502
8.5. TAYLOR POLYNOMIALS AND TAYLOR SERIES
for small values of x. This shows one way that a polynomial function can be used to
approximate a non-polynomial function; such approximations are one of the main themes
in this section and the next.
In Preview Activity 8.5, we begin our explorations of approximating non-polynomial
functions with polynomials, from which we will also develop ideas regarding infinite series
that involve a variable, x.
Preview Activity 8.5. Preview Activity 8.3 showed how we can approximate the number
e using linear, quadratic, and other polynomial functions; we then used similar ideas
in Preview Activity 8.4 to approximate ln(2). In this activity, we review and extend the
process to find the “best" quadratic approximation to the exponential function e x around
the origin. Let f (x) = e x throughout this activity.
(a) Find a formula for P 1 (x), the linearization of f (x) at x = 0. (We label this
linearization P 1 because it is a first degree polynomial approximation.) Recall that
P 1 (x) is a good approximation to f (x) for values of x close to 0. Plot f and P 1
near x = 0 to illustrate this fact.
(b) Since f (x) = e x is not linear, the linear approximation eventually is not a very
good one. To obtain better approximations, we want to develop a different
approximation that “bends” to make it more closely fit the graph of f near x = 0.
To do so, we add a quadratic term to P 1 (x). In other words, we let
P 2 (x) = P 1 (x) + c 2 x
2
for some real number c 2 . We need to determine the value of c 2 that makes the
graph of P 2 (x) best fit the graph of f (x) near x = 0.
Remember that P 1 (x) was a good linear approximation to f (x) near 0; this is
because P 1 (0) = f (0) and P ′
1 (0) = f ′ (0). It is therefore reasonable to seek a value
of c 2 so that
P 2 (0) = f (0),
P
′
2 (0) = f
′ (0), and
P
′′
2 (0) = f
′′ (0).
Remember, we are letting P 2 (x) = P 1 (x) + c 2 x 2 .
(i) Calculate P 2 (0) to show that P 2 (0) = f (0).
(ii) Calculate P ′
2 (0) to show that P ′
2 (0) = f ′ (0).
(iii) Calculate P ′′
2 (x). Then find a value for c 2 so that P ′′
2 (0) = f ′′ (0).
(iv) Explain why the condition P ′′
2 (0) = f ′′ (0) will put an appropriate “bend" in
the graph of P 2 to make P 2 fit the graph of f around x = 0.
⊲⊳
8.5. TAYLOR POLYNOMIALS AND TAYLOR SERIES
for small values of x. This shows one way that a polynomial function can be used to
approximate a non-polynomial function; such approximations are one of the main themes
in this section and the next.
In Preview Activity 8.5, we begin our explorations of approximating non-polynomial
functions with polynomials, from which we will also develop ideas regarding infinite series
that involve a variable, x.
Preview Activity 8.5. Preview Activity 8.3 showed how we can approximate the number
e using linear, quadratic, and other polynomial functions; we then used similar ideas
in Preview Activity 8.4 to approximate ln(2). In this activity, we review and extend the
process to find the “best" quadratic approximation to the exponential function e x around
the origin. Let f (x) = e x throughout this activity.
(a) Find a formula for P 1 (x), the linearization of f (x) at x = 0. (We label this
linearization P 1 because it is a first degree polynomial approximation.) Recall that
P 1 (x) is a good approximation to f (x) for values of x close to 0. Plot f and P 1
near x = 0 to illustrate this fact.
(b) Since f (x) = e x is not linear, the linear approximation eventually is not a very
good one. To obtain better approximations, we want to develop a different
approximation that “bends” to make it more closely fit the graph of f near x = 0.
To do so, we add a quadratic term to P 1 (x). In other words, we let
P 2 (x) = P 1 (x) + c 2 x
2
for some real number c 2 . We need to determine the value of c 2 that makes the
graph of P 2 (x) best fit the graph of f (x) near x = 0.
Remember that P 1 (x) was a good linear approximation to f (x) near 0; this is
because P 1 (0) = f (0) and P ′
1 (0) = f ′ (0). It is therefore reasonable to seek a value
of c 2 so that
P 2 (0) = f (0),
P
′
2 (0) = f
′ (0), and
P
′′
2 (0) = f
′′ (0).
Remember, we are letting P 2 (x) = P 1 (x) + c 2 x 2 .
(i) Calculate P 2 (0) to show that P 2 (0) = f (0).
(ii) Calculate P ′
2 (0) to show that P ′
2 (0) = f ′ (0).
(iii) Calculate P ′′
2 (x). Then find a value for c 2 so that P ′′
2 (0) = f ′′ (0).
(iv) Explain why the condition P ′′
2 (0) = f ′′ (0) will put an appropriate “bend" in
the graph of P 2 to make P 2 fit the graph of f around x = 0.
⊲⊳