Topic: Math tho
Mr. Bloody Lemonade !0JqexkOGBc started this discussion 3 weeks ago #136,487
You know how the distance formula is really just the square root of the dot product of a vector with itself, but you can define an inner product space using an integral since integration is linear and then find the distance between two polynomials?
What does that actually mean tho, conceptually? Like how does one imagine what that means for there to be a distance between 5x^2-8 and 8x^4+2x^2 that’s just one number?
What does that actually mean tho, conceptually? Like how does one imagine what that means for there to be a distance between 5x^2-8 and 8x^4+2x^2 that’s just one number?
Mr. Bloody Lemonade !0JqexkOGBc (OP) double-posted this 3 weeks ago, 7 minutes later[^] [v] #1,448,905
Like you do the thing and it works and that’s cool, and you get the answer. But what does the answer mean? I mean, it’s a number.
Anonymous B joined in and replied with this 3 weeks ago, 1 hour later, 1 hour after the original post[^] [v] #1,448,915
You don't know it, it's your opinion. Believe whatever you want, but don't tell me what we "know" when it just ain't so.
Mr. Bloody Lemonade !0JqexkOGBc joined in and replied with this 3 weeks ago, 13 minutes later, 1 hour after the original post[^] [v] #1,448,918
And what does the concept that two polynomials can be orthogonal within an inner product space mean as in, visually, if you picture the polynomials, what does two polynomials being orthogonal look like? Like what does that mean?
Mr. Bloody Lemonade !0JqexkOGBc double-posted this 3 weeks ago, 2 minutes later, 1 hour after the original post[^] [v] #1,448,919
It means something. It has to.Mr. Bloody Lemonade !0JqexkOGBc triple-posted this 3 weeks ago, 7 minutes later, 1 hour after the original post[^] [v] #1,448,920
And then you can make an orthonormal basis and from that bases you can make an infinite number of orthogonal polynomials. But a different inner product space would have a different orthonormal basis. But it’s all just curved squiggly lines.
Mr. Bloody Lemonade !0JqexkOGBc quadruple-posted this 3 weeks ago, 3 minutes later, 1 hour after the original post[^] [v] #1,448,921
This makes sense to me. This I can understand. Straight lines being orthogonal. That makes sense.Mr. Bloody Lemonade !0JqexkOGBc quintuple-posted this 3 weeks ago, 2 minutes later, 1 hour after the original post[^] [v] #1,448,922
Like I get the concept. That (0,0) and (1,1) have a distance. I get that. x^2 and 5x^3 have a distance… what does that mean exactly.
Mr. Bloody Lemonade !0JqexkOGBc sextuple-posted this 3 weeks ago, 1 minute later, 2 hours after the original post[^] [v] #1,448,923
And not a distance at all point, like, the functions have a distance.
Anonymous D joined in and replied with this 3 weeks ago, 9 minutes later, 2 hours after the original post[^] [v] #1,448,927
My parents beat me cause I couldn't do math
boof joined in and replied with this 3 weeks ago, 39 minutes later, 2 hours after the original post[^] [v] #1,448,941
@1,448,922 (Mr. Bloody Lemonade !0JqexkOGBc)
they have a distance? is that like, as simple as difference in y given same x? or something goofier
they have a distance? is that like, as simple as difference in y given same x? or something goofier
Mr. Bloody Lemonade !0JqexkOGBc joined in and replied with this 3 weeks ago, 9 hours later, 12 hours after the original post[^] [v] #1,448,999
@previous (boof)
(1 of 2: explaining the line thing)
So basically, you have the Pythagorean theorem right?
a^2 + b^2 = c^2
Well, that’s for a right angle triangle to find the hypotenuse c from the sides an and b. What else are at right angles? The x and y axes.
Since (-a)^2 = a^2 the order doesn’t matter. But you can come up with this formula for distance in 2D space:
d = sqrt((x1-x2)^2 + (y1-y2)^2)
That’s cool, but what if you want to do it for three dimensions?
d = sqrt((x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2)
So you still use sqrt no matter how many dimensions there are. What is squaring a number? It’s multiplying by itself so you could rewrite this as:
d = sqrt((x1-x2)(x1-x2) + (y1-y2)(y1-y2) + (z1-z2)(z1-z2))
Now what if instead of points, you think about two vectors in 3d space, x and y? u = <x1, y1, z1>, v = <x2, y2, z2>
Now obviously the notation is wrong usually they’re column vectors not row vectors (can’t do that on this website easily), and they’d be numbered u1, u2, etc. but this is for making it clear.
So given those two vectors, the operating is basically this:
d = sqrt((u-v).(u-v))
And that’s just true based on the definition of the dot product. If we say another vector, say g=u-v, then ||g|| (the magnitude of g) is the square root of the dot product of g with itself. It turns out this works for vectors dimensions of any dimension.
