we approximation these functions with an infinite sum of sines and cosines and not doing it just with sines?

dspig 10y ago
it will depend on the implementation and processor if it's faster to calculate simultaneous sin and cos of a given angle, or sin(w) and sin(w+pi/2), but you are right you could do it all with sin or cos only. But there is already a different thing called a discrete cosine transform :)
lutusp 10y ago
One reason is that in floating-point processing, in some cases adding a constant to an argument can truncate and lose the original argument, so using a single function will fail in a class of pathological edge cases: import numpy as np a = np.pi/2 print(a) > 1.5707963267948966 print(a == a + 1e-15) > False print(a == a + 1e-16) > True print(np.sin(1e-16)) > 1e-16 print(np.cos(np.pi/2+1e-16)) > 6.12323399574e-17 print(np.sin(1e-16) == np.cos(np.pi/2+1e-16)) > False
mjfl 10y ago
I believe the reason is due to cosine being even, i.e. cos(-x) = cos(x), and sine being odd, sin(-x) = -sin(x). The sum of even functions is still even, and the sum of odd functions is still odd. If you have a function that is symmetric about the origin, say f(x) = x^2, then you can approximate it with only cosines, but in general functions are not odd or even so you need to use a mix of sines and cosines.
wayn3 10y ago
Well, you may not like this answer, but: If cosine is just sine rotated by 90°, then we really just use sines. Except we write cosine instead of a sine with a 90° rotation because it looks less confusing :) The real goal is to arrive at the exponential representation, where the sines and cosines are absorbed into the exponential functions, which is later needed to write down the fourier integrals. Your question boils down to preference. The two things you name are, in fact, equal. Math people just choose to write cos(x) instead of sin(x+pi/2)
- Hongkongcunt 10y ago
  How do we calculate how big of a cunt you are?
- deleted 10y ago
  [deleted]
kmill 10y ago
The "true" Fourier transform is to write the function as an infinite sum of e^(2 pi i n x) with complex coefficients and n an integer. (This is the "true" Fourier transform because of a connection between periodic functions and circles, and because the representation theory of the circle group says that these are the so-called irreducible representations. This is extremely abridged -- the point is just to mention relevant words.) Due to Euler's formula, there is a correspondence between a linear combination of e^(2 pi i n x) and e^(-2 pi i n x) and a linear combination of cos(2 pi n x) and sin(2 pi n x). In other words, both pairs of functions span the same subspace (with complex coefficients). We need all of the exponentials to represent a periodic function, with positive and negative n, and so we need both cos and sin. If you leave out cos, you can only represent odd functions, and leaving out sin, even functions. But: if your question is just why not just use sin(x) and sin(x-pi/2) instead of sin(x) and cos(x), well, it's the same thing. Just because you write cos(x) like sin(x-pi/2) doesn't mean it's not still cos(x)! (Fun fact: cosine means "complementary sine," the sine of the complementary angle in the triangle.)
drinkwell 10y ago
Because with a Fourier series/transform you only get to play with the amplitude of the sine and cosine functions, not the offset. If you only had cosine functions in your sum you'd only be able to represent symmetric functions (and antisymmetric if only sine). Except actually I lied; if you sum both sine and cosine functions you do get to specify offsets. Think of this trig identity: sin(x + offset) = [cos(offset)] sin(x) + [sin(offset)] cos(x). Think about the terms in brackets as just being amplitudes. By having both sine and cosine in the sum you can represent any arbitrary offset by just changing their relative amplitude. This is one of those things which is nicer when you think about it in terms of complex numbers.
JadeNB 10y ago
The series \sum a_n\cos(n\theta) + \sum b_n\sin(n\theta) and \sum a_n\sin(n\theta + \pi/2) + \sum b_n\sin(n\theta) are literally, exactly the same; it is purely a matter of convention which one you choose to write. In more programming-type language, 'cosine' is just syntactic sugar (or 'sine' is, if you prefer; or, I would rather say, both are, for the real and complex parts of the exponential); but sometimes syntactic sugar is handy!

