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Find the first four terms of the binomial series for the functions in Exercises \(1-10 .\) $$ \left(1+x^{2}\right)^{-1 / 3} $$

Short Answer

Expert verified
The first four terms are 1, \(-\frac{1}{3}x^2\), \(\frac{2}{9}x^4\), and \(-\frac{14}{81}x^6\).

Step by step solution

01

Identify the Binomial Series Formula

The general binomial series formula for \((1+x)^n\) is given by: \[ (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots \] Here, the exponent \(n = -\frac{1}{3}\) and our function is \((1+x^2)^{-1/3}\), so replace \(x\) in the series expansion with \(x^2\).
02

Determine the First Term

The first term in the series is simply 1, irrespective of \(x\). So, the first term for \((1+x^2)^{-1/3}\) is: \[ 1 \]
03

Calculate the Second Term

The second term in the series is given by \(nx\). Substitute \(n = -\frac{1}{3}\) and \(x = x^2\) in the formula: \[-\frac{1}{3}x^2 \]
04

Calculate the Third Term

Using \(\frac{n(n-1)}{2!}x^2\), substitute \(n = -\frac{1}{3}\) and \(x = x^2\):\[\frac{-1/3(-1/3-1)}{2} (x^2)^2 = \frac{-1/3 \cdot -4/3}{2} x^4 = \frac{2}{9} x^4 \]
05

Calculate the Fourth Term

The fourth term uses \(\frac{n(n-1)(n-2)}{3!}x^3\), substituting \(n = -\frac{1}{3}\) and \(x = x^2\):\[\frac{-1/3(-1/3-1)(-1/3-2)}{6} (x^2)^3 = \frac{-1/3 \cdot -4/3 \cdot -7/3}{6} x^6 = -\frac{14}{81} x^6 \]
06

Combine the Terms for the Series

Combine all the terms to write the first four terms of the series:\[ 1 - \frac{1}{3}x^2 + \frac{2}{9}x^4 - \frac{14}{81}x^6 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Theorem
The Binomial Theorem is a powerful mathematical tool that allows us to expand expressions like \((1+x)^n\) into an infinite series. This is super handy when you want to evaluate expressions with powers and when the exponent \(n\) is a real number, not just a positive integer. For example, if you have an expression \((1+x^2)^{-1/3}\), you can use the binomial theorem to expand it.
The general formula for the binomial series is:
  • \((1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots \)
In this formula, each term of the series uses decreasing powers of \(n\), factorials in the denominator, and increasing powers of \(x\). With the binomial theorem, you can approximate complicated functions, which can be very useful in calculus and analysis. Remember, this expansion works best when \(|x| \lt 1\). In our example, the function was \((1+x^2)^{-1/3}\), so we'd replace \(x\) with \(x^2\) in the expansion steps, working term by term.
Series Expansion
Series expansion is just like breaking down complicated mathematical expressions into a series of simpler parts, making them much easier to work with. When you expand a series, you're essentially writing a function as a sum of its terms powered by some form of increment, like \(x, x^2, \) or \(x^3\), and so on.
In the context of our binomial series, we aim to find the first four terms of \((1+x^2)^{-1/3}\). We do this by utilizing the binomial theorem.
  • The first term in our case is always 1, easy enough! 💡
  • The second term is given by \(nx\), so it's \(-\frac{1}{3}x^2\).
  • The third term involves \(\frac{n(n-1)}{2!}x^2\) or \(-\frac{1}{3} \cdot -4/3 \cdot x^4 = \frac{2}{9} x^4\).
  • Last but not least, the fourth term involves \(\frac{n(n-1)(n-2)}{3!}x^3\) resulting in \(-\frac{14}{81}x^6\).
Combining these, you get the overall series expansion: \(1 - \frac{1}{3}x^2 + \frac{2}{9}x^4 - \frac{14}{81}x^6\). With each term calculated, we have made a complex exponential function simpler to understand and use.
Power Series
The concept of a power series is critical when dealing with functions that can be represented as an infinite sum of terms. Each term in a power series is a power of a variable multiplied by a coefficient. In this way, power series allow us to approximate even the most complex functions.
Power series look something like this:
  • \(a_0 + a_1x + a_2x^2 + a_3x^3 + \ldots \)
In the context of our example, the binomial series is a type of power series. Each term beyond the first involves a higher power of \(x^2\). Understanding the coefficient behavior and the pattern it's creating becomes important as the series goes on. Notice how each term in the series has both increasing powers of \(x^2\) and different coefficients multiplying those powers. This makes power series super versatile for approximating functions in calculus, especially when \(x\) is close to zero.
Mathematical Series
A mathematical series is simply the sum of the terms of a sequence. These sequences could be finite, like having a few terms, or infinite, extending endlessly. Series are fundamental in mathematics because they allow us to express functions and numbers as sums, which can be more convenient for calculations.
For example, in the binomial series of \((1+x^2)^{-1/3}\), we calculated:
  • The first term as \(1\)
  • The second term as \(-\frac{1}{3}x^2\)
  • And so on, up to the fourth term \(-\frac{14}{81}x^6\).
We take these individual terms and add them, resulting in the series: \(1 - \frac{1}{3}x^2 + \frac{2}{9}x^4 - \frac{14}{81}x^6\). In the grand scheme of things, a series can solve differential equations, model phenomena in physics, and approximate functions. Mastering the idea of a series is key to excelling in advanced mathematical concepts.

