4.08a Maclaurin series: find series for function

2 It is given that \(y = 2 ^ { x }\).

1. By differentiating \(\ln y\) with respect to \(x\), show that \(\frac { \mathrm { dy } } { \mathrm { dx } } = 2 ^ { \mathrm { x } } \ln 2\).
2. Write down \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\).
3. Hence find the first three terms in the Maclaurin's series for \(2 ^ { X }\).

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1. It is given that \(\mathrm { y } = \operatorname { sech } ^ { - 1 } \left( \mathrm { x } + \frac { 1 } { 2 } \right)\).
   Express cosh \(y\) in terms of \(x\) and hence show that \(\sinh y \frac { d y } { d x } = - \frac { 1 } { \left( x + \frac { 1 } { 2 } \right) ^ { 2 } }\).
2. Find the first three terms in the Maclaurin's series for \(\operatorname { sech } ^ { - 1 } \left( x + \frac { 1 } { 2 } \right)\) in the form $$\ln a + b x + c x ^ { 2 }$$ where \(a\), \(b\) and \(c\) are constants to be determined.

2 Find the Maclaurin's series for \(\ln \cosh x\) up to and including the term in \(x ^ { 4 }\).

5 The variables \(x\) and \(y\) are such that \(y = 0\) when \(x = 0\) and $$( x + 1 ) y + ( x + y + 1 ) ^ { 3 } = 1$$

1. Show that \(\frac { \mathrm { dy } } { \mathrm { dx } } = - \frac { 3 } { 4 }\) when \(x = 0\).
2. Find the Maclaurin's series for \(y\) up to and including the term in \(x ^ { 2 }\).

2

1. Find the coefficient of \(x ^ { 2 }\) in the Maclaurin's series for \(- \ln \cos x\).
2. Find the length of the arc of the curve with equation \(\mathrm { y } = - \operatorname { Incos } \mathrm { x }\) from the point where \(x = 0\) to the point where \(x = \frac { 1 } { 4 } \pi\).

1

1. Find the Maclaurin series for \(\sin ^ { - 1 } x\) up to and including the term in \(x ^ { 3 }\).
2. Deduce an approximation to \(\int _ { 0 } ^ { \frac { 1 } { 5 } } \frac { 1 } { \sqrt { 1 - u ^ { 2 } } } \mathrm {~d} u\), giving your answer as a fraction.

2 Find the Maclaurin's series for \(\mathrm { e } ^ { 1 + x ^ { 2 } } + \mathrm { e } ^ { 1 - x }\) up to and including the term in \(x ^ { 2 }\).

2 The curve \(C\) has parametric equations $$x = \cosh t , \quad y = \sinh t , \quad \text { for } 0 < t \leqslant \frac { 3 } { 5 }$$ The length of \(C\) is denoted by \(s\).

1. Show that \(s = \int _ { 0 } ^ { \frac { 3 } { 5 } } \sqrt { \cosh 2 t } \mathrm {~d} t\). \includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-04_2714_37_143_2008}
2. By finding the Maclaurin's series for \(\sqrt { \cosh 2 t }\) up to and including the term in \(t ^ { 2 }\) ,deduce an approximation to \(s\) .

1

1. By differentiating \(\mathrm { e } ^ { - x ^ { 2 } }\), find the Maclaurin's series for \(\mathrm { e } ^ { - x ^ { 2 } }\) up to and including the term in \(x ^ { 2 }\).
2. Deduce an approximation to \(\int _ { 0 } ^ { \frac { 1 } { 5 } } \mathrm { e } ^ { - x ^ { 2 } } \mathrm {~d} x\), giving your answer as a rational fraction in its lowest terms.

1 Find the Maclaurin's series for \(\tan \left( x + \frac { 1 } { 4 } \pi \right)\) up to and including the term in \(x ^ { 2 }\).

1 Find the Maclaurin's series for \(e ^ { x } \tan x\) from first principles up to and including the term in \(x ^ { 2 }\).

