# 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?

## 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

The Taylor series of $$\sqrt x$$ about x = 1 is given by

${a_0} + {a_1}(x – 1) + {a_2}{(x – 1)^2} + {a_3}{(x – 1)^3} + \ldots$

Find the values of $${a_0},{\text{ }}{a_1},{\text{ }}{a_2}$$ and $${a_3}$$.

[6]
a.

Hence, or otherwise, find the value of $$\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x – 1}}{{x – 1}}$$.

[3]
b.

## Markscheme

let $$f(x) = \sqrt x ,{\text{ }}f(1) = 1$$     (A1)

$$f'(x) = \frac{1}{2}{x^{ – \frac{1}{2}}},{\text{ }}f'(1) = \frac{1}{2}$$     (A1)

$$f”(x) = – \frac{1}{4}{x^{ – \frac{3}{2}}},{\text{ }}f”(1) = – \frac{1}{4}$$     (A1)

$$f”'(x) = \frac{3}{8}{x^{ – \frac{5}{2}}},{\text{ }}f”'(1) = \frac{3}{8}$$     (A1)

$${a_1} = \frac{1}{2} \cdot \frac{1}{{1!}},{\text{ }}{a_2} = – \frac{1}{4} \cdot \frac{1}{{2!}},{\text{ }}{a_3} = \frac{3}{8} \cdot \frac{1}{{3!}}$$     (M1)

$${a_0} = 1,{\text{ }}{a_1} = \frac{1}{2},{\text{ }}{a_2} = – \frac{1}{8},{\text{ }}{a_3} = \frac{1}{{16}}$$     A1

Note: Accept $$y = 1 + \frac{1}{2}(x – 1) – \frac{1}{8}{(x – 1)^2} + \frac{1}{{16}}{(x – 1)^3} + \ldots$$

[6 marks]

a.

METHOD 1

$$\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x – 1}}{{x – 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{2}(x – 1) – \frac{1}{8}{{(x – 1)}^2} + \ldots }}{{x – 1}}$$     M1

$$= \mathop {\lim }\limits_{x \to 1} \left( {\frac{1}{2} – \frac{1}{8}(x – 1) + \ldots } \right)$$     A1

$$= \frac{1}{2}$$     A1

METHOD 2

using l’Hôpital’s rule,     M1

$$\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x – 1}}{{x – 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{2}{x^{ – \frac{1}{2}}}}}{1}$$     A1

$$= \frac{1}{2}$$     A1

METHOD 3

$$\frac{{\sqrt x – 1}}{{x + 1}} = \frac{1}{{\sqrt x + 1}}$$     M1A1

$$\mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt x + 1}} = \frac{1}{2}$$     A1

[3 marks]

b.

## Examiners report

Many candidates achieved full marks on this question but there were still a large minority of candidates who did not seem familiar with the application of Taylor series. Whilst all candidates who responded to this question were aware of the need to use derivatives many did not correctly use factorials to find the required coefficients. It should be noted that the formula for Taylor series appears in the Information Booklet.

a.

Many candidates achieved full marks on this question but there were still a large minority of candidates who did not seem familiar with the application of Taylor series. Whilst all candidates who responded to this question were aware of the need to use derivatives many did not correctly use factorials to find the required coefficients. It should be noted that the formula for Taylor series appears in the Information Booklet.

b.

## Question

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

Show that $$f”(x) = 2\left( {f'(x) – f(x)} \right)$$.

[4]
a.

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]
b.

## 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.

[N/A]

a.

[N/A]

b.

## Question

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]
a.

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]
b.

## 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

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

[11]
a.

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

[1]
b.

By applying the ratio test, find the radius of convergence for this Maclaurin series.

[6]
c.

## 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.

[N/A]

a.

[N/A]

b.

[N/A]

c.

## Question

Let $$f(x)$$ be a function whose first and second derivatives both exist on the closed interval $$[0,{\text{ }}h]$$.

Let $$g(x) = f(h) – f(x) – (h – x)f'(x) – \frac{{{{(h – x)}^2}}}{{{h^2}}}\left( {f(h) – f(0) – hf'(0)} \right)$$.

State the mean value theorem for a function that is continuous on the closed interval $$[a,{\text{ }}b]$$ and differentiable on the open interval $$]a,{\text{ }}b[$$.

