133 lines
4.0 KiB
Typst
133 lines
4.0 KiB
Typst
#import "@preview/physica:0.9.6": *
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#import "lib.typ": ccases
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= problem 2
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== a)
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click the code to obtain the sketches for the expansions of $f(x)$.
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#link(
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"https://uiua.org/pad?src=0_18_0-dev_2__eJxTVnCtKEgtysxNzStJzFHkqlNQSs8ssVJIzyzJKE3SS87P1ffPzcvMTizSL80sTdQtyMkvUVKoU3BJLElUCMjJL-F61LHyUc_8R11Nj7p2HFr_qL3nUdeOR20zDBUeNU08vP3R_DmP2hcamkKUzXvUNV_X8PB0o8PbdQ1B6pbvPrThUVc_SFYDZBrE3Ee9nZpcAB-1SDo=",
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)[```uiua
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# Experimental!
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~ "git: github.com/Omnikar/uiua-plot" ~ Data Plot
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∩⌟⊂⊸¯⇌⊸↘1 ₑ÷⟜⇡15
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∩⌞⊟-1×2÷-1⊸⧻°⊏
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∩(Plot Data ⍉)
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```]
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#image("./2a_even_expansion.png")
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this is how the even expansion would look.
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#image("./2a_odd_expansion.png")
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this is how the odd expansion would look. it is a little off because i didn't
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bother to make a duplicate point at -1 to highlight the discrete jump.
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these are both created by evaluating $f(x) = exp(x)$ on $[0, 1]$, then taking
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those values in reverse from 0 to -1, one of them negated to obtain the odd
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property.
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== b)
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it makes sense that the even expansion is more practical in this case, since it
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is smoother and doesn't contain a discrete jump, only a point at which it isn't
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differentiable, but it is continuous.
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the gibbs artifacts will be smaller compared to the odd expansion, which will
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have large artifacts at the discrete jump, making for a less accurate
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approximation of $f(x)$ using a truncated fourier series.
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== c)
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let now $f(x) = x - x^2$ on $[0, 1]$.
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we only need to change our single function in the code, namely change from `ₑ`
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to `(-⊸°√)`.
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#link(
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"https://uiua.org/pad?src=0_18_0-dev_2__eJxTVnCtKEgtysxNzStJzFHkqlNQSs8ssVJIzyzJKE3SS87P1ffPzcvMTizSL80sTdQtyMkvUVKoU3BJLElUCMjJL-F61LHyUc_8R11Nj7p2HFr_qL3nUdeOR20zDBU0dEEiGx51zNI8vP3R_DmP2hcamkKUz3vUNV_X8PB0o8PbdQ1B6pfvPrThUVc_SFYDZCrE_Ee9nZpcAGxZTF0=",
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)[```uiua
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... ∩⌟⊂⊸¯⇌⊸↘1 (-⊸°√)÷⟜⇡15 ...
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```]
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#image("./2c_even_expansion.png")
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#table(
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image("./marine.png"), align(horizon)[this is how the even expansion looks.],
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stroke: none,
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columns: 2,
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)
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#image("./2c_odd_expansion.png")
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this is the odd expansion. very nice.
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we can see that the odd expansion already resembles a sine-wave. we can then
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guess that this will the more accurate one of the two.
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== d)
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denote the even and odd expansions as functions respectively
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$
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g(x) = ccases(x - x^2, x >= 0, -x - x^2)space, quad
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h(x) = ccases(x - x^2, x >= 0, -x + x^2)
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$
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we can find the coefficients for these by calculating the integrals from the
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definitions of the coefficients
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$
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a_0 (g) & = integral_(-1)^1 g(x) dd(x) \
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& = integral_(-1)^0 (-x - x^2) dd(x) + integral_0^1 (x - x^2) dd(x) \
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& = -1/2 [x^2]_(-1)^0 - 1/3 [x^3]_(-1)^0
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+ 1/2 [x^2]_0^1 - 1/3 [x^3]_0^1 \
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& = -1/2 - 1/3 + 1/2 - 1/3 = -2/3
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$
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$
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a_n (g) & = integral_(-1)^1 g(x) cos(2 pi n x) dd(x) \
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& = integral_(-1)^0 (-x - x^2) cos(2 pi n x) dd(x)
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+ integral_0^1 (x - x^2) cos(2 pi n x) dd(x) \
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& = 2 integral_0^1 (x - x^2) cos(2 pi n x) dd(x) \
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& = 2 [(x-x^2)/(2 pi n) sin(2 pi n x)
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+ (1 - 2x)/(4 pi^2 n^2) cos(2 pi n x)
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+ 2/(8 pi^3 n^3) sin(2 pi n x)]_0^1 \
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& = 2 [-1/(4 pi^2 n^2) - 1/(4 pi^2 n^2)] \
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& = -1/(pi^2 n^2)
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$
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$
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b_n (g) & = integral_(-1)^1 g(x) sin(2 pi n x) dd(x) = 0
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$
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$
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a_0 (h) & = integral_(-1)^1 h(x) dd(x) \
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& = integral_(-1)^0 (-x + x^2) dd(x) + integral_0^1 (x - x^2) dd(x) \
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& = -1/2 [x^2]_(-1)^0 + 1/3 [x^3]_(-1)^0
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+ 1/2 [x^2]_0^1 - 1/3 [x^3]_0^1 \
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& = -1/2 + 1/3 + 1/2 - 1/3 = 0
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$
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$
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a_n (h) & = integral_(-1)^1 h(x) cos(2 pi n x) dd(x) = 0
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$
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$
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b_n (h) & = integral_(-1)^1 h(x) sin(2 pi n x) dd(x) \
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& = integral_(-1)^0 (-x + x^2) sin(2 pi n x) dd(x)
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+ integral_0^1 (x - x^2) sin(2 pi n x) dd(x) \
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& = 2 integral_0^1 (x - x^2) sin(2 pi n x) dd(x) \
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& = 2 [(x^2 - x)/(2 pi n) cos(2 pi n x)
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+ (1 - 2x)/(4 pi^2 n^2) sin(2 pi n x)
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-2/(8 pi^3 n^3) cos(2 pi n x)]_0^1 \
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& = 0
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$
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well, that's definitely wrong.
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