Show that for any real constants c and d, the eqation x = c + dcos(x) has at least one root.
My professor gave the hint: Find an interval [a,b] on which
f(x) = x - c - dcos(x) changes sign.
Any advice?
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Show that for any real constants c and d, the eqation x = c + dcos(x) has at least one root.
My professor gave the hint: Find an interval [a,b] on which
f(x) = x - c - dcos(x) changes sign.
Any advice?
f(x) = x - c - dcos(x)
Case 1 dcos(c) > 0
f(c) = -dcos(c) < 0
f(c + pi) = pi -dcos(c+pi) = pi + dcos(c) > 0
We're done
Case 2 dcos(c) < 0
f(c) = -dcos(c) > 0
Let n be such that npi > dcos(c)
f(c-npi) = -npi -dcos(c-npi) = -npi - dcos(c)cos(npi)
cos(npi) = + 1
in either case f(c-npi) < 0