/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Expand \(\frac{1}{\sqrt{(1-2 t)}... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 14

Expand \(\frac{1}{\sqrt{(1-2 t)}}\) in ascending powers of \(t\) as far as the term in \(t^{3}\). State the limits of \(t\) for which the expression is valid. $$ \begin{aligned} &\frac{1}{\sqrt{(1-2 t)}} \\ &=(1-2 t)^{-\frac{1}{2}} \\ &=1+\left(-\frac{1}{2}\right)(-2 t)+\frac{(-1 / 2)(-3 / 2)}{2 !}(-2 t)^{2} \\ &=+\frac{(-1 / 2)(-3 / 2)(-5 / 2)}{3 !}(-2 t)^{3}+\cdots \end{aligned} $$ using the expansion for \((1+x)^{n}\) $$ =1+t+\frac{3}{2} t^{2}+\frac{5}{2} t^{3}+\cdots $$ The expression is valid when \(|2 t|<1\), i.e. \(|t|<\frac{1}{2}\) or \(-\frac{1}{2}

Short Answer

Expert verified
The expression expands to \(1 + t + \frac{3}{2}t^2 + \frac{5}{2}t^3\), valid for \(-\frac{1}{2} < t < \frac{1}{2}\).

Step by step solution

01

Identify the General Formula

We will use the binomial series expansion formula for \((1+x)^n\), which is given by \(1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots\). Here, we identify \(x = -2t\) and \(n = -\frac{1}{2}\).
02

Calculate the First Term

The first term of the expansion is simply 1, as according to the expansion formula, \((1 + x)^n = 1\) when expanded. Thus, the first term is 1.
03

Calculate the Second Term

For the second term, we use \(nx\) which is \(-\frac{1}{2}(-2t)\). Simplifying, this gives \(t\). So the second term is \(t\).
04

Calculate the Third Term

The third term is given by \(\frac{n(n-1)}{2!} x^2\). Plugging in the values, we have \(\frac{(-1/2)((-1/2) - 1)}{2!}(-2t)^2\). This simplifies to \(\frac{(-1/2)(-3/2)}{2} \times 4t^2\), which equals \(\frac{3}{2}t^2\).
05

Calculate the Fourth Term

The fourth term is \(\frac{n(n-1)(n-2)}{3!} x^3\). Substituting the values gives us \(\frac{(-1/2)(-3/2)(-5/2)}{3!}(-2t)^3\). This simplifies to \(\frac{-15}{48}\times -8t^3 = \frac{5}{2}t^3\).
06

Combine All Terms

Combine all terms to get the expanded series as far as the term in \(t^3\). Thus, the expression becomes \(1 + t + \frac{3}{2}t^2 + \frac{5}{2}t^3 + \cdots\).
07

State the Valid Range for \(t\)

For the binomial expansion to be valid, \(|x| < 1\) must hold true. As \(x = -2t\), we require \(|-2t| < 1\), thus it follows that \(|t| < \frac{1}{2}\), leading to the valid range \(-\frac{1}{2} < t < \frac{1}{2}\).