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Most popular questions from this chapter

Which of the sequences in Exercises \(101-106\) converge, and which diverge? Give reasons for your answers. $$ a_{n}=\left((-1)^{n}+1\right)\left(\frac{n+1}{n}\right) $$

a. Show that $$ \int_{2}^{\infty} \frac{d x}{x(\ln x)^{p}} \quad(p \text { a positive constant }) $$ converges if and only if \(p>1\) b. What implications does the fact in part (a) have for the convergence of the series $$ \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}} ? $$ Give reasons for your answer.

Use a CAS to perform the following steps for the sequences in Exercises \(129-140 .\) a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=n \sin \frac{1}{n} $$

Use series to evaluate the limits. $$ \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{x} $$

Compound interest, deposits, and withdrawals If you invest an amount of money \(A_{0}\) at a fixed annual interest rate \(r\) compounded \(m\) times per year, and if the constant amount \(b\) is added to the account at the end of each compounding period (or taken from the account if \(b<0 ),\) then the amount you have after \(n+1\) compounding periods is $$ A_{n+1}=\left(1+\frac{r}{m}\right) A_{n}+b $$ a. If \(A_{0}=1000, r=0.02015, m=12,\) and \(b=50\) , calculate and plot the first 100 points \(\left(n, A_{n}\right) .\) How much money is in your account at the end of 5 years? Does \(\left\\{A_{n}\right\\}\) converge? Is \(\left\\{A_{n}\right\\}\) bounded? b. Repeat part (a) with \(A_{0}=5000, r=0.0589, m=12,\) and \(b=-50 .\) c. If you invest 5000 dollars in a certificate of deposit (CD) that pays 4.5\(\%\) annually, compounded quarterly, and you make no further investments in the CD, approximately how many years will it take before you have \(20,000\) dollars? What if the CD earns 6.25\(\% ?\) d. It can be shown that for any \(k \geq 0\) , the sequence defined recursively by Equation \((1)\) satisfies the relation $$ A_{k}=\left(1+\frac{r}{m}\right)^{k}\left(A_{0}+\frac{m b}{r}\right)-\frac{m b}{r} $$ For the values of the constants \(A_{0}, r, m,\) and \(b\) given in part (a), validate this assertion by comparing the values of the first 50 terms of both sequences. Then show by direct substitution that the terms in Equation \((2)\) satisfy the recursion formula in Equation ( 1\()\) .

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