Tutorials › Calculus II › Evaluating Series Operations at Interior Points

Home  /  Tutorials  /  Power Series Operations

Power Series Operations · Tutorial 770 of 998

Evaluating Series Operations at Interior Points

Use coefficient formulas to evaluate power-series operations at interior points, with the convergence conditions that justify each step.

Intermediate 10 min read

What You'll Learn

  • Identify when two power series can be combined by adding or multiplying their coefficients.
  • Compute product coefficients using the Cauchy convolution formula.
  • Evaluate a differentiated power series at an interior point using its derivative coefficients.
  • Evaluate an integral of a power series by integrating its terms.
  • Check operation results by evaluating the original functions or differentiating an antiderivative.

Why Evaluate Operations at Interior Points?

A power series represents a function wherever it converges. When two such functions are added, multiplied, differentiated, or integrated, the result can often be represented by another power series whose coefficients follow a precise rule. Evaluating at an interior point is where these coefficient rules can be used safely: the series converge absolutely there, and the corresponding operations agree with the operations on the represented functions.

Here, power series are centered at the same real number \(c\). The central restriction is that the point \(x\) lies strictly inside the relevant radius of convergence. At a boundary point, convergence and the validity of termwise operations require separate analysis.

Definition: A point \(x\) is interior to the interval of convergence of a power series centered at \(c\) with radius \(R\) if \(|x-c|<R\). At such a point, the power series converges absolutely.

Coefficient Rules at a Glance

Suppose two power series are centered at \(c\):

$$ f(x)=\sum_{n=0}^{\infty}a_n(x-c)^n, \qquad g(x)=\sum_{n=0}^{\infty}b_n(x-c)^n. $$

At points interior to both radii of convergence, their sum and product are represented by coefficient operations. Differentiation and integration also produce power series at interior points, with the coefficients adjusted by powers and indices.

$$ \begin{aligned} f(x)+g(x) &=\sum_{n=0}^{\infty}(a_n+b_n)(x-c)^n,\\[4pt] f(x)g(x) &=\sum_{n=0}^{\infty} \left(\sum_{k=0}^{n}a_kb_{n-k}\right)(x-c)^n,\\[4pt] f'(x) &=\sum_{n=1}^{\infty}n a_n(x-c)^{n-1},\\[4pt] \int_c^x f(t)\,dt &=\sum_{n=0}^{\infty}\frac{a_n}{n+1}(x-c)^{n+1}. \end{aligned} $$

The product coefficient \(\sum_{k=0}^{n}a_kb_{n-k}\) is called the Cauchy product coefficient or convolution coefficient. For a given power \((x-c)^n\), it adds every product of coefficients whose indices sum to \(n\).

These formulas are valid where the original series converge in their common interior. For the product, it is enough to take \(|x-c|\) smaller than both radii. Differentiation and integration are valid at points interior to the radius of the series being operated on. These conditions do not assert anything about convergence at endpoints.

Worked Example: Adding and Multiplying Two Polynomials

Worked Example: Adding and Multiplying Two Polynomials

Let \(f(x)=1+2x-x^2\) and \(g(x)=2-x+3x^2\). Find the coefficients of their sum and product, then evaluate both at \(x=1\).

For the sum, add coefficients of like powers. Including zero coefficients where needed gives:

$$ \begin{aligned} f(x)+g(x) &=(1+2)+(2-1)x+(-1+3)x^2\\ &=3+x+2x^2. \end{aligned} $$

Thus \((f+g)(1)=3+1+2=6\). Direct evaluation checks this: \(f(1)=1+2-1=2\), and \(g(1)=2-1+3=4\), so \(f(1)+g(1)=2+4=6\).

