Question

Find the next two terms of the following Taylor series. $$\sqrt{1+x}: 1+\frac{1}{2} x-\frac{1}{2 \cdot 4} x^{2}+\frac{1 \cdot 3}{2 \cdot 4 \cdot 6} x^{3}-\cdots$$

Short answer

Answer: The next two terms of the Taylor series for $$\sqrt{1+x}$$ are: $$\frac{1}{384}x^4$$ and $$-\frac{1}{3840}x^5$$.

Step-by-step solution

Finding a pattern in the coefficients

We can rewrite the given Taylor series in a more compact form by organizing coefficients as fractions: $$1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3-\cdots$$ We can see that the coefficients are alternatively positive and negative, and the denominators are increasing: $$\frac{1}{1},\ -\frac{1}{8}, \ \frac{1}{16}, \ -\cdots$$ We can observe a pattern in the denominators: $$1, 2^3, 2^3\cdot2, 2^3\cdot2\cdot3, -\cdots$$ So the general coefficient of the Taylor series can be given as: $$(-1)^{n-1}\frac{1}{2^3(1)(2)(\cdots)(n-1)}$$

Finding the next two terms

Using the above general coefficient, we can find the next two terms of the Taylor series: The 5th term (n=5): $$(-1)^{5-1}\frac{1}{2^3(1)(2)(3)(4)} = \frac{1}{384}x^4$$ The 6th term (n=6): $$(-1)^{6-1}\frac{1}{2^3(1)(2)(3)(4)(5)} = -\frac{1}{3840}x^5$$

Write the Taylor series with the new terms

Finally, we write the Taylor series with the new terms included: $$\sqrt{1+x}: 1+\frac{1}{2} x-\frac{1}{2 \cdot 4} x^{2}+\frac{1 \cdot 3}{2 \cdot 4 \cdot 6} x^{3}-\frac{1}{384}x^4+\frac{1}{3840}x^5-\cdots$$

Key concepts

Series Expansion

Taylor series is a powerful tool used in mathematics to express functions as infinite sums of terms calculated from the values of their derivatives at a single point. When we talk about a series expansion, we basically mean that we are writing a function like a polynomial (think of it as a very long sum of power terms) that goes on indefinitely.

In this particular exercise, the focus is on the function \( \sqrt{1+x} \). We are expressing it as a Taylor series. The series expansion involves the initial term along with subsequent terms that account for higher powers of \( x \). This is extremely useful as it allows us to approximate functions in a simpler form, particularly when dealing with small values of \( x \). Through this process, we manage to capture the essential features of functions without solving them each time from scratch.

Overall, series expansion helps simplify complex functions, making them easier to analyze, especially close to a certain point in their domain.

Pattern Recognition

Recognizing patterns is crucial when working with Taylor series, as it's the key to understanding and predicting the sequence of coefficients. In this exercise, we've nicely observed a pattern in both the sign and magnitude of the coefficients.

• Alternating signs: The coefficients change their sign in an alternating positive-negative-positive manner. This behavior often stems from the powers of \( (-1) \), which is a tell-tale sign of alternating series.
• Fractional coefficients: Beyond the signs, there is a pattern in how the coefficients are structured as fractions. The denominators grow systematically in a factored form involving powers and products: \(2^3, 2^3 \cdot 2, 2^3 \cdot 2 \cdot 3, \ldots\).

Spotting these kinds of patterns not only aids in the completion of a given series but also builds mathematical intuition, allowing you to tackle similar problems in the future with more ease.

Coefficient Determination

The determination of the coefficients in a Taylor series is vital since it dictates the accuracy of the approximation. For our function \( \sqrt{1+x} \), this involves extracting these coefficients from the general formula we identified through pattern recognition. This general formula establishes how each term's coefficient is computed based on the term's order in the sequence.

Given that the coefficients follow a specific order:
\[(-1)^{n-1}\frac{1}{2^3 \cdot (1)(2)\cdots(n-1)}\]we can deduce each coefficient smoothly.

• For the 5th term, the coefficient is calculated as \( \frac{1}{384} \) due to the contributors \( 2^3, 1, 2, 3, \) and \( 4 \).
• The coefficient for the 6th term is determined to be \( -\frac{1}{3840} \), extending the pattern to include the factor 5.

This structured approach to coefficient determination ensures the Taylor series correctly represents the function and provides generalizable skills for mathematical problem-solving.

Sequence Convergence

A Taylor series is particularly favored due to its convergence properties — the way its sum approaches the function it represents as more terms are added. In essence, sequence convergence analyses how effectively the series expansion imitates the original function.

Sequence convergence is especially important when using series to approximate real-world scenarios. The rate of convergence dictates how quickly a Taylor series provides a reliable approximation and determines the number of terms needed for sufficient accuracy.

• When discussing convergence, higher degree terms in the series tend to have diminishing contributions, allowing for increasing accuracy even when truncating the series.
• Such properties are crucial in determining applications, such as engineering calculations and numerical approximations, where precision and computational efficiency are requisite.

The convergence of a Taylor series inspires confidence in its use for approximating functions across various practical and theoretical applications.