MATH101 Calculus Assessed Exercise 4 Solutions

[10] total for exercise

A4.1. (i) Let $P_{n}$ be the statement

$${{d^{n}}\over{dx^{n}}}\log(1+x)={{(-1)^{n-1}(n-1)!}\over{(1+x)^{n}}}\qquad(-1<x).$$

Basis of induction:$P_{1}$ asserts that

$${{d}\over{dx}}\log(1+x)={{1}\over{1+x}},$$

which is true.

Induction step: Assume that $P_{n}$ is true, so

$${{d^{n}}\over{dx^{n}}}\log(1+x)={{(-1)^{n-1}(n-1)!}\over{(1+x)^{n}}}\qquad(-1<x).$$

Now differentiate both sides to obtain

$${{d}\over{dx}}{{d^{n}}\over{dx^{n}}}\log(1+x)={{(-n)(-1)^{n-1}(n-1)!}\over{(1+x)^{n+1}}}\qquad(-1<x);$$

so

$${{d^{n+1}}\over{dx^{n+1}}}\log(1+x)={{(-1)^{n}(n)!}\over{(1+x)^{n+1}}}\qquad(-1<x);$$

hence $P_{n+1}$ is true and hence result by induction.

[2] marks

(ii) The Maclaurin series is

$$f(x)=f(0)+f^{\prime}(0)x+{{f^{\prime\prime}(0)}\over{2!}}x^{2}+\dots+{{f^{(n)}(0)}\over{n!}}x^{n}+\dots.$$

so here we have

$$\eqalignno{\log(1+x)&=\log 1+x-{{1!x^{2}}\over{2!}}+{{2!x^{3}}\over{3!}}-\dots+{{(-1)^{n-1}(n-1)!x^{n}}\over{n!}}+\dots\cr&=x-{{x^{2}}\over{2}}+{{x^{3}}\over{3}}-\dots+{{(-1)^{n-1}x^{n}}\over{n}}+\dots.\cr}$$

[2] marks

(iii) Substituting $-x$ in place of $x$, we obtain the series

$$\log(1+x)=-x-{{x^{2}}\over{2}}-{{x^{3}}\over{3}}-\dots-{{x^{n}}\over{n}}-\dots.$$

[1] marks

Total [5] marks for question

A4.2 (i) Let $P_{n}$ be the statement

$$1+e^{i\theta}+\dots+e^{i(n-1)\theta}={{1-e^{in\theta}}\over{1-e^{i\theta}}}.$$

Basis of induction.$P_{1}$ asserts that $1=1$, which holds.

Induction step Suppose that $P_{n}$ holds for some $n$, and consider $P_{n+1}$; then

$$\eqalignno{\bigl(1+e^{i\theta}+\dots+e^{i(n-1)\theta}\bigr)+e^{in\theta}&={{1-e^{in\theta}}\over{1-e^{i\theta}}}+e^{in\theta}\cr&={{1-e^{in\theta}+e^{in\theta}-e^{-i(n+1)\theta}}\over{1-e^{i\theta}}}\cr&={{1-e^{i(n+1)\theta}}\over{1-e^{i\theta}}};\cr}$$

hence $P_{n+1}$ holds and hence result by induction.

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(ii) The complex conjugate of $e^{i\theta}-1=\cos\theta+i\sin\theta-1$ is $\cos\theta-i\sin\theta-1=e^{-i\theta}-1$.

(iii) We have

$${{1-e^{in\theta}}\over{1-e^{i\theta}}}={{(1-e^{in\theta})(1-e^{-i\theta})}\over{(1-e^{i\theta})(1-e^{-i\theta})}}={{1+e^{i(n-1)\theta}-e^{in\theta}-e^{-i\theta}}\over{2-e^{i\theta}-e^{-i\theta}}}$$

[1]

and $e^{ik\theta}=\cos k\theta+i\sin k\theta$, so the real parts are

$$1+\cos\theta+\cos 2\theta+\dots+\cos(n-1)\theta={{1+\cos(n-1)\theta-\cos n\theta-\cos\theta}\over{2-2\cos\theta}}.$$

The imaginary parts are

$$\sin\theta+\sin 2\theta+\dots+\sin(n-1)\theta={{\sin(n-1)\theta-\sin n\theta+\sin\theta}\over{2-2\cos\theta}}$$

[1]

(One can reduce this further by considering addition rules with $\sin\theta/2$).

Total [3] marks for question

A4.3 (i) The derivative $f^{\prime}(x)$ is a polynomial of degree $n-1$, so the equation $f^{\prime}(x)=0$ has at most $n-1$ real roots by the remainder theorem.

[1]

(ii) Every local maximum, local minimum and inflexion gives a stationary point of $f$; that is a real root of $f^{\prime}(x)=0$. Hence by (i),

$$a+b+c\leq n-1.$$

[1]

[2] marks for question