Why Small Index Errors Matter
A power-series calculation can look convincing while containing a one-step index shift, an omitted term, or a missing constant. These errors are consequential: they can change every coefficient after an operation, even when the underlying rule is familiar.
We will use a power series centered at \(c\), written as \(\sum_{n=0}^{\infty}a_n(x-c)^n\). Its radius of convergence is \(R\). Inside its interval of convergence, differentiation and integration can be performed term by term. A product can be calculated by the Cauchy product when both series converge absolutely; in particular, this applies inside the common interval of convergence. The errors in this tutorial arise from misapplying these rules or claiming more than they establish.
Shifting Indices After Differentiation
Termwise differentiation lowers each power by one. The term with \(n=0\) is constant, so its derivative is zero. The safest formula first leaves the surviving terms indexed from \(n=1\):
If the result must be written as a power series in powers \((x-c)^k\) starting at \(k=0\), reindex with \(k=n-1\). Then \(n=k+1\). The new coefficient is therefore \((k+1)a_{k+1}\), not \(ka_k\).
Worked Example: Reindexing a First Derivative
Differentiate \(f(x)=3+2(x-1)-4(x-1)^2+5(x-1)^3\), and express the result in powers of \(x-1\).
Set \(t=x-1\). Differentiating each term gives
Using the shifted coefficient rule gives the same terms: \(a_1=2\) contributes \(1a_1=2\); \(a_2=-4\) contributes \(2a_2=-8\); and \(a_3=5\) contributes \(3a_3=15\). The constant coefficient contributes nothing.
Check by differentiating the proposed answer: the derivative of \(2-8(x-1)+15(x-1)^2\) is \(-8+30(x-1)\), which is the second derivative of the original polynomial. More directly, the first derivative calculation above differentiates each original term, including the constant, and agrees with the reindexed formula.
A common wrong calculation is to write \(\sum_{k=0}^{\infty}ka_k(x-c)^k\) after differentiating. That expression uses the original coefficient index and the original power, rather than the index shift forced by the exponent \(n-1\). For example, it would assign zero as the constant term of the derivative, even though the original linear term \(a_1(x-c)\) differentiates to \(a_1\).
Convolution Limits in a Product
For two power series centered at the same point, write their terms as \(a_i(x-c)^i\) and \(b_j(x-c)^j\). A product contributes to the power \((x-c)^n\) exactly when the exponents satisfy \(i+j=n\). Thus \(i\) can only range from \(0\) through \(n\), and \(j=n-i\).
These coefficients describe the product within the common interval where the Cauchy product applies. A frequent error is to use an infinite sum for the coefficient of a fixed power, or to multiply only coefficients with matching indices. Neither operation accounts for all and only the pairs of exponents that add to \(n\).
Worked Example: Finding Product Coefficients
Multiply \(A(x)=1+2t+3t^2\) and \(B(x)=4-t+2t^2\), where \(t=x-c\), using convolution coefficients.
The constant coefficient is \(a_0b_0=1\cdot4=4\). The coefficient of \(t\) is \(a_0b_1+a_1b_0=1\cdot(-1)+2\cdot4=7\). For \(t^2\), the limits are \(k=0,1,2\):
For \(t^3\), the contributing pairs are \((0,3),(1,2),(2,1)\). Since \(b_3=0\), the coefficient is \(1\cdot0+2\cdot2+3\cdot(-1)=1\). For \(t^4\), only \((2,2)\) contributes, giving \(3\cdot2=6\). Therefore
Check by direct multiplication: \(1(4-t+2t^2)=4-t+2t^2\), \(2t(4-t+2t^2)=8t-2t^2+4t^3\), and \(3t^2(4-t+2t^2)=12t^2-3t^3+6t^4\). Adding gives \(4+7t+12t^2+t^3+6t^4\), as calculated.
Falling Factorials in Repeated Derivatives
Each differentiation multiplies by the current exponent and lowers the power by one. After three derivatives, a term with exponent \(n\geq3\) has the factor \(n(n-1)(n-2)\) and the power \((x-c)^{n-3}\). Terms with \(n=0,1,2\) vanish after three derivatives.
The factor is a falling factorial. It is not generally \(n!\): \(n!\) includes every positive integer from \(1\) through \(n\), whereas three derivatives supply only three descending factors.