3 Find the first three terms in the Maclaurin's series for \(\tanh ^ { - 1 } \left( \frac { 1 } { 2 } e ^ { x } \right)\) in the form \(\frac { 1 } { 2 } \ln a + b x + c x ^ { 2 }\), giving the exact values of the constants \(a , b\) and \(c\).

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1. State the sum of the series \(1 + z + z ^ { 2 } + \ldots + z ^ { n - 1 }\), for \(z \neq 1\).
2. By letting \(z = \cos \theta + i \sin \theta\), where \(\cos \theta \neq 1\), show that $$1 + \cos \theta + \cos 2 \theta + \ldots + \cos ( n - 1 ) \theta = \frac { 1 } { 2 } \left( 1 - \cos n \theta + \frac { \sin n \theta \sin \theta } { 1 - \cos \theta } \right)$$ \includegraphics[max width=\textwidth, alt={}, center]{dffdf588-eb26-4d08-b1a3-a0226f5e7763-15_833_785_214_680} The diagram shows the curve with equation \(\mathrm { y } = \cos \mathrm { x }\) for \(0 \leqslant x \leqslant 1\), together with a set of \(n\) rectangles of width \(\frac { 1 } { n }\).
3. By considering the sum of the areas of these rectangles, show that $$\int _ { 0 } ^ { 1 } \cos x d x < \frac { 1 } { 2 n } \left( 1 - \cos 1 + \frac { \sin 1 \sin \frac { 1 } { n } } { 1 - \cos \frac { 1 } { n } } \right)$$
4. Use a similar method to find, in terms of \(n\), a lower bound for \(\int _ { 0 } ^ { 1 } \cos x d x\).
   If you use the following page to complete the answer to any question, the question number must be clearly shown.

1 Find the Maclaurin's series for \(\ln ( x + 2 ) + \ln \left( x ^ { 2 } + 5 \right)\) up to and including the term in \(x ^ { 2 }\).

6 \includegraphics[max width=\textwidth, alt={}, center]{374b91df-926d-4f7f-a1d3-a54c70e8ff0e-12_533_1532_278_264} The diagram shows the curve with equation \(y = \left( \frac { 1 } { 2 } \right) ^ { x }\) for \(0 \leqslant x \leqslant 1\), together with a set of \(N\) rectangles each of width \(\frac { 1 } { N }\).

1. By considering the sum of the areas of these rectangles, show that \(\int _ { 0 } ^ { 1 } \left( \frac { 1 } { 2 } \right) ^ { x } \mathrm {~d} x > L _ { N }\), where $$L _ { N } = \frac { 1 } { 2 N \left( 2 ^ { \frac { 1 } { N } } - 1 \right) }$$ \includegraphics[max width=\textwidth, alt={}, center]{374b91df-926d-4f7f-a1d3-a54c70e8ff0e-12_2717_38_109_2009}
2. Use a similar method to find, in terms of \(N\), an upper bound \(U _ { N }\) for \(\int _ { 0 } ^ { 1 } \left( \frac { 1 } { 2 } \right) ^ { x } \mathrm {~d} x\).
3. Find the least value of \(N\) such that \(U _ { N } - L _ { N } \leqslant 10 ^ { - 3 }\).
4. Given that \(\int _ { 0 } ^ { 1 } \left( \frac { 1 } { 2 } \right) ^ { x } \mathrm {~d} x = \frac { 1 } { 2 \ln 2 }\) ,use the value of \(N\) found in part(c)to find upper and lower bounds for \(\ln 2\) .

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1. Starting from the definition of tanh in terms of exponentials, prove that \(\tanh ^ { - 1 } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)\). [3]
2. Given that \(y = \tanh ^ { - 1 } \left( \frac { 1 - x } { 2 + x } \right)\), show that \(( 2 x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } + 1 = 0\).
3. Hence find the first three terms in the Maclaurin's series for \(\tanh ^ { - 1 } \left( \frac { 1 - x } { 2 + x } \right)\) in the form $$a \ln 3 + b x + c x ^ { 2 } ,$$ where \(a , b\) and \(c\) are constants to be determined.

