Math 104 - Calculus I

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Math 104 - Calculus I

• Part 8
• Power series, Taylor series

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Series

• First .. a review of what we have done so far
• 1. We examined series of constants and learned
  that we can say everything there is to say about
  geometric and telescoping series.
• 2. We developed tests for convergence of series
  of constants.
• 3. We considered power series, derived formulas
  and other tricks for finding them, and know them
  for a few functions.
• 4. We used the ratio tests to determine
  intervals on which power series converge, and use
  the other tests to check convergence at the
  endpoints of the intervals.

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Geometric series

• a/(1-r) a ar ar2 ar3 ...
• provided rlt 1.
• We often use partial fractions to detect
  telescoping series, for which we can calculate
  explicitly the partial sums sn

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Tests for convergence for series of constants

• Fundamental divergence test (nth term must go to
  zero for convergence to be possible)
• Integral test
• Comparison and limit comparison tests
• Ratio test
• Root test
• Alternating series test

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Power series

• f(x) a0 a1 x a2 x2 a3 x3 ...
• where an f(n)(0)/n!
• We know the series for ex, sin(x), cos(x),
• 1/(1-x), and a few related functions.

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Convergence of power series

• Before we get too excited about finding series,
  let's make sure that, at the very least, the
  series converge.
• Next week, we'll deal with the question of
  whether they converge to the function we expect.
  But for now, we'll assume that if they converge,
  they converge to the function they "came from".
• (Strictly speaking, this is not always true --
  but it is true for a large class of functions,
  which includes nearly all the ones encountered in
  basic science and mathematics. This fact was not
  fully appreciated until the early part of the
  twentieth century.)
• Fortunately, most of the question of whether
  power series converge is answered fairly directly
  by the ratio test.

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Recall that...

• for a series of constants , we have that
  the series converges (absolutely) if the the
  limit of the absolute value of is less than one,
  diverges if
• the limit is greater than one, and the test is
  indeterminate if the limit equals one.
• To use the ratio test on power series, just leave
  the x there and calculate the limit for each
  value of x. This will give an inequality that x
  must satisfy in order for the series to converge.
  

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For the series for the exponential function...
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Your turn...

• Calculate the series for the function sin(x) and
  determine for which x the series converges.

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Heres a more interesting example
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What remains...
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Final Conclusion
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OK, your turn...
A. -1 lt x lt 1 B. -2 lt x lt 2 C. 1/2 lt x lt 1/2
D. -2 lt x lt 2 E. -1/2 lt x lt 1/2
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One more...
A. -1 lt x lt 1 B. -1 lt x lt 1 C. 1 lt x lt 1 D. -1
lt x lt 1 E. 0 lt x lt 1
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From these examples,

• ...it should be apparent that power series
  converge for values of x in an interval that is
  centered at zero, i.e., an interval of the form
  -a, a , (-a, a, -a, a) or (-a, a) (where a
  might be either zero or infinity). The interval
  is called the interval of convergence and the
  number a is called the radius of convergence .

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Lets go back
To finding series of functions
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The other way
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Try this...

• Take the derivative of the series for sin(x) to
  get

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Integrate both sides of the geometric series from
0 to x to get
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Negate both sides and replace x by (-x)
everywhere to get
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Start from the geometric series again...
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And substitute x for x everywhere it appears to
get
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A challenge to think about...

• How to get the other one from previously

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Application of Series

• 1. Limits Series give a good idea of the
  behavior of functions in the neighborhood of 0
• We know for other reasons that
• We could do this by series

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This can be used on complicated limits...

• Calculate the limitA. 0
• B. 1/6
• C. 1
• D. 1/12
• E. does not exist

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Application of series (continued)

• 2. Approximate evaluation of integrals Many
  integrals that cannot be evaluated in closed form
  (i.e., for which no elementary anti-derivative
  exists) can be approximated using series (and we
  can even estimate how far off the approximations
  are).
• Example Calculate to the nearest 0.001.

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We begin by...
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According to Maple...

• The last series is an alternating series with
  decreasing terms. We need to find the first one
  that is less than 0.0005 to ensure that the error
  will be less than 0.001. According to Maple
• evalf(1/(7factorial(3))), evalf(1/(9factorial(4)
  )),evalf( 1/(11factorial(5)))
• evalf(1/(13factorial(6)))

.02380952381, .004629629630, .0007575757576
.0001068376068
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Keep going...

• So it's enough to go out to the 5! term. We do
  this as follows
• Sum((-1)n/((2n1)factorial(n)),n0..5)
  sum((-1)n/((2n1) factorial(n)),n0..5)
• evalf()

.7467291967.7467291967
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and finally...

• So we get that to the nearest
  thousandth.
• Again, according to Maple, the actual answer (to
  10 places) is
• evalf(int(exp(-x2),x0..1))

.74669241330
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Try this...

• Sum the first four nonzero terms to approximate
• A. 0.7635
• B. 0.5637
• C. 0.3567
• D. 0.6357
• E. 0.6735