w11_16c1_lec_3_4&3_7_11
TRANSCRIPT
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10.4: Power Series and Taylor’s Theorem
A power series is like an infinite polynomial. It has the form
∞∑n=0
an(x− c)n = a0 + a1(x− c)+ a2(x− c)2 + ...+ an(x− c)n + ...
here c is any real number and a series of this form is called a powerseries centered at c. Note that c = 0 is ok and then the powerseries will look like a0 + a1x + a2x
2 + a3x3 + .... Note that a power
series is just like any other series except that it depends on x.
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Radius of Convergence of a power series
Let
f(x) =∞∑
n=0
an(x− c)n
be the function defined by this power series. Note that f(x) is onlydefined if the power series converges so we will consider thedomain of the function f to be the set of x values for which theseries converges. There are three possible cases:
1 The power series converges at x = c (note f(c) = a0)
2 The power series converges for all x, i.e. (−∞,∞)3 There is a number R called the Radius of convergence such
that the series converges for all c−R < x < c + R and theseries diverges outside this interval.
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Find the Radius of Convergence of∞∑
n=0
xn
n!:
First note that this is a power series centered at c = 0, and thecoefficients an = 1
n! . We will use the ratio test to find the radius ofconvergence. We look at the terms of the series
limn→∞
∣∣∣∣an+1xn+1
anxn
∣∣∣∣ = limn→∞
∣∣∣∣xn+1/(n + 1)!xn/n!
∣∣∣∣= lim
n→∞
∣∣∣∣ x
n + 1
∣∣∣∣= 0.
Since the ratio test implies that this converges and final answer i.e.0 does not depend on x we see that this series will converge for allx meaning that the Radius of Convergence is infinite.
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Find the Radius of Convergence of∞∑
n=0
(−1)n(x− 2)n
3n:
First note that this is a power series centered at c = 2 and that thecoefficients an = (−1)n
3n . We use the ratio test and have
limn→∞
∣∣∣∣an+1xn+1
anxn
∣∣∣∣ = limn→∞
∣∣∣∣(−1)n+1(x− 2)n+1/3n+1
(−1)n(x− 2)n/3n
∣∣∣∣= lim
n→∞
∣∣∣∣(−1)(x− 2)3
∣∣∣∣= lim
n→∞
∣∣∣∣x− 23
∣∣∣∣ .Using the ratio test we know that this will converge if |x− 2|/3 < 1or if |x− 2| < 3. This gives the Radius of Convergence is R = 3and the interval is (−1, 5) since the center is c = 2.
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Taylor and Maclaurin series
If you start with a function f(x) and want to find the power seriesrepresentation for it, there is a nice formula, called the TaylorSeries. (If c = 0 it is called the Maclaurin Series). If f(x) isrepresented by a power series centered at c, then
f(x) =∞∑
n=0
f (n)(c)n!
(x− c)n
This can be written out the long way as
f(x) = f(c) + f ′(c)(x− c) +f ′′(c)
2(x− c)2 +
f ′′′(c)3!
(x− c)3 + ...
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Finding a Maclaurin Series
Find the power series for f(x) = ex centered at x = 0. This is anice and easy one since all the derivatives of ex are also ex. Ingeneral you start by writing the derivatives out at x = c andfinding a pattern.
f(x) = ex f(0) = 1f ′(x) = ex f ′(0) = 1f ′′(x) = ex f ′′(0) = 1
...
So in general f (n)(0) = 1 so
ex = 1 + x +x2
2!+
x3
3!+ ... =
∞∑n=0
xn
n!.
We saw from earlier that this series converges for all values of x,because the radius of convergence was infinite.
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Finding a Taylor Series
Find the power series for f(x) = lnx centered at x = 1.
f(x) = ln x f(1) = ln 1 = 0
f ′(x) =1x
f ′(1) = 1/1 = 1
f ′′(x) = −x−2 f ′′(1) = −1
f (3)(x) = 2x−3 f (3)(1) = 2
f (4)(x) = 2(−3)x−4 f (4)(1) = −6
f (5)(x) = 2(−3)(−4)x−5 f (4)(1) = 4 · 3 · 2...
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Continued
In general we see that
f (n)(1)=(−1)nn!
so the Taylor Series
ln x = f(1) + f ′(1)(x− 1) +f ′′(1)
2(x− 1)2 +
f ′′′(1)3!
(x− 1)3 + ...
= 0 + (x− 1)− 12(x− 1)2 +
13(x− 1)3 − 1
4(x− 1)4 + ..
We can write this out as
ln x =∞∑
n=1
(−1)n+1
n(x− 1)n.
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Using known power series to help find new ones...
Find the Maclaurin series for e−x, and ex2.
We know that
ex =∞∑
n=0
xn
n!= 1 + x +
x2
2+
x3
3!+ ...
so we can plug in −x for x to get the series for e−x which gives
e−x = 1 + (−x) +(−x)2
2+
(−x)3
3!+ ...
so the negative sign will cancel on even terms leaving
e−x = 1− x +x2
2− x3
3!+
x4
4!+ ... =
∞∑n=0
(−1)n
n!xn.
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Using known power series to help find new ones...
Find the Maclaurin series for e−x, and ex2.
We know that
ex =∞∑
n=0
xn
n!= 1 + x +
x2
2+
x3
3!+ ...
so we can plug in x2 for x to get the series for ex2which gives
ex2= 1 + (x2) +
(x2)2
2+
(x2)3
3!+ ...
so the square will return only even terms giving
ex = 1 + x2 +x4
2+
x6
3!+
x8
4!+ ... =
∞∑n=0
1n!
x2n.
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Basic List of Power Series
(I will give you this or something like it on the test)
ex = 1 + x + x2
2 + x3
3! + ... + xn
n! + ..., −∞ < x <∞
ln x = (x− 1)− (x−1)2
2 + ... + (−1)n(x−1)n
n + ... 0 < x ≤ 2
1x = 1− (x− 1) + (x− 1)2 + ... + (−1)n(x− 1)n + ... 0 < x < 2
1x+1 = 1− x + x2 − x3 + ...(−1)nxn... −1 < x < 1
(1 + x)k = 1 + kx + k(k−1)x2
2! + k(k−1)(k−2)x3
3! + ... −1 < x < 1
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Using the basic list:
Find the power series for f(x) =√
1 + x centered at c = 0. Wecan use the binomial series (1 + x)k with k = 1/2 :
(1 + x)k = 1 + kx +k(k − 1)x2
2!+
k(k − 1)(k − 2)x3
3!+ ...
and so for k = 1/2 we have
(1+x)1/2 = 1+x
2+(1/2)(1/2−1)
x2
2!+(1/2)(1/2−1)(1/2−2)
x3
3!+...
Which gives
√1 + x = 1 +
x
2− x2
8+
3x3
48− ...
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Using the basic list:
Find the power series for 2xx+1 We can use the basic list to tell us
the power series for 11+x and multiply each term by 2x :
1x + 1
= 1− x + x2 − x3 + ...(−1)nxn + ...
So2x
x + 1= 2x
[1− x + x2 − x3 + ...(−1)nxn + ...
]2x
x + 1= 2x− 2x2 + 2x3 − 2x4 + ...(−1)n2xn+1 + ...
or2x
x + 1=∞∑
n=0
(−1)n2xn+1