dumbish question: if cosine is simply sine but offset by pi/2 radians, why are we approximation these functions with an infinite sum of sines and cosines and not doing it just with sines?

dspig 10y ago

it will depend on the implementation and processor if it's faster to calculate simultaneous sin and cos of a given angle, or sin(w) and sin(w+pi/2), but you are right you could do it all with sin or cos only. But there is already a different thing called a discrete cosine transform :)

lutusp 10y ago

One reason is that in floating-point processing, in some cases adding a constant to an argument can truncate and lose the original argument, so using a single function will fail in a class of pathological edge cases: import numpy as np a = np.pi/2 print(a) > 1.5707963267948966 print(a == a + 1e-15) > False print(a == a + 1e-16) > True print(np.sin(1e-16)) > 1e-16 print(np.cos(np.pi/2+1e-16)) > 6.12323399574e-17 print(np.sin(1e-16) == np.cos(np.pi/2+1e-16)) > False

mjfl 10y ago

I believe the reason is due to cosine being even, i.e. cos(-x) = cos(x), and sine being odd, sin(-x) = -sin(x). The sum of even functions is still even, and the sum of odd functions is still odd. If you have a function that is symmetric about the origin, say f(x) = x^2, then you can approximate it with only cosines, but in general functions are not odd or even so you need to use a mix of sines and cosines.

wayn3 10y ago

Well, you may not like this answer, but: If cosine is just sine rotated by 90°, then we really just use sines. Except we write cosine instead of a sine with a 90° rotation because it looks less confusing :) The real goal is to arrive at the exponential representation, where the sines and cosines are absorbed into the exponential functions, which is later needed to write down the fourier integrals. Your question boils down to preference. The two things you name are, in fact, equal. Math people just choose to write cos(x) instead of sin(x+pi/2)

Hongkongcunt 10y ago
How do we calculate how big of a cunt you are?
deleted 10y ago
[deleted]

kmill 10y ago

The "true" Fourier transform is to write the function as an infinite sum of e^(2 pi i n x) with complex coefficients and n an integer. (This is the "true" Fourier transform because of a connection between periodic functions and circles, and because the representation theory of the circle group says that these are the so-called irreducible representations. This is extremely abridged -- the point is just to mention relevant words.) Due to Euler's formula, there is a correspondence between a linear combination of e^(2 pi i n x) and e^(-2 pi i n x) and a linear combination of cos(2 pi n x) and sin(2 pi n x). In other words, both pairs of functions span the same subspace (with complex coefficients). We need all of the exponentials to represent a periodic function, with positive and negative n, and so we need both cos and sin. If you leave out cos, you can only represent odd functions, and leaving out sin, even functions. But: if your question is just why not just use sin(x) and sin(x-pi/2) instead of sin(x) and cos(x), well, it's the same thing. Just because you write cos(x) like sin(x-pi/2) doesn't mean it's not still cos(x)! (Fun fact: cosine means "complementary sine," the sine of the complementary angle in the triangle.)

drinkwell 10y ago

Because with a Fourier series/transform you only get to play with the amplitude of the sine and cosine functions, not the offset. If you only had cosine functions in your sum you'd only be able to represent symmetric functions (and antisymmetric if only sine). Except actually I lied; if you sum both sine and cosine functions you do get to specify offsets. Think of this trig identity: sin(x + offset) = [cos(offset)] sin(x) + [sin(offset)] cos(x). Think about the terms in brackets as just being amplitudes. By having both sine and cosine in the sum you can represent any arbitrary offset by just changing their relative amplitude. This is one of those things which is nicer when you think about it in terms of complex numbers.

JadeNB 10y ago

The series \sum a_n\cos(n\theta) + \sum b_n\sin(n\theta) and \sum a_n\sin(n\theta + \pi/2) + \sum b_n\sin(n\theta) are literally, exactly the same; it is purely a matter of convention which one you choose to write. In more programming-type language, 'cosine' is just syntactic sugar (or 'sine' is, if you prefer; or, I would rather say, both are, for the real and complex parts of the exponential); but sometimes syntactic sugar is handy!