Composite and implicit series
Combine series through a chosen order, derive local coefficients by repeated differentiation, and verify implicit approximations.
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Choose the final order before multiplying
A coefficient can receive contributions from several pairs of terms. To expand e^x cos 2x through x³, use
The quadratic coefficient is 1/2-2 = -3/2. The cubic coefficient combines (x³/6) · 1 with x · (-2x²), giving 1/6-2 = -11/6. Hence
Multiplying only terms with the same degree would miss those mixed contributions. Retain every pair whose degrees add to a requested power. Higher-degree terms cannot contribute to these lower-degree coefficients in this product.
Check your understanding 1 Include the mixed contribution
Find terms through x³ in e^2x cos x. Show which terms contribute to the cubic coefficient, and explain why a term of degree four in either factor is unnecessary for this product.
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Use (1 + 2x + 2x² + (4/3)x³ + …)(1-x²/2 + …). The result is 1 + 2x + (3/2)x² + (1/3)x³ + …. The cubic coefficient is 4/3 + 2(-1/2) = 1/3. With only nonnegative powers in these factors, a fourth-degree term cannot contribute to a coefficient through degree three.
Substitution also changes powers and convergence conditions. For a composition such as ln(1 + 2x), transform the input in each retained term and in the original logarithm-series interval. If several expansions are combined, use their common valid interval before making a series-equality claim.
Logarithm subtraction can expose a symmetry
For |x| < 1, both 1 + x and 1-x are positive, so
Substitute into the two standard series. The linear, cubic and fifth-degree terms add, while the even-degree terms cancel:
The function is odd: replacing x by -x reverses the logarithm’s sign. That is an independent check on the absence of even powers. At either endpoint one logarithm is undefined, so neither endpoint belongs to this combined interval.
Check your understanding 2 Transform every retained odd power
Find the first three nonzero terms of h(x) = ln [(1 + 2x)/(1-2x)] and its real convergence interval. Explain why replacing only the first x by 2x gives wrong higher coefficients. Is the real function defined at x = 0.6?
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Substitution throughout gives h(x) = 4x + (16/3)x³ + (64/5)x⁵ + …, with |2x| < 1, or -1/2 < x < 1/2. The cubic input contributes (2x)³, and the fifth contributes (2x)⁵; both change their coefficients.
At x = 0.6, the ratio (1 + 2x)/(1-2x) is negative, so its real logarithm is undefined. A finite polynomial evaluated there would not restore that missing function value.
Repeated differentiation gives the first few terms
For y = sec x near zero, y(0) = 1 and y' = y tan x, so y'(0) = 0. Differentiating again and using tan² x = sec² x-1 gives
Thus y''(0) = 1. Two more differentiations give
At zero these are 0 and 5. Dividing the derivative values by their factorials yields
locally. The odd powers vanish consistently with the function being even. These few derivatives determine the displayed terms; they do not establish a convergence interval merely by being finite.
Coefficient matching gives a different check. Write sec x = 1 + ax² + bx⁴ + … and multiply by cos x = 1-x²/2 + x⁴/24 + …. Since the product is one, its quadratic and fourth-degree coefficients must be zero:
These give a = 1/2 and b = 5/24, agreeing with repeated differentiation. This check uses the original reciprocal relationship, rather than just comparing the appearance of coefficients.
Check your understanding 3 Turn derivative values into polynomial coefficients
For y = ln(cos x) near zero, start with y' = - tan x and use further differentiation to find its first two nonzero Maclaurin terms. State why the real function requires a positive cosine and check your coefficient factorials.
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We have y'' = - sec² x, y⁽³⁾ = -2 sec² x tan x and y⁽⁴⁾ = -4 sec² x tan² x-2 sec⁴ x. At zero, y = 0, y' = 0, y'' = -1, y⁽³⁾ = 0 and y⁽⁴⁾ = -2. Thus
near zero. The fourth-degree coefficient is -2/4! = -1/12, not -2. The logarithm needs cos x > 0, which holds on -π/2 < x < π/2 around zero. This real-domain statement does not by itself prove a full series convergence interval.
