Calculus Of One Real Variable By Pheng Kim Ving
Chapter 15: Representations Of Functions By Power Series Section 15.4: Applications Of Taylor Series


15.4
Applications Of Taylor Series

 

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1. Approximations Of Values Of Functions

 

In Section 8.3 we approximate the value of f at a + h by the relation:

 

 

Approximation [1.3] approximates the value of f at x by using the values of f itself and of its derivatives of all orders themselves
at c and the signed distance from x to c. It's of course an extension of and improvement on approximation [1.2].

 

Fig. 1.1

 

 

Fig. 1.2

 

 

If a function that's more complicated than polynomials has a Taylor- or Maclaurin-series representation and x1 is a point in the
interval of representation, then naturally an approximate sum, which is usually a partial sum, of the series obtained at x1 is an
approximate value of the function at x1. This polynomial approximation of values of functions is used by calculators and
computers to approximately evaluate the transcendental functions.

 

A Geometric Series As An Error Bound

 

Example 1.1

 

 

Note

We're asked to find an approximate value of the function ex at x = 1/2 by using the Maclaurin series of ex obtained at x = 1/2.
See Fig. 1.3.

Fig. 1.3

 

 

Solution

 

We have:

 

 

EOS

 

Remarks 1.1

 

1. As (d n/dxn) ex = ex and thus (d n/dxn) ex|x=0 = e0 = 1 for all n = 0, 1, 2, ..., the series:

 

 

is the Maclaurin series of ex obtained at x = 1/2.

 

2. Recall that sn is the sum of the first n terms of the series, so, if the subscripts of the terms start from 0 up, then sn is the
sum of the terms from subscript 0 up thru to subscript n 1. To make it clear that the subscripts start from 0 up, we
express the series in the sigma notation, where the subscript is at the sigma symbol.

 

3. The error is the size of a tail of the series; here it's just the tail because all the terms are positive and thus any tail is
positive. In this case we find a geometric series to be a bound for the error. The absolute value of the common ratio
1/(2(n +1)) of the geometric series is less than 1 for any one n = 0, 1, 2, ..., so we can apply the formula for the sum of
such a series to it.

 

For any particular value n1 of n we have:

 

 

which is a geometric series with first term 1/2n(n!) and common ratio 1/2(n + 1), and, as 0 < 1/2(n + 1) < 1 for all n = 0, 1,
2, ..., has a sum of:

 

 

etc. We see that g(n)(0) doesn't exist for any n. The square root function has no Maclaurin-series representation.

 

6. For a function represented by a Taylor or Maclaurin series on an interval, we can use a partial sum sn of the series obtained
at a point x in that interval to approximate the value of the function at x. Recall that for the function ex, that interval is the
entire real line.

 

The Magnitude Of The Next Term As An Error Bound

 

Example 1.2

 

Find an approximation of sin 42o accurate to 4 decimal places by using a Maclaurin series.

Notes

1. We have to convert degrees to radians. We're asked to find an approximate value of the function sin x at x = x1, where
x1 rad = 42o.

 

2. For an accuracy of 4 decimal places, the error bound will be 0.00005; see Section 14.7 Part 1.

 

3. When an alternating series satisfies the hypotheses of the AST (Alternating-Series Test) and when we approximate its sum
by one of its partial sums, the error is less than or equal to the magnitude of its next term; see Section 14.7 Part 4.

 

Solution

 

Clearly the series satisfies the conditions of the AST (Alternating-Series Test). The accuracy of 4 decimal places requires that
the error bound is 0.00005. So:

 


EOS

 

Remark 1.2

 

 

Using A Taylor Series Centered At A Point Other Than 0

 

Example 1.3

 

 

Note

 

Fig. 1.4

 

 

Solution

 


EOS


Remarks 1.3

 

1. As (d n/dxn) ex = ex and thus (d n/dxn) ex|x=1 = e1 = e for all n = 0, 1, 2, ..., the series:

 

 

3. Generally, when employing a Taylor series centered at a point c other than 0 to approximate the value f(x1) of a function
f(x) at a point x1, we choose as c a non-0 point that's closest to x1 and where the value f(c) of f is known or readily
calculated. See Fig. 1.5. This is because the series will involve f(c). Usually the values of the derivatives of all orders of f at
c are obtained from the algebraic manipulation of f(x1) to get the difference x1 c.