For the product, calculate each coefficient by convolution:

$$ \begin{aligned} [x^0]:\;&1\cdot2=2,\\ [x^1]:\;&1\cdot(-1)+2\cdot2=3,\\ [x^2]:\;&1\cdot3+2\cdot(-1)+(-1)\cdot2=-1,\\ [x^3]:\;&2\cdot3+(-1)\cdot(-1)=7,\\ [x^4]:\;&(-1)\cdot3=-3. \end{aligned} $$

Therefore \(f(x)g(x)=2+3x-x^2+7x^3-3x^4\), and its value at \(1\) is \(2+3-1+7-3=8\). Independently, \(f(1)g(1)=2\cdot4=8\), which verifies the coefficient calculation.

Why the Product Uses Convolution

The multiplication rule follows from multiplying the two series and grouping terms with the same power. The essential condition is absolute convergence at the chosen point: it permits the terms in the resulting double sum to be regrouped without changing its value.

Theorem: If both power series centered at \(c\) converge at \(x\) with \(|x-c|\) strictly inside both radii of convergence, then their product is represented there by the Cauchy product series. Its coefficient of \((x-c)^n\) is \(\sum_{k=0}^{n}a_kb_{n-k}\).

To justify this, set \(h=x-c\), \(u_k=a_kh^k\), and \(v_j=b_jh^j\). Since \(x\) is interior to both radii, \(\sum_{k=0}^{\infty}|u_k|\) and \(\sum_{j=0}^{\infty}|v_j|\) are finite. Consequently, the sum of the absolute values of all terms in the product array is finite:

$$ \sum_{k=0}^{\infty}\sum_{j=0}^{\infty}|u_kv_j| = \left(\sum_{k=0}^{\infty}|u_k|\right) \left(\sum_{j=0}^{\infty}|v_j|\right) <\infty. $$

Because the double sum is absolutely convergent, its terms can be grouped by the value \(n=k+j\). For each fixed \(n\), there are only \(n+1\) pairs of nonnegative indices with \(k+j=n\). Thus the grouped sum is

$$ \sum_{n=0}^{\infty}\sum_{k=0}^{n}u_kv_{n-k} = \sum_{n=0}^{\infty} \left(\sum_{k=0}^{n}a_kb_{n-k}\right)h^n. $$

This is the value of the original product, since the absolute convergence ensures that grouping has not changed the sum. The proof explains both the convolution formula and why its convergence condition matters.

Worked Example: Squaring an Infinite Series

Worked Example: Squaring an Infinite Series

Consider \(A(x)=\sum_{n=0}^{\infty}x^n/n!\). The ratio test gives infinite radius of convergence: for \(x\ne0\), the ratio of successive term magnitudes is \(|x|/(n+1)\), which tends to zero. Find the coefficients of \(A(x)^2\), and relate its value at \(x=1\) to values of \(A\).

The coefficient of \(x^n\) in the square is

$$ \sum_{k=0}^{n}\frac{1}{k!(n-k)!} = \frac{1}{n!}\sum_{k=0}^{n}\binom{n}{k} = \frac{2^n}{n!}. $$

Here the first equality uses \(\binom{n}{k}=n!/(k!(n-k)!)\), and the second uses the binomial theorem. The first few coefficients are \(1\) for \(n=0\), \(2\) for \(n=1\), \(2\) for \(n=2\), and \(4/3\) for \(n=3\). Hence

$$ A(x)^2=\sum_{n=0}^{\infty}\frac{2^n}{n!}x^n=A(2x). $$

At \(x=1\), the coefficient series gives \(A(1)^2=\sum_{n=0}^{\infty}2^n/n!=A(2)\). This is an exact evaluation in terms of the defining series \(A\). It is checked directly by the product rule: the product of the two convergent sums \(A(1)\) and \(A(1)\) is precisely their Cauchy product at \(1\), whose terms are \(2^n/n!\). No endpoint issue arises because both points are interior to the infinite radius.

Differentiation and Integration at a Point

For differentiation, the coefficient of \((x-c)^{n-1}\) becomes \(n a_n\): differentiating \((x-c)^n\) brings down the factor \(n\) and reduces the exponent by one. The termwise differentiation theorem guarantees that this operation represents \(f'\) at each point strictly inside the radius of convergence.

