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Discussion questions · Section 2 of 2

Questions 7–12 in this collection

Recover a logarithmic series from a differential identity · Build a shifted-tangent expansion and measure its error · +4 more

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  1. 7

    Let y=ln⁡(1+sin⁡x)y=\ln(1+\sin x).

    (i) Show that eyd2ydx2=1−ey[(dydx)2+1]e^y\dfrac{d^2y}{dx^2}=1-e^y[(\dfrac{dy}{dx})^2+1].

    (ii) Find its Maclaurin series through x2x^2.

    (iii) Verify it using the standard series for sin⁡x\sin x and ln⁡(1+x)\ln(1+x).

    [2007/CJC/I/7]

    CJC · Discussion · Question 7

  2. 8

    Given y=tan⁡(π/4+x)y=\tan(\pi/4+x), show that d2ydx2=2ydydx\dfrac{d^2y}{dx^2}=2y\dfrac{dy}{dx}. Find its series through x4x^4. When x=0.1x=0.1, y=1.22304888y=1.22304888 correct to 8 decimal places. Verify that the fourth-degree estimate has error about 0.004%0.004\%.

    CJC · Discussion · Question 8

  3. 9

    The curve y=f(x)y=f(x) passes through (0,π/4)(0,\pi/4) and satisfies y′tan⁡y=ln⁡((ex+1)/2)y^\prime\tan y=\ln((e^x+1)/2). Find the first three nonzero Maclaurin terms of yy. Deduce the expansion of (cot⁡y)ln⁡((ex+1)/2)(\cot y)\ln((e^x+1)/2) through x2x^2.

    [2007/YJC/II/1]

    CJC · Discussion · Question 9

  4. 10

    Let y=e1/(x+1)y=e^{1/(x+1)}. Show that yy′′−(y′)2=2y2(x+1)3yy^{\prime\prime}-(y^\prime)^2=\dfrac{2y^2}{(x+1)^3}. Hence find its Maclaurin expansion through x2x^2, and deduce the expansion of e−x/(x+1)e^{-x/(x+1)} through x2x^2.

    [2007/HCI/I/8]

    CJC · Discussion · Question 10

  5. 11

    Let f(x)=asec⁡x+btan⁡xf(x)=a\sec x+b\tan x for real distinct positive constants a,ba,b.

    (i) Show that f′(x)=f(x)tan⁡x+bf^\prime(x)=f(x)\tan x+b.

    (ii) Hence find the first three nonzero Maclaurin terms in terms of a,ba,b.

    (iii) These terms agree with the first three terms of e2xcos⁡xe^{\sqrt2 x}\cos x for small xx. Using MF27, find the exact values of a,ba,b.

    [2022/RVHS/II/4(modified)]

    CJC · Discussion · Question 11

  6. 12

    In triangle ABCABC, AB=1AB=1, ∠BAC=θ\angle BAC=\theta radians and ∠ABC=3π/4\angle ABC=3\pi/4 radians (see diagram).

    (i) Show that AC=1cos⁡θ−sin⁡θAC=\dfrac1{\cos\theta-\sin\theta}.

    (ii) For sufficiently small θ\theta, show that AC≈1+aθ+bθ2AC\approx1+a\theta+b\theta^2, determining a,ba,b.

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    Triangle ABC: A is at the lower left, B above and to its right, and C at the lower right. AB has length 1, angle BAC is theta, and angle ABC is 3 pi/4 radians. The third angle is pi/4 minus theta.
    Triangle ABC: A is at the lower left, B above and to its right, and C at the lower right. AB has length 1, angle BAC is theta, and angle ABC is 3 pi/4 radians. The third angle is pi/4 minus theta.
    Triangle ABC: A is at the lower left, B above and to its right, and C at the lower right. AB has length 1, angle BAC is theta, and angle ABC is 3 pi/4 radians. The third angle is pi/4 minus theta.

    [N2012/I/4]

    CJC · Discussion · Question 12

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