4.08a Maclaurin series: find series for function

\(y = \sec^2 x\)

1. Show that \(\frac{d^2 y}{dx^2} = 6 \sec^4 x - 4 \sec^2 x\). [4]
2. Find a Taylor series expansion of \(\sec^2 x\) in ascending powers of \(\left(x - \frac{\pi}{4}\right)\), up to and including the term in \(\left(x - \frac{\pi}{4}\right)^3\). [6]

The displacement \(x\) metres of a particle at time \(t\) seconds is given by the differential equation $$\frac{d^2 x}{dt^2} + x + \cos x = 0.$$ When \(t = 0\), \(x = 0\) and \(\frac{dx}{dt} = \frac{1}{2}\). Find a Taylor series solution for \(x\) in ascending powers of \(t\), up to and including the term in \(t^4\). [5]

$$\frac{d^2 y}{dx^2} = e^x \left(2x \frac{dy}{dx} + y^2 + 1\right).$$

1. Show that $$\frac{d^4 y}{dx^4} = e^x \left[2x \frac{d^3 y}{dx^3} + 4 \frac{d^2 y}{dx^2} + 6y \frac{dy}{dx} + y^2 + 1\right],$$ where \(k\) is a constant to be found. [3]

Given that, at \(x = 0\), \(y = 1\) and \(\frac{dy}{dx} = 2\),

1. find a series solution for \(y\) in ascending powers of \(x\), up to and including the term in \(x^4\). [4]

$$x \frac{dy}{dx} = 3x + y^2.$$

1. Show that $$\frac{d^2 y}{dx^2} + (1 - 2y) \frac{dy}{dx} = 3.$$ [2]

Given that \(y = 1\) at \(x = 1\),

1. find a series solution for \(y\) in ascending powers of \((x - 1)\), up to and including the term in \((x - 1)^3\). [8]

Given that $$y \frac{d^3 y}{dx^3} + \left(\frac{dy}{dx}\right)^2 + 5y = 0$$

1. find \(\frac{d^3 y}{dx^3}\) in terms of \(\frac{d^2 y}{dx^2}\), \(\frac{dy}{dx}\) and \(y\). [4]

Given that \(y = 2\) and \(\frac{dy}{dx} = 2\) at \(x = 0\),

1. find a series solution for \(y\) in ascending powers of \(x\), up to and including the term in \(x^3\). [5]

$$\frac{d^2 y}{dx^2} + 4y - \sin x = 0$$ Given that \(y = \frac{1}{2}\) and \(\frac{dy}{dx} = \frac{1}{8}\) at \(x = 0\), find a series expansion for \(y\) in terms of \(x\), up to and including the term in \(x^5\). [5]

$$(x^2 + 1)\frac{d^2y}{dx^2} = 2y^2 + (1 - 2x)\frac{dy}{dx}$$ (I)

1. By differentiating equation (I) with respect to \(x\), show that $$(x^2 + 1)\frac{d^3y}{dx^3} = (1 - 4x)\frac{d^2y}{dx^2} + (4y - 2)\frac{dy}{dx}.$$ (3) Given that \(y = 1\) and \(\frac{dy}{dx} = 1\) at \(x = 0\),
2. find the series solution for \(y\), in ascending powers of \(x\), up to and including the term in \(x_3\).(4)
3. Use your series to estimate the value of \(y\) at \(x = -0.5\), giving your answer to two decimal places.(1)

$$y\frac{d^2 y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + y = 0.$$

1. Find an expression for \(\frac{d^3 y}{dx^3}\). [5] Given that \(y = 1\) and \(\frac{dy}{dx} = 1\) at \(x = 0\),
2. find the series solution for \(y\), in ascending powers of \(x\), up to an including the term in \(x^3\). [5]
3. Comment on whether it would be sensible to use your series solution to give estimates for \(y\) at \(x = 0.2\) and at \(x = 50\). [2]

