IB DP Maths Topic 9.6 Taylor polynomials; the Lagrange form of the error term. HL Paper 3

Question

The function f is defined by

\[f(x) = \ln \left( {\frac{1}{{1 – x}}} \right).\]

(a)     Write down the value of the constant term in the Maclaurin series for \(f(x)\) .

(b)     Find the first three derivatives of \(f(x)\) and hence show that the Maclaurin series for \(f(x)\) up to and including the \({x^3}\) term is \(x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3}\).

(c)     Use this series to find an approximate value for ln 2 .

(d)     Use the Lagrange form of the remainder to find an upper bound for the error in this approximation.

(e)     How good is this upper bound as an estimate for the actual error?

▶️Answer/Explanation

Markscheme

(a)     Constant term = 0     A1

[1 mark]

 

(b)     \(f'(x) = \frac{1}{{1 – x}}\)     A1

\(f”(x) = \frac{1}{{{{(1 – x)}^2}}}\)     A1

\(f”'(x) = \frac{2}{{{{(1 – x)}^3}}}\)     A1

\(f'(0) = 1;{\text{ }}f”(0) = 1;{\text{ }}f”'(0) = 2\)     A1

Note: Allow FT on their derivatives.

 

\(f(x) = 0 + \frac{{1 \times x}}{{1!}} + \frac{{1 \times {x^2}}}{{2!}} + \frac{{2 \times {x^3}}}{{3!}} + …\)     M1A1

\( = x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3}\)     AG

[6 marks]

 

(c)     \(\frac{1}{{1 – x}} = 2 \Rightarrow x = \frac{1}{2}\)     (A1)

\(\ln 2 \approx \frac{1}{2} + \frac{1}{8} + \frac{1}{{24}}\)     M1

\( = \frac{2}{3}{\text{ (0.667)}}\)     A1

[3 marks]

 

(d)     Lagrange error \({\text{ = }}\frac{{{f^{(n + 1)}}(c)}}{{(n + 1)!}} \times {\left( {\frac{1}{2}} \right)^{n + 1}}\)     (M1)

\( = \frac{6}{{{{(1 – c)}^4}}} \times \frac{1}{{24}} \times {\left( {\frac{1}{2}} \right)^4}\)     A1

\( < \frac{6}{{{{\left( {1 – \frac{1}{2}} \right)}^4}}} \times \frac{1}{{24}} \times \frac{1}{{16}}\)     A2

giving an upper bound of 0.25.     A1

[5 marks]

 

(e)     Actual error \( = \ln 2 – \frac{2}{3} = 0.0265\)     A1

The upper bound calculated is much larger that the actual error therefore cannot be considered a good estimate.     R1

[2 marks]

Total [17 marks]

Examiners report

In (a), some candidates appeared not to understand the term ‘constant term’. In (b), many candidates found the differentiation beyond them with only a handful realising that the best way to proceed was to rewrite the function as \(f(x) = – \ln (1 – x)\). In (d), many candidates were unable to use the Lagrange formula for the upper bound so that (e) became inaccessible.

 

Question

Let \(f(x) = {{\text{e}}^x}\sin x\).

a.Show that \(f”(x) = 2\left( {f'(x) – f(x)} \right)\).[4]

b.By further differentiation of the result in part (a) , find the Maclaurin expansion of \(f(x)\), as far as the term in \({x^5}\).[6]

▶️Answer/Explanation

Markscheme

\(f'(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x\)     M1A1

\(f”(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x – {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x = 2{{\text{e}}^x}\cos x\)     A1

\( = 2\left( {{{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x – {{\text{e}}^x}\sin x} \right)\)     M1

\( = 2\left( {f'(x) – f(x)} \right)\)     AG

[4 marks]

a.

\(f(0) = 0,{\text{ }}f'(0) = 1,{\text{ }}f”(0) = 2(1 – 0) = 2\)     (M1)A1

Note:     Award M1 for attempt to find \(f(0)\), \(f'(0)\) and \(f”(0)\).

\(f”'(x) = 2\left( {f”(x) – f'(x)} \right)\)     (M1)

\(f”'(0) = 2(2 – 1) = 2,{\text{ }}{f^{IV}}(0) = 2(2 – 2) = 0,{\text{ }}{f^V}(0) = 2(0 – 2) =  – 4\)     A1

so \(f(x) = x + \frac{2}{{2!}}{x^2} + \frac{2}{{3!}}{x^3} – \frac{4}{{5!}} +  \ldots \)     (M1)A1

\( = x + {x^2} + \frac{1}{3}{x^3} – \frac{1}{{30}}{x^5} +  \ldots \)

[6 marks]

Total [10 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Question

a.Given that \(f(x) = \ln x\), use the mean value theorem to show that, for \(0 < a < b\), \(\frac{{b – a}}{b} < \ln \frac{b}{a} < \frac{{b – a}}{a}\).[7]

b.Hence show that \(\ln (1.2)\) lies between \(\frac{1}{m}\) and \(\frac{1}{n}\), where \(m\), \(n\) are consecutive positive integers to be determined.[2]