An implicit relationship supplies derivative values
Suppose y³ + y² + y = x²-2x on the branch through (0,0). Differentiating the entire relationship gives
so y'(0) = -2. Differentiating again,
At (0,0) this gives 2(-2)² + y''(0) = 2, hence y''(0) = -6. A third differentiation gives
Therefore 6(-2)³ + 3(2)(-2)(-6) + y⁽³⁾(0) = 0, giving y⁽³⁾(0) = -24. The resulting local expansion is
The branch point supplies the starting y-value, and the derivative coefficient there is nonzero, so each displayed derivative can be obtained. Divide by the factorials before writing the polynomial. Substituting -2x-3x²-4x³ into the original equation matches the coefficients through x³ but leaves a term -11x⁴ and higher powers: it is not the exact curve.
Try it yourself 1
Approximate an implicitly defined value
Problem
A local branch satisfies (1 + y)e^y = 1 + x and passes through (0,0). Find its Maclaurin terms through x³ by repeated implicit differentiation. Verify the coefficients by substituting a cubic polynomial into the original relationship and using the exponential series. Use your polynomial to estimate y at x = 0.1, then compare its left-hand side with 1.1. Explain what this check establishes about the approximation.
Write your answer, then compare it with the solution
Hints
Hint 1
Differentiate the whole y-dependent product
Every occurrence of y changes with x. After each differentiation, evaluate at the given point using the derivatives already found. For the coefficient check, decide which powers of y can affect terms through x³ near this branch point.
Show full solution
Differentiate through the implicit relationship
Method
Keep the product and chain-rule contributions together.Reason
The rate of change of both factors depends on y'.Working
Answer
The first derivative is e^y(y + 2)y' = 1, so y'(0) = 1/2. Differentiating again gives e^y(y + 3)(y')² + e^y(y + 2)y'' = 0. At zero, 3/4 + 2y''(0) = 0, hence y''(0) = -3/8.Obtain the cubic coefficient
Method
Differentiate the second-derivative relationship once more.Reason
The cubic coefficient needs the third derivative divided by 3!.Working
Answer
The third derivative equation is e^y(y + 4)(y')³ + 3e^y(y + 3)y'y'' + e^y(y + 2)y⁽³⁾ = 0. At zero it gives 1/2-27/16 + 2y⁽³⁾(0) = 0, so y⁽³⁾(0) = 19/32. Thus y = x/2-(3/16)x² + (19/192)x³ + … locally.Check coefficients in the original equation
Method
Expand the left-hand side in powers of y, then substitute y = ax + bx² + cx³.Reason
Agreement through the requested order checks the derivative calculation by another method.Working
Answer
Through cubic order, (1 + y)e^y = 1 + 2y + (3/2)y² + (2/3)y³ + …. Matching 1 + x gives 2a = 1, 2b + (3/2)a² = 0 and 2c + 3ab + (2/3)a³ = 0. These give a = 1/2, b = -3/16, c = 19/192, agreeing with the derivative method.Interpret the numerical estimate
Method
Evaluate the polynomial and test it in the original relationship.Reason
A small residual supports a local approximation without making the finite polynomial an exact solution.Working
Answer
At x = 0.1, the polynomial gives y ≈ 0.0482239583. Substitution gives (1 + y)e^y-1.1 ≈ 1.21 × 10⁻⁵ before rounding. The residual is nonzero, so the estimated point does not lie exactly on the implicit curve. Its small size is a consistency check on this local approximation. Matching three coefficients does not establish a global convergence interval or guarantee a requested number of correct decimal places.
Quick check
Compare the chosen branch, each chain-rule factor, factorials and the original-relationship check. Finite-order agreement is the claim supported here. A nearby numerical value or an algebraically matching polynomial does not prove that the truncation is exact.
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