 

Fig. 1.5

 

c is non-0 and is closest to x1 such that f(c) is known or
readily calculated.

 

 

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2. Approximations Of Definite Integrals

 

Finding Taylor Or Maclaurin Series Of Integral Functions

 

For the definition of integral functions see Section 9.4. They're functions defined by integrals.

 

Example 2.1

 

Find the Maclaurin series of the integral function:

 

 

and write the series in the sigma notation.

 

Note

We can try to integrate (sin t)/t first, obtaining a function say F(x), which is an antiderivative of (sin x)/x, then find the series
of F(x). We can also try to find the series of (sin t)/t first, then integrate that series to obtain the series of the given integral
function. Because the series of (sin t)/t represents (sin t)/t, the integral of that series represents the integral of (sin t)/t. Since
integrating a series is easier than integrating (sin t)/t, we'll adopt the second approach.

 

Solution

EOS

 

Approximations Of Definite Integrals

 

Example 2.2

 

Approximate the definite integral:

 

 

with error < 0.001, by using series.

 

Note

 

Solution

 


EOS

 

Remark 2.1

 

1. In Example 2.2, the series:

 

 

2. For many functions that are expressible as simple combinations of elementary functions, their antiderivatives aren't simple
combinations of elementary functions. They can't be antidifferentiated by elementary techniques. However we can often find
their Taylor series, antidifferentiate these series to obtain the series of the antiderivatives, and employ the latter series to
approximate definite integrals of the original functions.

 

Example 2.3

 

1. Determine the Maclaurin series of the integral function:

 

 

with error < 0.001.

 

Solution


EOS

 

Remark 2.2

 

 

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3. Indeterminate Forms

 

 

{3.1} Section 8.6 Theorem 2.1.

 

Example 3.1

 

Evaluate the following limit:

 

 

Solution

EOS

 

Remark 3.1

 

In Example 3.1, we can use the series representation of cos x because it's valid for all x, and thus for all x near 0.

 

Example 3.2

 

Calculate this limit by using series:

 

 

Solution

EOS

 

Remark 3.2

 

 

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Problems & Solutions

 

1. Approximate the value of e0.1 with error < 0.0001 by using a Maclaurin series.

 

Solution

 

 

 

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2. Estimate the value of cos 10o with error < 0.0001 by using a Maclaurin series.

 

Solution

 

 

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3. Use a Maclaurin series to find an approximate value of ln 0.6 accurate to 2 decimal places.

 

Solution

 

 

 

The estimated value is accurate to 2 decimal places if error < 0.005, which is true if:

 

 

Notes

 

 

2. The subscripts of the terms of the series start from 1 up. We express the series in the sigma notation to make this clear. So
sn is the sum of the terms from subscripts 1 up thru to n. The corresponding tail of the series starts from term with
subscript n + 1 up.

 

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4. Find the Maclaurin series of the integral function:

 

 

and write the series in the sigma notation.

 

Solution

 

 

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5. Approximate the definite integral:

 

 

with error < 1/100,000, by using series.

 

Solution

 

 

 

We can get by with n = 2. Thus:

 

 

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6.

a. Determine the Maclaurin series of the integral function:

 

 

with error < 0.001.

 

Solution

 

 

 

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7. Evaluate this limit:

 

 

Solution

 

 

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8. Consider the system of equations:

 

b. Determine all the coefficients of the series in terms of a0 and a1.

c. Solve the system of equations.

 

Solution

 

y' = a1 + 2a2x + 3a3x2 + 4a4x3 + 5a5x4 + ...,

y'' = 2a2 + 6a3x + 12a4x2 + 20a5x3 + ...,

xy' = x(a1 + 2a2x + 3a3x2 + 4a4x3 + 5a5x4 + ...) = a1x + 2a2x2 + 3a3x3 + 4a4x4 + 5a5x5 + ...,

 

0 = y'' + xy' + y,

= (2a2 + 6a3x + 12a4x2 + 20a5x3 + ...) + (a1x + 2a2x2 + 3a3x3 + ...) + (a0 + a1x + a2x2 + a3x3 + ...)
= (a0 + 2a2) + 2(a1 + 3a3)x + 3(a2 + 4a4)x2 + 4(a3 + 5a5)x3 + ...,

 

an + (n + 2)an+2 = 0, for all n = 0, 1, 2, ... .

 

 

and if n is odd then:

 

 

 

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