Q41P f(x)=ex,聽a=3.... [FREE SOLUTION]

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Chapter 1: Q41P (page 32)

$f (x) = e^{x}, a = 3 .$

Short Answer

Expert verified

The first few terms of Taylor鈥檚 series expansion are:

$e^{x} = e^{3} 路 [1 + (x - 3) + \frac{(x - 3)^{2}}{2} + \frac{(x - 3)^{3}}{6} + 鈥�]$ .

Step by step solution

Given Information

The given function.

$f (x) = e^{x} .$

Definition of Taylor’s series

The Taylor鈥檚 series is a power series that yields the expansion of a function in the vicinity of a point if the function is continuous, all of its derivatives exist, and the series converges.

Find the Taylor’s series for ex

Use the Maclaurin series of and replace $x$ by $x - 3$ ,

localid="1657388907873" $e^{x} = {e^{3 + (x - 3}}^{)} e^{x} = e^{3} 路 e^{(x - 3)} e^{x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + 鈥� e^{(x - 3)} = 1 + (x - 3) + \frac{(x - 3)^{2}}{2} + \frac{(x - 3)^{3}}{6} + 鈥�$

Write first few terms of Taylor鈥檚 series about localid="1657388920739" $a = 3$

localid="1657388930350" $e^{x} = e^{3} 路 e^{(x - 3)} = e^{3} 路 [1 + (x - 3) + \frac{(x - 3)^{2}}{2} + \frac{(x - 3)^{3}}{6} + 鈥�]$

Hence, first few terms of Taylor鈥檚 series expansion are:

localid="1657388838541" $e^{x} = e^{3} 路 [1 + (x - 3) + \frac{(x - 3)^{2}}{2} + \frac{(x - 3)^{3}}{6} + 鈥�]$

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91影视

$f (x) = e^{x}, a = 3 .$

Short Answer

Step by step solution

Given Information

Definition of Taylor’s series

Find the Taylor’s series for ex

One App. One Place for Learning.

Most popular questions from this chapter

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91影视

f(x)=ex,a=3.

Short Answer

Step by step solution

Given Information

Definition of Taylor’s series

Find the Taylor’s series for ex

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Radiation

Engineering Physics

Linear Momentum

Nuclear Physics

Thermodynamics

Quantum Physics

Study anywhere. Anytime. Across all devices.

$f (x) = e^{x}, a = 3 .$