(1 of 2: explaining the line thing)
So basically, you have the Pythagorean theorem right?
a^2 + b^2 = c^2
Well, that’s for a right angle triangle to find the hypotenuse c from the sides an and b. What else are at right angles? The x and y axes.
Since (-a)^2 = a^2 the order doesn’t matter. But you can come up with this formula for distance in 2D space:
d = sqrt((x1-x2)^2 + (y1-y2)^2)
That’s cool, but what if you want to do it for three dimensions?
d = sqrt((x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2)
So you still use sqrt no matter how many dimensions there are. What is squaring a number? It’s multiplying by itself so you could rewrite this as:
d = sqrt((x1-x2)(x1-x2) + (y1-y2)(y1-y2) + (z1-z2)(z1-z2))
Now what if instead of points, you think about two vectors in 3d space, x and y? u = <x1, y1, z1>, v = <x2, y2, z2>
Now obviously the notation is wrong usually they’re column vectors not row vectors (can’t do that on this website easily), and they’d be numbered u1, u2, etc. but this is for making it clear.
So given those two vectors, the operating is basically this:
d = sqrt((u-v).(u-v))
And that’s just true based on the definition of the dot product. If we say another vector, say g=u-v, then ||g|| (the magnitude of g) is the square root of the dot product of g with itself. It turns out this works for vectors dimensions of any dimension.
Mr. Bloody Lemonade !0JqexkOGBc double-posted this 3 weeks ago, 12 minutes later, 12 hours after the original post[^] [v] #1,449,001
(2 of 2: explaining the polynomial thing)
Instead of defining the dot product this way:
u.v = (u1*v1) + (u2.v2) …
What if you could arbitrarily define your own inner product space? Well it turns out you can with integration because integration is just the inverse of a derivative and derivatives are linear transformations and we’re dealing with vectors which are basically lines that have a start and an end point and a direction, but are still sorta technically just lines.
So what if you define an inner product space like this?
<a(x), b(x)> = integral of a(x)b(x) from 0 to 1
Say c(x) is x^2 + 2x and d(x) = x^2 as just random polynomial examples.
An inner product can just be used in place of a dot product so you could do:
d = sqrt(<c(x)-d(x), c(x)-d(x)>)
To make it easier, just ignoring the square root for now:
<c(x)-d(x), c(x)-d(x)> = integral of from 0 to 1
< x^2 + 2x - x^2, x^2 + 2x - x^2> = integral of (2x)(2x) from 0 to 1
= integral of 4x^2 from 0 to 1
= ((4/3)x^3 where x = 1) - ((4/3)x^3 where x = 0)
The right side of the subtraction is obviously 0 so you can just pretend it doesn’t exist
= (4/3)x^3 where x = 1)
= 4/3
So now if you add the square root back in, the distance between x^2 + 2x and x^2 is sqrt(4/3)
But the thing is, what does the number sqrt(4/3) actually mean?
Instead of defining the dot product this way:
u.v = (u1*v1) + (u2.v2) …
What if you could arbitrarily define your own inner product space? Well it turns out you can with integration because integration is just the inverse of a derivative and derivatives are linear transformations and we’re dealing with vectors which are basically lines that have a start and an end point and a direction, but are still sorta technically just lines.
So what if you define an inner product space like this?
<a(x), b(x)> = integral of a(x)b(x) from 0 to 1
Say c(x) is x^2 + 2x and d(x) = x^2 as just random polynomial examples.
An inner product can just be used in place of a dot product so you could do:
d = sqrt(<c(x)-d(x), c(x)-d(x)>)
To make it easier, just ignoring the square root for now:
<c(x)-d(x), c(x)-d(x)> = integral of from 0 to 1
< x^2 + 2x - x^2, x^2 + 2x - x^2> = integral of (2x)(2x) from 0 to 1
= integral of 4x^2 from 0 to 1
= ((4/3)x^3 where x = 1) - ((4/3)x^3 where x = 0)
The right side of the subtraction is obviously 0 so you can just pretend it doesn’t exist
= (4/3)x^3 where x = 1)
= 4/3
So now if you add the square root back in, the distance between x^2 + 2x and x^2 is sqrt(4/3)
But the thing is, what does the number sqrt(4/3) actually mean?
Mr. Bloody Lemonade !0JqexkOGBc triple-posted this 3 weeks ago, 3 minutes later, 12 hours after the original post[^] [v] #1,449,002
And if <c(x), d(x)> is 0, that means that in the inner product space, c(x) and d(x) are vectors that are perpendicular at a right angle (orthogonal), except c(x) and d(x) aren’t actually straight lines, they’re polynomials.