Given that \(y = \tan x\),

1. find \(\frac{dy}{dx}\), \(\frac{d^2 y}{dx^2}\) and \(\frac{d^3 y}{dx^3}\). [3]
2. Find the Taylor series expansion of \(\tan x\) in ascending powers of \(\left(x - \frac{\pi}{4}\right)\) up to and including the term in \(\left(x - \frac{\pi}{4}\right)^3\). [3]
3. Hence show that \(\tan \frac{3\pi}{10} \approx 1 + \frac{\pi}{10} + \frac{\pi^2}{200} + \frac{\pi^3}{3000}\). [2]

$$I_n = \int \sec^n x \, dx \quad n \geq 0$$

1. Prove that for \(n \geq 2\) $$(n-1)I_n = \tan x \sec^{n-2} x + (n-2)I_{n-2}$$ [6]
2. Hence, showing each step of your working, find the exact value of $$\int_0^{\pi/4} \sec^6 x \, dx$$ [4]

$$I_n = \int_{0}^{\ln 2} \tanh^{2n} x \, dx, \quad n \geq 0$$

1. Show that, for \(n \geq 1\) $$I_n = I_{n-1} - \frac{1}{2n-1}\left(\frac{3}{5}\right)^{2n-1}$$ [5]
2. Hence show that $$\int_{0}^{\ln 2} \tanh^{-1} x \, dx = p + \ln 2$$ where \(p\) is a rational number to be found. [5]

$$I_n = \int_1^e x^2 (\ln x)^n dx, \quad n \geq 0$$

1. Prove that, for \(n \geq 1\), $$I_n = \frac{e^3}{3} - \frac{n}{3} I_{n-1}$$ [4]
2. Find the exact value of \(I_3\). [4]

$$I_n = \int (x^2 + 1)^{-n} dx, \quad n > 0$$

1. Show that, for \(n > 0\) $$I_{n+1} = \frac{x(x^2 + 1)^{-n}}{2n} + \frac{2n - 1}{2n}I_n$$ [5]
2. Find \(I_2\) [3]

$$I_n = \int_0^{\frac{\pi}{2}} x^n \cos x \, dx, \quad n \geq 0.$$

1. Prove that \(I_n = \left(\frac{\pi}{2}\right)^n - n(n-1)I_{n-2}\), \(n \geq 2\). [5]
2. Find an exact expression for \(I_6\). [4]

$$I_n = \int_0^{\pi} x^n \sin x \, dx$$

1. Show that for \(n \geq 2\) $$I_n = n \left( \frac{\pi}{2} \right)^{n-1} - n(n-1)I_{n-2}$$ [4]
2. Hence obtain \(I_3\), giving your answers in terms of \(\pi\). [4]

(Total 8 marks)

It is given that \(f(x) = \tan^{-1}(1 + x)\).

1. Find \(f(0)\) and \(f'(0)\), and show that \(f''(0) = -\frac{1}{2}\). [4]
2. Hence find the Maclaurin series for \(f(x)\) up to and including the term in \(x^2\). [2]

Given that \(f(x) = \ln(\cos 3x)\), find \(f'(0)\) and \(f''(0)\). Hence show that the first term in the Maclaurin series for \(f(x)\) is \(ax^2\), where the value of \(a\) is to be found. [4]

The first two non-zero terms of the Maclaurin series expansion of \(f(x)\) are \(x\) and \(-\frac{1}{2}x^3\) Which one of the following could be \(f(x)\)? Circle your answer. [1 mark] \(xe^{\frac{1}{2}x^2}\) \quad \(\frac{1}{2}\sin 2x\) \quad \(x \cos x\) \quad \((1 + x^3)^{-\frac{1}{2}}\)

Use l'Hôpital's rule to prove that $$\lim_{x \to \pi} \frac{x \sin 2x}{\cos\left(\frac{x}{2}\right)} = -4\pi$$ [5 marks]

Which one of the following statements is correct? Tick (\(\checkmark\)) one box. [1 mark] \(\lim_{x \to 0}(x^2 \ln x) = 0\) \(\square\) \(\lim_{x \to 0}(x^2 \ln x) = 1\) \(\square\) \(\lim_{x \to 0}(x^2 \ln x) = 2\) \(\square\) \(\lim_{x \to 0}(x^2 \ln x)\) is not defined. \(\square\)

The integral \(I_n\) is defined by $$I_n = \int_0^{\frac{\pi}{4}} \cos^n x \, dx \quad\quad (n \geq 0)$$

1. Show that $$I_n = \left(\frac{n-1}{n}\right)I_{n-2} + \frac{1}{n\left(2^{\frac{n}{2}}\right)} \quad\quad (n \geq 2)$$ [6 marks]
2. Use the result from part (a) to show that $$\int_0^{\frac{\pi}{4}} \cos^6 x \, dx = \frac{a\pi + b}{192}$$ where \(a\) and \(b\) are integers to be found. [3 marks]

Given that \(I_n = \int_0^{\frac{\pi}{2}} \sin^n x \, dx\) \quad \(n \geq 0\) show that \(n I_n = (n-1)I_{n-2}\) \quad \(n \geq 2\) [5 marks]