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Calculus · real student question

Find the Maclaurin series of tanh(10⁻¹³x)/x up to and including the degree-10 term.

Question

Find the Maclaurin series of

tanh⁡ ⁣(10−13x)x\frac{\tanh\!\left(10^{-13}x\right)}{x}

up to and including the term of degree 1010.

Step-by-step solution

  1. Start from the standard expansion of tanh⁡z\tanh z. The hyperbolic tangent is odd, so its Maclaurin series contains only odd powers:

    tanh⁡z=z−z33+2z515−17z7315+62z92835−1382z11155925+⋯\tanh z=z-\frac{z^{3}}{3}+\frac{2z^{5}}{15}-\frac{17z^{7}}{315}+\frac{62z^{9}}{2835}-\frac{1382z^{11}}{155925}+\cdots

    The coefficients come from tanh⁡z=∑n≥122n(22n−1)B2n(2n)!z2n−1\tanh z=\sum_{n\ge1}\frac{2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}z^{2n-1}, but for a degree-10 answer the six listed terms are all that is needed.

  2. Decide how far to expand before substituting. Dividing by xx lowers every degree by one, so an odd power z2k+1z^{2k+1} becomes an even power x2kx^{2k}. To reach degree 1010 in the final answer we therefore need tanh⁡z\tanh z through z11z^{11} — one term further than a first glance suggests. This is the step most often cut short.

  3. Substitute z=10−13xz=10^{-13}x. Each zmz^{m} contributes a factor 10−13m10^{-13m}:

    tanh⁡(10−13x)=10−13x−10−393x3+2⋅10−6515x5−17⋅10−91315x7+62⋅10−1172835x9−1382⋅10−143155925x11+⋯\tanh(10^{-13}x)=10^{-13}x-\frac{10^{-39}}{3}x^{3}+\frac{2\cdot 10^{-65}}{15}x^{5}-\frac{17\cdot 10^{-91}}{315}x^{7}+\frac{62\cdot 10^{-117}}{2835}x^{9}-\frac{1382\cdot 10^{-143}}{155925}x^{11}+\cdots

  4. Divide by xx term by term. Every exponent drops by one, turning odd powers into even ones:

    tanh⁡(10−13x)x=10−13−10−393x2+2⋅10−6515x4−17⋅10−91315x6+62⋅10−1172835x8−1382⋅10−143155925x10+⋯\frac{\tanh(10^{-13}x)}{x}=10^{-13}-\frac{10^{-39}}{3}x^{2}+\frac{2\cdot 10^{-65}}{15}x^{4}-\frac{17\cdot 10^{-91}}{315}x^{6}+\frac{62\cdot 10^{-117}}{2835}x^{8}-\frac{1382\cdot 10^{-143}}{155925}x^{10}+\cdots

    The series has no odd-degree terms at all, and the removable singularity at x=0x=0 is filled by the constant 10−1310^{-13}.

  5. Read off what the tiny parameter does. Successive terms shrink by a factor of about 10−26x210^{-26}x^{2}, so unless ∣x∣|x| is of order 101310^{13} the function is numerically indistinguishable from the constant 10−1310^{-13}. That is the practical content of the expansion: the series is convergent for ∣10−13x∣<π/2|10^{-13}x|<\pi/2, i.e. ∣x∣<1.5708×1013|x|<1.5708\times 10^{13}, which is exactly where the nearest singularity of tanh⁡\tanh sits, at z=iπ/2z=i\pi/2.

Answer

10−13−10−393x2+2⋅10−6515x4−17⋅10−91315x6+62⋅10−1172835x8−1382⋅10−143155925x10+O(x12)10^{-13}-\frac{10^{-39}}{3}x^{2}+\frac{2\cdot 10^{-65}}{15}x^{4}-\frac{17\cdot 10^{-91}}{315}x^{6}+\frac{62\cdot 10^{-117}}{2835}x^{8}-\frac{1382\cdot 10^{-143}}{155925}x^{10}+O(x^{12})

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