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| == A graphical illustration of $\lim_{x -> 0} sin(x)/x =1$ == | == A graphical illustration of sin(x)/x -> 1 as x-> 0 == by Wai Yan Pong |
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| html('Below is a unit circle, so the length of the <font color=red>red line</font> is |sin(x)|') | html('Below is the unit circle, so the length of the <font color=red>red line</font> is |sin(x)|') |
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| html('is |sin(x)| and |tan(x)|, respectively where x is length of the arc.') html('From the picture, we see that |sin(x)| $\le$ |x| $\le$ |tan(x)|') html('and it follows easily from this that') html('cos(x) $\le$ sin(x)/x $\le$ 1 when x is near 0.') |
html('From the picture, we see that |sin(x)| $\le$ |x| $\le$ |tan(x)|.') html('It follows easily from this that cos(x) $\le$ sin(x)/x $\le$ 1 when x is near 0.') |
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| pretty_print("The limit of sin(x)/x as x tends to 0 is 1") | pretty_print("The limit of sin(x)/x as x tends to 0 is 1.") |
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== Root Finding Using Bisection == by William Stein {{{ def bisect_method(f, a, b, eps): try: f = f._fast_float_(f.variables()[0]) except AttributeError: pass intervals = [(a,b)] two = float(2); eps = float(eps) while True: c = (a+b)/two fa = f(a); fb = f(b); fc = f(c) if abs(fc) < eps: return c, intervals if fa*fc < 0: a, b = a, c elif fc*fb < 0: a, b = c, b else: raise ValueError, "f must have a sign change in the interval (%s,%s)"%(a,b) intervals.append((a,b)) html("<h1>Double Precision Root Finding Using Bisection</h1>") @interact def _(f = cos(x) - x, a = float(0), b = float(1), eps=(-3,(-16..-1))): eps = 10^eps print "eps = %s"%float(eps) try: time c, intervals = bisect_method(f, a, b, eps) except ValueError: print "f must have opposite sign at the endpoints of the interval" show(plot(f, a, b, color='red'), xmin=a, xmax=b) else: print "root =", c print "f(c) = %r"%f(c) print "iterations =", len(intervals) P = plot(f, a, b, color='red') h = (P.ymax() - P.ymin())/ (1.5*len(intervals)) L = sum(line([(c,h*i), (d,h*i)]) for i, (c,d) in enumerate(intervals) ) L += sum(line([(c,h*i-h/4), (c,h*i+h/4)]) for i, (c,d) in enumerate(intervals) ) L += sum(line([(d,h*i-h/4), (d,h*i+h/4)]) for i, (c,d) in enumerate(intervals) ) show(P + L, xmin=a, xmax=b) }}} attachment:bisect.png == Newton's Method == Note that there is a more complicated Newton's method below. by William Stein {{{ def newton_method(f, c, eps, maxiter=100): x = f.variables()[0] fprime = f.derivative(x) try: g = f._fast_float_(x) gprime = fprime._fast_float_(x) except AttributeError: g = f; gprime = fprime iterates = [c] for i in xrange(maxiter): fc = g(c) if abs(fc) < eps: return c, iterates c = c - fc/gprime(c) iterates.append(c) return c, iterates html("<h1>Double Precision Root Finding Using Newton's Method</h1>") @interact def _(f = x^2 - 2, c = float(0.5), eps=(-3,(-16..-1)), interval=float(0.5)): eps = 10^(eps) print "eps = %s"%float(eps) time z, iterates = newton_method(f, c, eps) print "root =", z print "f(c) = %r"%f(z) n = len(iterates) print "iterations =", n html(iterates) P = plot(f, z-interval, z+interval, rgbcolor='blue') h = P.ymax(); j = P.ymin() L = sum(point((w,(n-1-float(i))/n*h), rgbcolor=(float(i)/n,0.2,0.3), pointsize=10) + \ line([(w,h),(w,j)],rgbcolor='black',thickness=0.2) for i,w in enumerate(iterates)) show(P + L, xmin=z-interval, xmax=z+interval) }}} |
attachment:sinelimit.png |
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A graphical illustration of sin(x)/x -> 1 as x-> 0
by Wai Yan Pong
x=var('x')
@interact
def _(x = slider(-7/10,7/10,1/20,1/2)):
html('<h3>A graphical illustration of $\lim_{x -> 0} \sin(x)/x =1$</h3>')
html('Below is the unit circle, so the length of the <font color=red>red line</font> is |sin(x)|')
html('and the length of the <font color=blue>blue line</font> is |tan(x)| where x is the length of the arc.')
html('From the picture, we see that |sin(x)| $\le$ |x| $\le$ |tan(x)|.')
html('It follows easily from this that cos(x) $\le$ sin(x)/x $\le$ 1 when x is near 0.')
html('As $\lim_{x ->0} \cos(x) =1$, we conclude that $\lim_{x -> 0} \sin(x)/x =1$.')
if not (x == 0):
pretty_print("sin(x)/x = "+str(sin(float(x))/float(x)))
elif x == 0:
pretty_print("The limit of sin(x)/x as x tends to 0 is 1.")
C=circle((0,0),1, rgbcolor='black')
mvp = (cos(x),sin(x));tpt = (1, tan(x))
p1 = point(mvp, pointsize=30, rgbcolor='red'); p2 = point((1,0), pointsize=30, rgbcolor='red')
line1 = line([(0,0),tpt], rgbcolor='black'); line2 = line([(cos(x),0),mvp], rgbcolor='red')
line3 = line([(0,0),(1,0)], rgbcolor='black'); line4 = line([(1,0),tpt], rgbcolor='blue')
result = C+p1+p2+line1+line2+line3+line4
result.show(aspect_ratio=1, figsize=[3,3], axes=False)attachment:sinelimit.png