Worked Example: Checking a Third Derivative
Find the third derivative of \(f(x)=5x^5\). Differentiating successively gives
The general factor gives the same result: \(5\cdot4\cdot3\cdot5x^{5-3}=300x^2\). Check by differentiating the answer twice in reverse order: an antiderivative step from \(300x^2\) gives \(100x^3\) up to a constant, and another gives \(25x^4\) up to a linear term, consistent with the successive derivatives of \(5x^5\).
A wrong calculation would use \(5!=120\) as the multiplier and report \(600x^2\). The correction is to count the three actual exponent factors, \(5\), \(4\), and \(3\); their product is \(60\), which multiplies the original coefficient \(5\) to give \(300\).
Integration Requires a Constant
Termwise integration raises each exponent by one and divides the coefficient by the new exponent. For a power series, the arbitrary constant must also be included:
This formula holds on the interval of convergence, and \(C\) is an arbitrary constant. Omitting it gives one antiderivative, not the full family of antiderivatives; if a value such as \(F(c)\) is specified, it determines \(C\).
Worked Example: Integrating and Using a Value
Find an antiderivative of \(g(x)=6(x-2)+4(x-2)^2-3\), and choose the constant so that \(F(2)=7\). Integrating the terms gives
At \(x=2\), every displayed power of \(x-2\) is zero, so \(F(2)=C\). The condition \(F(2)=7\) therefore gives \(C=7\). The requested antiderivative is
Check by differentiation: the four terms give \(6(x-2)\), \(4(x-2)^2\), \(-3\), and \(0\), respectively. Their sum is \(g(x)\), and substitution of \(x=2\) gives \(F(2)=7\).
Radius Is Not an Endpoint Guarantee
Termwise differentiation and integration preserve the radius of convergence of a power series. This concerns the radius, not automatic convergence at either boundary point. At a boundary point, the modified series must be tested separately. Moreover, the termwise differentiation theorem guarantees the derivative representation inside the open interval of convergence; it does not, by itself, settle what happens at an endpoint.
Worked Example: A Boundary Claim That Fails
Consider \(\sum_{n=1}^{\infty}\frac{x^n}{n^2}\), which has radius \(1\). At \(x=1\), it converges because \(\sum 1/n^2\) is a convergent \(p\)-series. Differentiating term by term for \(|x|<1\) gives
This differentiated series also has radius \(1\), but at \(x=1\) it becomes the harmonic series \(\sum 1/n\), which diverges. At \(x=-1\), it becomes \(\sum_{n=1}^{\infty}(-1)^{n-1}/n\), which converges by the alternating-series test, but not absolutely because \(\sum 1/n\) diverges. Thus the differentiated series has different endpoint behavior at its two boundary points.
The correction to the mistaken claim “the radius is unchanged, so the derivative series converges at every point where the original series converges” is to test each endpoint after the operation. Here the original series converges at \(x=1\), while its differentiated series diverges there. This calculation does not assert whether the original function has an endpoint derivative; that requires a separate argument.
A Reliable Error-Checking Routine
When an operation on a power series produces an unexpected result, work through the relevant checks rather than relying on visual pattern matching.
Keep its coefficient and exponent together as \(a_n(x-c)^n\).
For differentiation it is \(n-1\); for three derivatives it is \(n-3\); for integration it is \(n+1\).
When changing an index, change the coefficient, power, and bounds consistently.
Differentiate an antiderivative, multiply a product, or test convergence at each endpoint being discussed.
These checks matter because a correct formal calculation and a valid convergence claim are separate parts of an answer. A shifted coefficient can be wrong even when the radius is right; an antiderivative can differentiate correctly while having the wrong specified constant; and a power series can retain its radius while changing its endpoint behavior.
Check Your Understanding
For each question, identify the index or hypothesis that must be checked.
- After differentiating \(\sum_{n=0}^{\infty}a_n(x-c)^n\), what coefficient multiplies \((x-c)^k\) in the reindexed series?
- For the coefficient of \((x-c)^4\) in a product, what are the convolution limits?
- What factor multiplies \((x-c)^{n-3}\) after three derivatives when \(n\geq3\)?
- Why must an arbitrary constant be included when finding an indefinite integral?
- Does an unchanged radius of convergence determine convergence at a boundary point? Explain what must be checked instead.