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Key Concepts

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

Power Series
A power series is a way to represent a function as an infinite sum of terms involving powers of a variable. The series we've dealt with here, \[ \frac{1}{\sqrt{(1-2t)}} = (1-2t)^{-\frac{1}{2}} \], is a power series. It's a powerful tool because it transforms a complex expression into a summation which often is easier to handle.
  • Key idea: We rewrite the function in terms of powers of a variable, usually making calculus easier.
  • This specific power series is extended using the Binomial Theorem, which applies when the exponent is not necessarily a whole number.
The general form of the Binomial Series we used 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 exercise, since \[ x = -2t \] and \[ n = -\frac{1}{2} \], terms of increasing powers of \( t \) are systematically derived to a desired order.
Valid Range of t
When using power series expansions, determining the range for which the expansion holds true is crucial. This is called the valid range of the series. For the binomial series expansion to converge (or be valid), the quantity \[ |x| < 1 \] must always hold.
  • Here, since we have \[ x = -2t \], this means \[ |-2t| < 1 \].
  • Thus, simplifying gives us \[ |t| < \frac{1}{2}. \]
Breaking it down on a number line, the valid values for \( t \) satisfy \[-\frac{1}{2} < t < \frac{1}{2}. \] This valid range is important to ensure any approximation made using this expansion remains accurate.
Step-by-Step Solution
Understanding the step-by-step process in deriving the expansion can deepen comprehension. Here's a recap of the steps involved:
  • **Step 1**: We started by identifying the necessary components for the binomial series: \[ (1+x)^n \] where \[ x = -2t \] and \[ n = -\frac{1}{2}. \]
  • **Step 2**: The first term, taken directly from the formula, is 1.
  • **Step 3**: For the second term, we calculated \[ nx \]and found \[ t. \]
  • **Step 4**: The third term required finding \[ \frac{n(n-1)}{2!} x^2, \] which simplified to \[ \frac{3}{2}t^2. \]
  • **Step 5**: The fourth term needed \[ \frac{n(n-1)(n-2)}{3!} x^3 \]to arrive at \[ \frac{5}{2}t^3. \]
  • The final expression condensed into \[ 1 + t + \frac{3}{2} t^2 + \frac{5}{2} t^3 + \cdots. \]
This methodical process helps ensure each term is correctly computed and contributes to the understanding of the series expansion.

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Most popular questions from this chapter

Evaluate \((0.97)^{6}\) correct to 4 significant figures using the binomial expansion. \((0.97)^{6}\) is written as \((1-0.03)^{6}\) Using the expansion of \((1+x)^{n}\) where \(n=6\) and \(x=-0.03\) gives: $$ \begin{aligned} (1-0.03)^{6}=& 1+6(-0.03)+\frac{(6)(5)}{(2)(1)}(-0.03)^{2} \\ &+\frac{(6)(5)(4)}{(3)(2)(1)}(-0.03)^{3} \\ &+\frac{(6)(5)(4)(3)}{(4)(3)(2)(1)}(-0.03)^{4}+\cdots \\ =& 1-0.18+0.0135-0.00054 \\ & \quad+0.00001215-\cdots \\ \approx 0.83297215 \end{aligned} $$ i.e. \((0.97)^{6}=0.8330\), correct to 4 significant

Use the binomial series to determine the expansion of \((2 a-3 b)^{5}\) From equation (1), the binomial expansion is given by: $$ \begin{array}{r} (a+x)^{n}=a^{n}+n a^{n-1} x+\frac{n(n-1)}{2 !} a^{n-2} x^{2} \\ +\frac{n(n-1)(n-2)}{3 !} a^{n-3} x^{3}+\cdots \end{array} $$ When \(a=2 a, x=-3 b\) and \(n=5\) : $$ \begin{aligned} &(2 a-3 b)^{5}=(2 a)^{5}+5(2 a)^{4}(-3 b) \\ &+\frac{(5)(4)}{(2)(1)}(2 a)^{3}(-3 b)^{2} \\ &+\frac{(5)(4)(3)}{(3)(2)(1)}(2 a)^{2}(-3 b)^{3} \\ &+\frac{(5)(4)(3)(2)}{(4)(3)(2)(1)}(2 a)(-3 b)^{4} \\ &+\frac{(5)(4)(3)(2)(1)}{(5)(4)(3)(2)(1)}(-3 b)^{5} \\ &\text { i.e. }(2 a-3 b)^{5}=32 a^{5}-240 a^{4} b+720 a^{3} b^{2} \\ &-1080 a^{2} b^{3}+810 a b^{4}-243 b^{5} \end{aligned} $$