5. Given that $$\left( 2 - x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 x \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } = 3 y$$

1. show that $$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = \frac { 1 } { \left( 2 - x ^ { 2 } \right) } \left( 2 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } \left( 1 - 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) - 5 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 3 \frac { \mathrm {~d} y } { \mathrm {~d} x } \right)$$ Given also that \(y = 3\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 4 }\) at \(x = 0\)
2. obtain a series solution for \(y\) in ascending powers of \(x\) with simplified coefficients, up to and including the term in \(x ^ { 3 }\)

5. $$y = \sqrt { 4 + \ln x } \quad x > \frac { 1 } { 2 }$$

1. Show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = - \frac { 9 + 2 \ln x } { 4 x ^ { 2 } ( 4 + \ln x ) ^ { \frac { 3 } { 2 } } }$$
2. Hence, or otherwise, determine the Taylor series expansion about \(x = 1\) for \(y\), in ascending powers of ( \(x - 1\) ), up to and including the term in \(( x - 1 ) ^ { 2 }\), giving each coefficient in simplest form.

4. $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ^ { 2 } - x$$

1. Show that $$\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } = A y \frac { \mathrm {~d} ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } + B \frac { \mathrm {~d} y } { \mathrm {~d} x } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$$ where \(A\) and \(B\) are integers to be determined. Given that \(y = 1\) at \(x = - 1\)
2. determine the Taylor series solution for \(y\), in ascending powers of \(( x + 1 )\) up to and including the term in \(( x + 1 ) ^ { 4 }\), giving each coefficient in simplest form.

1. In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

1. Determine, in ascending powers of \(\left( x - \frac { \pi } { 6 } \right)\) up to and including the term in \(\left( x - \frac { \pi } { 6 } \right) ^ { 3 }\), the Taylor series expansion about \(\frac { \pi } { 6 }\) of $$y = \tan \left( \frac { 3 x } { 2 } \right)$$ giving each coefficient in simplest form.
2. Hence show that $$\tan \frac { 3 \pi } { 8 } \approx 1 + \frac { \pi } { 4 } + \frac { \pi ^ { 2 } } { A } + \frac { \pi ^ { 3 } } { B }$$ where \(A\) and \(B\) are integers to be determined.

5. $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 0$$

1. Show that $$\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } = \left( a x ^ { 2 } + b \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$$ where \(a\) and \(b\) are constants to be found. Given that \(y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3\) at \(x = 0\)
2. find a series solution for \(y\) in ascending powers of \(x\) up to and including the term in \(x ^ { 4 }\)
3. use your series to estimate the value of \(y\) at \(x = - 0.2\), giving your answer to four decimal places.

4. $$f ( x ) = \sin \left( \frac { 3 } { 2 } x \right)$$

1. Find the Taylor series expansion for \(\mathrm { f } ( x )\) about \(\frac { \pi } { 3 }\) in ascending powers of \(\left( x - \frac { \pi } { 3 } \right)\) up to and including the term in \(\left( x - \frac { \pi } { 3 } \right) ^ { 4 }\)
2. Hence obtain an estimate of \(\sin \frac { 1 } { 2 }\), giving your answer to 4 decimal places.

5. $$y = \mathrm { e } ^ { \cos ^ { 2 } x }$$

1. Show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \mathrm { e } ^ { \cos ^ { 2 } x } \left( \sin ^ { 2 } 2 x - 2 \cos 2 x \right)$$
2. Hence find the Maclaurin series expansion of \(\mathrm { e } ^ { \cos ^ { 2 } x }\) up to and including the term in \(x ^ { 2 }\)

4. Given that $$y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 3 y = 0$$

1. show that $$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = \frac { 28 } { y ^ { 2 } } \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 3 } - \frac { 24 } { y } \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right)$$ Given also that \(y = 8\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\) at \(x = 0\)
2. find a series solution for \(y\) in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\), simplifying the coefficients where possible.