63 lines
1.6 KiB
Typst
63 lines
1.6 KiB
Typst
#import "@preview/physica:0.9.6": *
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= problem 2
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== a)
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$h(x) := f(x) g(x)$ is odd for odd $f$ and even $g$, since
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$
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h(x) = -f(-x) g(-x) = - (f(-x) g(-x)) = -h(-x)
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$
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== b)
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if $f$ and $g$ are both odd or even, then $h(x) := f(x) g(x)$ is even
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1. for the first case: $f$ and $g$ are both even
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$
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h(x) = f(x) g(x) = f(-x) g(-x) = h(-x)
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$
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2. for the second case: $f$ and $g$ are both odd
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$
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h(x) & = f(x) g(x) \
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& = (-f(-x)) (-g(-x)) = f(-x) g(-x) = h(-x)
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$
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== c)
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if $f$ is odd and $g$ is even then both $f compose g$ and $g compose f$ are
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even, since
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1. $f(g(x)) = f(g(-x))$, since g is even
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2. $g(f(x)) = g(-f(-x)) = g(f(-x))$
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== d)
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if $f$ is odd and $L > 0$,
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$
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integral_(-L)^L f(x) dd(x) & = integral_(-L)^0 f(x) dd(x)
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+ integral_0^L f(x) dd(x) \
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& = integral_0^L f(-x) (-1) dd(x)
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+ integral_0^L f(x) dd(x) \
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& = -integral_0^L f(-x) dd(x)
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+ integral_0^L f(x) dd(x) \
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& = -integral_0^L f(x) dd(x)
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+ integral_0^L f(x) dd(x) \
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& = 0
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$
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== e)
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if $f$ is even and $L > 0$,
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$
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integral_(-L)^L f(x) dd(x) & = integral_(-L)^0 f(x) dd(x)
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+ integral_0^L f(x) dd(x) \
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& = integral_0^L f(-x) dd(x)
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+ integral_0^L f(x) dd(x) \
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& = integral_0^L f(x) dd(x)
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+ integral_0^L f(x) dd(x) \
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& = 2 integral_0^L f(x) dd(x)
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$
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