For integration from the center to an interior point, each term is integrated using \(\int_c^x(t-c)^n\,dt=(x-c)^{n+1}/(n+1)\). Termwise integration is valid on any closed interval from \(c\) to an interior point; the resulting antiderivative has the indicated coefficients. A general antiderivative may also include an arbitrary constant, but the definite integral from \(c\) fixes that constant by making the value at \(c\) zero.

Takeaway: To evaluate a derivative series at \(x_0\), use the derivative coefficients and powers of \(x_0-c\). To evaluate an integral from \(c\) to \(x_0\), divide each coefficient by its new exponent and use the power \((x_0-c)^{n+1}\). Both operations require \(x_0\) to be interior to the original radius.

Worked Example: Differentiating and Integrating at an Interior Point

Worked Example: Differentiating and Integrating at an Interior Point

Let \(f(x)=1+2x+3x^2+4x^3\), a power series centered at \(0\) with infinite radius. Find \(f'(1/2)\) from the derivative coefficients and \(\int_0^{1/2}f(t)\,dt\) from the integrated coefficients.

Differentiating term by term gives

$$ f'(x)=2+6x+12x^2. $$

At \(x=1/2\), the coefficient evaluation is \(2+6(1/2)+12(1/2)^2=2+3+3=8\). Directly differentiating the polynomial gives the same derivative, so \(f'(1/2)=8\).

Integrating each term from \(0\) to \(x\) gives \(x+x^2+x^3+x^4\). Therefore

$$ \int_0^{1/2}f(t)\,dt = \frac12+\frac14+\frac18+\frac1{16} = \frac{15}{16}. $$

To check, differentiate the proposed antiderivative: \(\frac{d}{dx}(x+x^2+x^3+x^4)=1+2x+3x^2+4x^3=f(x)\). Its value at \(0\) is zero, so its change from \(0\) to \(1/2\) is exactly the stated definite integral.

A Common Mistake: Forgetting the Derivative Factor

A frequent error is to evaluate a derivative series by substituting into the original series instead of first changing its coefficients. For the preceding polynomial, this incorrect calculation would give \(f'(1/2)=f(1/2)=1+1+3/4+1/2=13/4\). That is wrong: the original series evaluates \(f\), not \(f'\).

The correction is to differentiate each term before substituting. In particular, \(a_n(x-c)^n\) contributes \(n a_n(x-c)^{n-1}\), not \(a_n(x-c)^n\). Here that gives \(2+6(1/2)+12(1/2)^2=8\), consistent with direct differentiation.

Using the Rules Reliably

When evaluating an operation at an interior point, first identify the center and the radii of the series involved. Then apply the coefficient rule, and only after that substitute the point. Addition uses like powers; multiplication uses convolution; differentiation multiplies coefficients by their indices and lowers powers; integration divides by the new exponent and raises powers. Finally, check the result by evaluating the original functions, differentiating an antiderivative, or comparing with a direct calculation when one is available.

Interior convergence is not a technical decoration. It supplies the absolute convergence needed to regroup a product and the validity needed for termwise differentiation and integration. A point on the boundary may behave differently, so these interior formulas should not be extended to endpoints without an additional argument.

Check Your Understanding

Use the coefficient rules and state the convergence condition needed in each case.

  1. For two power series centered at \(c\), what coefficient gives the power \((x-c)^n\) in their product?
  2. Why does absolute convergence justify grouping product terms by their total exponent?
  3. If \(f(x)=\sum_{n=0}^{\infty}a_n(x-c)^n\), what is the coefficient attached to \((x-c)^{n-1}\) after differentiation?
  4. Write the power-series formula for \(\int_c^x f(t)\,dt\).
  5. Why do the interior-point formulas not automatically settle what happens at a boundary point?

Previous

Next

Back to Tutorials