[2]
a.

(i)     Find $$g(0)$$.

(ii)     Find $$g(h)$$.

(iii)     Apply the mean value theorem to the function $$g(x)$$ on the closed interval $$[0,{\text{ }}h]$$ to show that there exists $$c$$ in the open interval $$]0,{\text{ }}h[$$ such that $$g'(c) = 0$$.

(iv)     Find $$g'(x)$$.

(v)     Hence show that $$– (h – c)f”(c) + \frac{{2(h – c)}}{{{h^2}}}\left( {f(h) – f(0) – hf'(0)} \right) = 0$$.

(vi)     Deduce that $$f(h) = f(0) + hf'(0) + \frac{{{h^2}}}{2}{\text{ }}f”(c)$$.

[9]
b.

Hence show that, for $$h > 0$$

$$1 – \cos (h) \leqslant \frac{{{h^2}}}{2}$$.

[5]
c.

## Markscheme

there exists $$c$$ in the open interval $$]a,{\text{ }}b[$$ such that     A1

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

Note: Open interval is required for the A1.

[2 marks]

a.

(i)     $$g(0) = f(h) – f(0) – hf'(0) – \frac{{{h^2}}}{{{h^2}}}\left( {{\text{ }}f(h) – f(0) – hf'(0)} \right)$$

$$= 0$$    A1

(ii)     $$g(h) = f(h) – f(h) – 0 – 0$$

$$= 0$$    A1

(iii)     ($$g(x)$$ is a differentiable function since it is a combination of other differentiable functions $$f$$, $${f’}$$ and polynomials.)

there exists $$c$$ in the open interval $$]0,{\text{ }}h[$$ such that

$$\frac{{g(h) – g(0)}}{h} = g'(c)$$    A1

$$\frac{{g(h) – g(0)}}{h} = 0$$    A1

hence $$g'(c) = 0$$     AG

(iv)     $$g'(x) = – f'(x) + f'(x) – (h – x)f”(x) + \frac{{2(h – x)}}{{{h^2}}}\left( {f(h) – f(0) – hf'(0)} \right)$$     A1A1

Note: A1 for the second and third terms and A1 for the other terms (all terms must be seen).

$$= – (h – x)f”(x) + \frac{{2(h – x)}}{{{h^2}}}\left( {f(h) – f(0) – hf'(0)} \right)$$

(v)     putting $$x = c$$ and equating to zero     M1

$$– (h – c)f”(c) + \frac{{2(h – c)}}{{{h^2}}}\left( {f(h) – f(0) – hf'(0)} \right) = g'(c) = 0$$    AG

(vi)     $$– f”(c) + \frac{2}{{{h^2}}}\left( {f(h) – f(0) – hf'(0)} \right) = 0$$     A1

since $$h – c \ne 0$$     R1

$$\frac{{{h^2}}}{2}f”(c) = f(h) – f(0) – hf'(0)$$

$$f(h) = f(0) + hf'(0) + \frac{{{h^2}}}{2}f”(c)$$    AG

[9 marks]

b.

letting $$f(x) = \cos (x)$$     M1

$$f'(x) = – \sin (x)$$    $$f”(x) = – \cos (x)$$     A1

$$\cos (h) = 1 + 0 – \frac{{{h^2}}}{2}\cos (c)$$     A1

$$1 – \cos (h) = \frac{{{h^2}}}{2}\cos (c)$$    (A1)

since $$\cos (c) \leqslant 1$$     R1

$$1 – \cos (h) \leqslant \frac{{{h^2}}}{2}$$    AG

Note: Allow $$f(x) = a \pm b\cos x$$.

[5 marks]

c.

[N/A]

a.

[N/A]

b.

[N/A]

c.

## Question

The function $$f$$ is defined by $$f(x){\text{ }}={\text{ }}{(\arcsin{\text{ }}x)^2},{\text{ }} – 1 \leqslant x \leqslant 1$$.

The function $$f$$ satisfies the equation $$\left( {1 – {x^2}} \right)f”\left( x \right) – xf’\left( x \right) – 2 = 0$$.

Show that $$f’\left( 0 \right) = 0$$.

[2]
a.

