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    IB Math AAHL
    /
    Maclaurin
    /

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    Not your average video:

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    Exam Preparation: Complete unit reviews designed for final exam preparation with all key concepts covered systematically.

    Expert Teaching: High-quality instruction from Perplex co-founder James Mullen with clear explanations, worked examples, and exam tips.

    Maclaurin

    Video Reviews

    Watch comprehensive video reviews for Maclaurin, designed for final exam preparation. Each video includes integrated problems you can solve alongside detailed solutions.

    Not your average video:

    Interactive Problems: Solve problems alongside the video with step-by-step guidance and detailed solutions.

    Exam Preparation: Complete unit reviews designed for final exam preparation with all key concepts covered systematically.

    Expert Teaching: High-quality instruction from Perplex co-founder James Mullen with clear explanations, worked examples, and exam tips.

    Not your average video:

    Interactive Problems: Solve problems alongside the video with step-by-step guidance and detailed solutions.

    Exam Preparation: Complete unit reviews designed for final exam preparation with all key concepts covered systematically.

    Expert Teaching: High-quality instruction from Perplex co-founder James Mullen with clear explanations, worked examples, and exam tips.

    The video will automatically pause when it reaches a problem.

    Division of Maclaurin Series

    AHL 5.19

    We can perform "division" of Maclaurin series by assuming that the result of the division is a series, and working backwards to find it's coefficients. For example, let:

    xln(x+1)​=a0​+a1​x+a2​x2+⋯

    for x=0.


    Multiplying through by x and using the series expansion of ln(x+1):

    x−2x2​+3x3​−⋯=a0​x+a1​x2+a2​x3+⋯

    Equating coefficients we find a0​=1, a1​=−21​, and a2​=31​. So the series is:

    xln(x+1)​=1−2x​+3x2​+⋯

    Division of Maclaurin Series

    AHL 5.19

    We can perform "division" of Maclaurin series by assuming that the result of the division is a series, and working backwards to find it's coefficients. For example, let:

    xln(x+1)​=a0​+a1​x+a2​x2+⋯

    for x=0.


    Multiplying through by x and using the series expansion of ln(x+1):

    x−2x2​+3x3​−⋯=a0​x+a1​x2+a2​x3+⋯

    Equating coefficients we find a0​=1, a1​=−21​, and a2​=31​. So the series is:

    xln(x+1)​=1−2x​+3x2​+⋯