Chapter 10: Problem 40
a. Find the first four nonzero terms of the Taylor series centered at 0 for the given finction. b. Use the first four terms of the series to approximate the given quantity. $$f(x)=\sqrt{1+x} ; \text { approximate } \sqrt{1.06}$$
Short Answer
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Answer: The Taylor series (up to the fourth term) of the function $$f(x) = \sqrt{1+x}$$ centered at 0 is $$1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{64}x^4$$. Using this series, the approximation of $$\sqrt{1.06}$$ is approximately $$1.0296$$.
Step by step solution
01
Find the first four derivatives of the function
First, we need to find the derivatives of the function $$f(x) = \sqrt{1+x}$$. We'll find the first four derivatives:
1. $$f'(x) = \frac{1}{2\sqrt{1+x}}$$
2. $$f''(x) = -\frac{1}{4(1+x)^{\frac{3}{2}}}$$
3. $$f^{(3)}(x) = \frac{3}{8(1+x)^{\frac{5}{2}}}$$
4. $$f^{(4)}(x) = -\frac{15}{16(1+x)^{\frac{7}{2}}}$$
02
Evaluate the derivatives at x = 0
Next, we need to evaluate the derivatives at x = 0, since the center of the Taylor series is at 0:
1. $$f'(0) = \frac{1}{2\sqrt{1+0}} = \frac{1}{2}$$
2. $$f''(0) = -\frac{1}{4(1+0)^{\frac{3}{2}}} = -\frac{1}{4}$$
3. $$f^{(3)}(0) = \frac{3}{8(1+0)^{\frac{5}{2}}} = \frac{3}{8}$$
4. $$f^{(4)}(0) = -\frac{15}{16(1+0)^{\frac{7}{2}}} = -\frac{15}{16}$$
03
Write out the Taylor series formula
The Taylor series formula up to the fourth term is as follows:
$$T_4(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2} + \frac{f^{(3)}(0)x^3}{6} + \frac{f^{(4)}(0)x^4}{24}$$
04
Plug in the derivatives and simplify
Now we will plug in the derivatives that we found in the previous steps and simplify:
$$T_4(x) = \sqrt{1} + \frac{1}{2}x - \frac{1}{4}\frac{x^2}{2} + \frac{3}{8}\frac{x^3}{6} - \frac{15}{16}\frac{x^4}{24}$$
$$T_4(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{64}x^4$$
05
Use the Taylor series to approximate the square root of 1.06
Finally, we will use the Taylor series to approximate $$\sqrt{1.06}$$. We will plug in 0.06 for x in the series:
$$T_4(0.06) = 1 + \frac{1}{2}(0.06) - \frac{1}{8}(0.06)^2 + \frac{1}{16}(0.06)^3 - \frac{5}{64}(0.06)^4$$
Calculating this expression, we get:
$$T_4(0.06) \approx 1.0296$$
So, the first four terms of the Taylor series centered at 0 for the function $$f(x)=\sqrt{1+x}$$ are:
$$1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{64}x^4$$
And the approximation of $$\sqrt{1.06}$$ using these terms is:
$$\sqrt{1.06} \approx 1.0296$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Derivatives
Derivatives are fundamental in calculus and represent the rate of change of a function with respect to a variable. In simple terms, they tell us how a function is behaving at a particular point. When constructing a Taylor series, we rely on derivatives to capture the behavior of a function at the center, which in our case is 0. Consider the function \( f(x) = \sqrt{1+x} \). The first derivative, \( f'(x) \), gives us the slope or the gradient of the function at any point. The subsequent derivatives, \( f''(x), f^{(3)}(x), \) and so on, provide more information about the function's curvature and other higher-order changes.
- The first derivative, \( f'(x) = \frac{1}{2\sqrt{1+x}} \), tells us how rapidly the function is increasing or decreasing.
- The second derivative, \( f''(x) = -\frac{1}{4(1+x)^{\frac{3}{2}}} \), gives information about the concavity of the function.
- Each higher derivative adds more nuance to our understanding of the function's behavior.
Importance of Approximation
Approximation is a powerful mathematical technique that helps us estimate the value of a function near a specific point without having to compute its exact value. The Taylor series offers one such method of approximation by utilizing derivatives.With a Taylor series, we start from a point close to where we want to measure—here, at 0—and use derivatives to predict the function's behavior at a point nearby, like 1.06 in this case. This series allows us to approximate functions that are often difficult to compute directly. For \( \sqrt{1+x} \), we can avoid a direct computation and instead use the series to estimate \( \sqrt{1.06} \).
- Approximations are crucial in fields like physics and engineering where exact values may not be necessary or possible.
- Such techniques can also simplify complex calculations, making them more manageable and faster to perform.
Taylor Series Centered at 0
A Taylor series is a type of infinite series used to approximate smooth functions. It is centered at a particular point \( a \), which serves as the expansion point for the series. In this problem, we have the series centered at \( a = 0 \), also known as the Maclaurin series.By using a Maclaurin series, the terms in the expansion only depend on derivatives evaluated at 0. This simplifies calculating the series as no additional terms related to \( a \) need to be added.
- The function \( \sqrt{1+x} \) becomes quite handy to expand around 0 due to simple evaluation of the derivatives.
- This form of series is beneficial when approximating values close to 0, like our interest in 1.06, which can be expressed as 1 + 0.06.
The Concept of a Mathematical Series
A mathematical series, in general, is the sum of the terms of a sequence. The Taylor series is a specific type of mathematical series that attempts to express a function as an infinite sum of terms calculated from the values of its derivatives at a single point.The standard form is:\[T_n(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f^{(3)}(0)x^3}{3!} + \cdots + \frac{f^{(n)}(0)x^n}{n!}\]Here, \( n! \) is the factorial of \( n \), and each successive term in the series provides a finer approximation of the function around the chosen center.
- The terms \( f(0), f'(0)x, .. \) signify the function's behavior captured progressively at higher levels of precision.
- This series provides more accurate approximations as more terms are included.