By differentiating the above equation twice, show that

$\left( {1 – {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) – 5x{f^{\left( 3 \right)}}\left( x \right) – 4f”\left( x \right) = 0$

where $${f^{\left( 3 \right)}}\left( x \right)$$ and $${f^{\left( 4 \right)}}\left( x \right)$$ denote the 3rd and 4th derivative of $$f\left( x \right)$$ respectively.

[4]
b.

Hence show that the Maclaurin series for $$f\left( x \right)$$ up to and including the term in $${x^4}$$ is $${x^2} + \frac{1}{3}{x^4}$$.

[3]
c.

Use this series approximation for $$f\left( x \right)$$ with $$x = \frac{1}{2}$$ to find an approximate value for $${\pi ^2}$$.

[2]
d.

## Markscheme

$$f’\left( x \right) = \frac{{2\,{\text{arcsin}}\,\left( x \right)}}{{\sqrt {1 – {x^2}} }}$$     M1A1

Note: Award M1 for an attempt at chain rule differentiation.
Award M0A0 for $$f’\left( x \right) = 2\,{\text{arcsin}}\,\left( x \right)$$.

$$f’\left( 0 \right) = 0$$     AG

[2 marks]

a.

differentiating gives $$\left( {1 – {x^2}} \right){f^{\left( 3 \right)}}\left( x \right) – 2xf”\left( x \right) – f’\left( x \right) – xf”\left( x \right)\left( { = 0} \right)$$      M1A1

differentiating again gives $$\left( {1 – {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) – 2x{f^{\left( 3 \right)}}\left( x \right) – 3f”\left( x \right) – 3x{f^{\left( 3 \right)}}\left( x \right) – f”\left( x \right)\left( { = 0} \right)$$     M1A1

Note: Award M1 for an attempt at product rule differentiation of at least one product in each of the above two lines.
Do not penalise candidates who use poor notation.

$$\left( {1 – {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) – 5x{f^{\left( 3 \right)}}\left( x \right) – 4f”\left( x \right) = 0$$      AG

[4 marks]

b.

attempting to find one of $$f”\left( 0 \right)$$, $${f^{\left( 3 \right)}}\left( 0 \right)$$ or $${f^{\left( 4 \right)}}\left( 0 \right)$$ by substituting $$x = 0$$ into relevant differential equation(s)       (M1)

Note: Condone $$f”\left( 0 \right)$$ found by calculating $$\frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{{2\,{\text{arcsin}}\,\left( x \right)}}{{\sqrt {1 – {x^2}} }}} \right)$$ at $$x = 0$$.

$$\left( {f\left( 0 \right) = 0,\,f’\left( 0 \right) = 0} \right)$$

$$f”\left( 0 \right) = 2$$ and $${f^{\left( 4 \right)}}\left( 0 \right) – 4f”\left( 0 \right) = 0 \Rightarrow {f^{\left( 4 \right)}}\left( 0 \right) = 8$$      A1

$${f^{\left( 3 \right)}}\left( 0 \right) = 0$$ and so $$\frac{2}{{2{\text{!}}}}{x^2} + \frac{8}{{4{\text{!}}}}{x^4}$$     A1

Note: Only award the above A1, for correct first differentiation in part (b) leading to $${f^{\left( 3 \right)}}\left( 0 \right) = 0$$ stated or $${f^{\left( 3 \right)}}\left( 0 \right) = 0$$ seen from use of the general Maclaurin series.
Special Case: Award (M1)A0A1 if $${f^{\left( 4 \right)}}\left( 0 \right) = 8$$ is stated without justification or found by working backwards from the general Maclaurin series.

so the Maclaurin series for $$f\left( x \right)$$ up to and including the term in $${x^4}$$ is $${x^2} + \frac{1}{3}{x^4}$$     AG

[3 marks]

c.

substituting $$x = \frac{1}{2}$$ into $${x^2} + \frac{1}{3}{x^4}$$      M1

the series approximation gives a value of $$\frac{{13}}{{48}}$$

so $${\pi ^2} \simeq \frac{{13}}{{48}} \times 36$$

$$\simeq 9.75\,\,\left( { \simeq \frac{{39}}{4}} \right)$$     A1

Note: Accept 9.76.

[2 marks]

d.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.