(a) Expand \(\frac{1}{(1+2 x)^{3}}\) in ascending powers of \(x\) as far as the term in \(x^{3}\), using the binomial series. (b) State the limits of \(x\) for which the expansion is valid. (a) Using the binomial expansion of \((1+x)^{n}\), where \(n=-3\) and \(x\) is replaced by \(2 x\) gives: $$ \begin{aligned} \frac{1}{(1+2 x)^{3}}=&(1+2 x)^{-3} \\ =1+(-3)(2 x)+\frac{(-3)(-4)}{2 !}(2 x)^{2} \\ &+\frac{(-3)(-4)(-5)}{3 !}(2 x)^{3}+\cdots \\ =& 1-6 x+24 x^{2}-80 x^{3}+\cdots \end{aligned} $$ (b) The expansion is valid provided \(|2 x|<1\), i.e. \(|x|<\frac{1}{2}\) or \(-\frac{1}{2}

The resonant frequency of a vibratung shatt is given by: \(f=\frac{1}{2 \pi} \sqrt{\frac{k}{I}}\), where \(k\) is the stiffness and \(I\) is the inertia of the shaft. Use the binomial theorem to determine the approximate percentage error in determining the frequency using the measured values of \(k\) and \(I\) when the measured value of \(k\) is \(4 \%\) too large and the measured value of \(I\) is \(2 \%\) too small. Let \(f, k\) and \(I\) be the true values of frequency, stiffness and inertia respectively. Since the measured value of stiffness, \(k_{1}\), is \(4 \%\) too large, then $$ k_{1}=\frac{104}{100} k=(1+0.04) k $$ The measured value of inertia, \(I_{1}\), is \(2 \%\) too small, hence $$ I_{1}=\frac{98}{100} I=(1-0.02) I $$ The measured value of frequency, $$ \begin{aligned} f_{1} &=\frac{1}{2 \pi} \sqrt{\frac{k_{1}}{I_{1}}}=\frac{1}{2 \pi} k_{1}^{\frac{1}{2}} I_{1}^{-\frac{1}{2}} \\ &=\frac{1}{2 \pi}[(1+0.04) k]^{\frac{1}{2}}[(1-0.02) I]^{-\frac{1}{2}} \\ &=\frac{1}{2 \pi}(1+0.04)^{\frac{1}{2}} k^{\frac{1}{2}}(1-0.02)^{-\frac{1}{2}} I^{-\frac{1}{2}} \\ &=\frac{1}{2 \pi} k^{\frac{1}{2}} I^{-\frac{1}{2}}(1+0.04)^{\frac{1}{2}}(1-0.02)^{-\frac{1}{2}} \end{aligned} $$ $$ \text { i.e. } f_{1}=f(1+0.04)^{\frac{1}{2}}(1-0.02)^{-\frac{1}{2}} $$ $$ \begin{aligned} &\approx f\left[1+\left(\frac{1}{2}\right)(0.04)\right]\left[1+\left(-\frac{1}{2}\right)(-0.02)\right] \\\ &\approx f(1+0.02)(1+0.01) \end{aligned} $$ Neglecting the products of small terms, $$ f_{1} \approx(1+0.02+0.01) f \approx 1.03 f $$ Thus the percentage error in \(f\) based on the measured values of \(k\) and \(I\) is approximately \([(1.03)(100)-100]\), i.e. \(3 \%\) too large.

Evaluate (1.002) \(^{9}\) using the binomial theorem correct to (a) 3 decimal places and (b) 7 significant figures. $$ \begin{aligned} &(1+x)^{n}=1+n x+\frac{n(n-1)}{2 !} x^{2} \\ &+\frac{n(n-1)(n-2)}{3 !} x^{3}+\cdots \\ &(1.002)^{9}=(1+0.002)^{9} \end{aligned} $$ Substituting \(x=0.002\) and \(n=9\) in the general expansion for \((1+x)^{n}\) gives: $$ \begin{gathered} (1+0.002)^{9}=1+9(0.002)+\frac{(9)(8)}{(2)(1)}(0.002)^{2} \\ +\frac{(9)(8)(7)}{(3)(2)(1)}(0.002)^{3}+\cdots \\ =1+0.018+0.000144 \\ +0.000000672+\cdots \\ =1.018144672 \ldots \end{gathered} $$ Hence \((1.002)^{9}=1.018\), correct to 3 decimal places \(=1.018145\), correct to 7 significant figures.
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