Mr. Bloody Lemonade !0JqexkOGBc quadruple-posted this 3 weeks ago, 1 minute later, 12 hours after the original post[^] [v] #1,449,003
And you could use a different integral over a different range (I just made it easier on purpose), so there are an infinite number of ways you could define distance.
Mr. Bloody Lemonade !0JqexkOGBc quintuple-posted this 3 weeks ago, 2 minutes later, 12 hours after the original post[^] [v] #1,449,004
And then if you find a set of vectors that are orthogonal to each other, that’s an orthonormal basis, and you can multiply a scalar value to any of those vectors and get an infinite number of alternative orthonormal vectors that are all perpendicular.
Mr. Bloody Lemonade !0JqexkOGBc sextuple-posted this 3 weeks ago, 4 minutes later, 12 hours after the original post[^] [v] #1,449,006
Like it’s not the distance between x and y, it’s the square root of the area under the curve of the square of one function minus another function.
Engineer II joined in and replied with this 3 weeks ago, 15 minutes later, 13 hours after the original post[^] [v] #1,449,008
@1,448,905 (Mr. Bloody Lemonade !0JqexkOGBc)
> Like you do the thing and it works and that’s cool, and you get the answer. But what does the answer mean? I mean, it’s a number.
Means a great deal when designing filters for electronic devices. Different slopes produce minimal sidebands or a sharper filter that produces more sidebands.
> Like you do the thing and it works and that’s cool, and you get the answer. But what does the answer mean? I mean, it’s a number.
Means a great deal when designing filters for electronic devices. Different slopes produce minimal sidebands or a sharper filter that produces more sidebands.
Mr. Bloody Lemonade !0JqexkOGBc replied with this 3 weeks ago, 2 minutes later, 13 hours after the original post[^] [v] #1,449,009
Mr. Bloody Lemonade !0JqexkOGBc double-posted this 3 weeks ago, 2 minutes later, 13 hours after the original post[^] [v] #1,449,010
Also, tangent: but if everything works regardless of how many dimensions you have why do we have 3?
Mr. Bloody Lemonade !0JqexkOGBc joined in and replied with this 3 weeks ago, 7 minutes later, 13 hours after the original post[^] [v] #1,449,011
Or I guess maybe you could say 4, but time only goes one way for some reason.
(Edited 13 seconds later.)
Engineer II replied with this 3 weeks ago, 1 minute later, 13 hours after the original post[^] [v] #1,449,012
@1,449,009 (Mr. Bloody Lemonade !0JqexkOGBc)
They even have names
Butterworth filter
Chebyshev filter
Elliptic (Cauer) filter
Bessel filter
Gaussian filter
Optimum "L" (Legendre) filter
Linkwitz–Riley filter
They even have names
Butterworth filter
Chebyshev filter
Elliptic (Cauer) filter
Bessel filter
Gaussian filter
Optimum "L" (Legendre) filter
Linkwitz–Riley filter
Mr. Bloody Lemonade !0JqexkOGBc joined in and replied with this 3 weeks ago, 1 hour later, 15 hours after the original post[^] [v] #1,449,022
boof replied with this 3 weeks ago, 3 minutes later, 15 hours after the original post[^] [v] #1,449,023
oh, a square root of a difference in areas under squared functions' curves. hmmm
Engineer II replied with this 3 weeks ago, 1 hour later, 16 hours after the original post[^] [v] #1,449,043
@1,449,022 (Mr. Bloody Lemonade !0JqexkOGBc)
Electronics - Hardware - Prior Imaging Radar and now Satellites.
Electronics - Hardware - Prior Imaging Radar and now Satellites.
Anonymous J joined in and replied with this 3 weeks ago, 4 hours later, 21 hours after the original post[^] [v] #1,449,083
nobody in this thread has ever raped a nigguh, and you can just smell the Lstink wafting off their poasts
Anonymous K joined in and replied with this 3 weeks ago, 48 minutes later, 22 hours after the original post[^] [v] #1,449,091
The IQ gap within this thread is fascinating.
@previous (J)
> nobody in this thread has ever raped a nigguh, and you can just smell the Lstink wafting off their poasts
@previous (J)
> nobody in this thread has ever raped a nigguh, and you can just smell the Lstink wafting off their poasts
Anonymous J replied with this 3 weeks ago, 13 minutes later, 22 hours after the original post[^] [v] #1,449,095
Anonymous K replied with this 3 weeks ago, 1 minute later, 22 hours after the original post[^] [v] #1,449,097
Anonymous J replied with this 3 weeks ago, 13 minutes later, 22 hours after the original post[^] [v] #1,449,104
@previous (K)
yeah, they gotta be beaten over the head with a certificate of acheivement, and their lack of one
yeah, they gotta be beaten over the head with a certificate of acheivement, and their lack of one