 
▶️Answer/Explanation

Markscheme

\(f'(x) = \frac{1}{x}\)    (A1)

using the MVT \(f'(c) = \frac{{f(b) – f(a)}}{{b – a}}\) (where \(c\) lies between \(a\) and \(b\))     (M1)

\(f'(c) = \frac{{\ln b – \ln a}}{{b – a}}\)    A1

\(\ln \frac{b}{a} = \ln b – \ln a\)    (M1)

\(f'(c) = \frac{{\ln \frac{b}{a}}}{{b – a}}\)

since \(f'(x)\) is a decreasing function or \(a < c < b \Rightarrow \frac{1}{b} < \frac{1}{c} < \frac{1}{a}\)     R1

\(f'(b) < f'(c) < f'(a)\)    (M1)

\(\frac{1}{b} < \frac{{\ln \frac{b}{a}}}{{b – a}} < \frac{1}{a}\)    A1

\(\frac{{b – a}}{b} < \ln \frac{b}{a} < \frac{{b – a}}{a}\)    AG

[7 marks]

a.

putting \(b = 1.2,{\text{ }}a = 1\), or equivalent     M1

\(\frac{1}{6} < \ln 1.2 < \frac{1}{5}\)    A1

\((m = 6,{\text{ }}n = 5)\)

[2 marks]

b.

Examiners report

Although many candidates achieved at least a few marks in this question, the answers revealed difficulties in setting up a proof. The Mean value theorem was poorly quoted and steps were often skipped. The conditions under which the Mean value theorem is valid were largely ignored, as were the reasoned steps towards the answer.

a.

There were inequalities everywhere, without a great deal of meaning or showing progress. A number of candidates attempted to work backwards and presented the work in a way that made it difficult to follow their reasoning; in part (b) many candidates ignored the instruction ‘hence’ and just used GDC to find the required values; candidates that did notice the link to part a) answered this question well in general. A number of candidates guessed the answer and did not present an analytical derivation as required.

b.

Question

a.By successive differentiation find the first four non-zero terms in the Maclaurin series for \(f(x) = (x + 1)\ln (1 + x) – x\).[11]

b.Deduce that, for \(n \geqslant 2\), the coefficient of \({x^n}\) in this series is \({( – 1)^n}\frac{1}{{n(n – 1)}}\).[1]

c.By applying the ratio test, find the radius of convergence for this Maclaurin series.[6]

▶️Answer/Explanation

Markscheme

\(f(x) = (x + 1)\ln (1 + x) – x\)     \(f(0) = 0\)    A1

\(f'(x) = \ln (1 + x) + \frac{{x + 1}}{{1 + x}} – 1{\text{ }}\left( { = \ln (1 + x)} \right)\)     \(f'(0) = 0\)    M1A1A1

\(f”(x) = {(1 + x)^{ – 1}}\)     \(f”(0) = 1\)    A1A1

\(f”'(x) =  – {(1 + x)^{ – 2}}\)     \(f”'(0) =  – 1\)    A1

\({f^{(4)}}(x) = 2{(1 + x)^{ – 3}}\)     \({f^{(4)}}(0) = 2\)    A1

\({f^{(5)}}(x) =  – 3 \times 2{(1 + x)^{ – 4}}\)     \({f^{(5)}}(0) =  – 3 \times 2\)    A1

\(f(x) = \frac{{{x^2}}}{{2!}} – \frac{{1{x^3}}}{{3!}} + \frac{{2{x^4}}}{{4!}} – \frac{{6{x^5}}}{{5!}} \ldots \)    M1A1

\(f(x) = \frac{{{x^2}}}{{1 \times 2}} – \frac{{{x^3}}}{{2 \times 3}} + \frac{{{x^4}}}{{3 \times 4}} – \frac{{{x^5}}}{{4 \times 5}} \ldots \)

\(f(x) = \frac{{{x^2}}}{2} – \frac{{{x^3}}}{6} + \frac{{{x^4}}}{{12}} – \frac{{{x^5}}}{{20}} \ldots \)

Note: Allow follow through from the first error in a derivative (provided future derivatives also include the chain rule), no follow through after a second error in a derivative.

[11 marks]

a.

\({f^{(n)}}(0) = {( – 1)^n}(n – 2)!\) So coefficient of \({x^n} = {( – 1)^n}\frac{{(n – 2)!}}{{n!}}\)     A1

coefficient of \({x^n}\) is \({( – 1)^n}\frac{1}{{n(n – 1)}}\)     AG

[1 mark]

b.

applying the ratio test to the series of absolute terms

\(\mathop {\lim }\limits_{n \to \infty } \frac{{\frac{{{{\left| x \right|}^{n + 1}}}}{{(n + 1)n}}}}{{\frac{{{{\left| x \right|}^n}}}{{n(n – 1)}}}}\)    M1A1

\( = \mathop {\lim }\limits_{n \to \infty } \left| x \right|\frac{{(n – 1)}}{{(n + 1)}}\)    A1

\( = \left| x \right|\)    A1

so for convergence \(\left| x \right| < 1\), giving radius of convergence as 1     (M1)A1

[6 marks]

c.

Examiners report

[N/A]

a.

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b.

[N/A]

c.
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