Recover a series from a relation

Obtain coefficients from a differential relation

1. Record the given value at zero.
2. Differentiate the relation before substituting zero; retain every product-rule term.
3. Evaluate each new derivative at zero using values already found.
4. Divide by the matching factorial and stop at the requested power.

As an independent check, compare coefficients directly in the original relation when possible. Differentiating a truncated series lowers the known order by one.

To form a series through $x^r$ by this direct method, obtain derivatives through order r. If the equation already supplies the first derivative, fewer additional differentiations are needed. Known series or a recurrence may also provide the required coefficients.

Example 6 starts from $y^{\prime}=2xy-1$ and y(0)=1. The Teacher completes every intermediate product-rule line and zero value, then supplies the quartic approximation and 0.90938 at x=0.1. Factorials convert derivative values into coefficients. Rounding that polynomial value is distinct from bounding the omitted remainder.

Example 7 uses $y=\tan2x$ and integrates the shared callout into its final deduction: relate the requested $\sec^2(2x)$ to the derivative of the tangent series. Keep the chain-rule factor 2, then divide by it. Termwise differentiation applies within the power series’ convergence radius, rather than being an unrestricted operation on an arbitrary approximation.

Verify a series by a genuinely different route

Example 8 is marked Self-Reading and attributed to 2022/ACJC/I/7(modified). Both copies already print the full implicit-derivative solution and its MF27 sine/logarithm verification. Comparing their cubic coefficients checks the product-rule calculation independently. All four asks and marks 2, 3, 3, 2 are retained.

For its final quotient, read the denominator as $\sqrt{1-x}$, giving exponent $-1/2$ in the reciprocal binomial factor. Truncate both factors only after identifying the requested quadratic order, and retain the linear-times-linear contribution; it cancels the numerator’s non-logarithmic quadratic term. The final answer is $\ln2+(1+\tfrac12\ln2)x+(\tfrac38\ln2)x^2$.