Truncating Series Before and After Operations

Why the Order of Truncation Matters

A power series is often replaced by a polynomial when only its first few terms are needed. But truncating before an operation is not always the same as performing the operation first and then truncating. Multiplication combines coefficients from both factors, while differentiation shifts each term down by one degree.

The key question is not simply whether a series has been truncated. It is: through what degree are the resulting coefficients still exact? Keeping track of the indices gives a precise answer.

Truncation and Exact Coefficients

Let \(f(x)=\sum_{n=0}^{\infty}a_nx^n\) be a power series. Its degree-\(N\) truncation is the polynomial formed from the terms through \(x^N\).

Definition: For an integer \(N\geq 0\), the degree-\(N\) truncation of \(f\) is \(T_Nf(x)=\sum_{n=0}^{N}a_nx^n\).

To say that a result is exact through degree \(N\) means that its coefficients of \(x^0,x^1,\ldots,x^N\) agree with those of the full power series. It does not mean that all higher-degree terms are correct, or that the truncated polynomial equals the full function.

When we use power series as functions, the operations below are understood at points inside the relevant radius of convergence, where the power-series operations are valid. The coefficient calculations themselves follow directly from the series formulas.

Multiplication: Truncating Both Factors

Suppose \(f(x)=\sum_{n=0}^{\infty}a_nx^n\) and \(g(x)=\sum_{n=0}^{\infty}b_nx^n\). Their product has coefficients \(c_n=\sum_{k=0}^{n}a_kb_{n-k}\). Each coefficient of degree \(n\) uses only coefficients of the two factors up to degree \(n\).

f(x)g(x)=\sum_{n=0}^{\infty}c_nx^n, \qquad c_n=\sum_{k=0}^{n}a_kb_{n-k}.

Consequently, truncating both factors at degree \(N\) preserves every product coefficient through degree \(N\). Terms of degree greater than \(N\) in either factor cannot contribute to a product term of degree at most \(N\): all powers in the other factor are nonnegative.

Theorem: For power series \(f\) and \(g\), the coefficients of \(T_Nf\,T_Ng\) through degree \(N\) equal the coefficients of \(fg\) through degree \(N\). In general, coefficients above degree \(N\) need not agree.

This result compares the two procedures: multiply the full series and then truncate at degree \(N\), or truncate both factors at degree \(N\) and multiply. Their results agree through degree \(N\); the product of the truncated factors may also contain higher-degree terms, but those terms are not necessarily the correct ones.

Worked Example: Multiplying Truncated Factors

Let \(f(x)=1+2x+3x^2+4x^3+\cdots\) and \(g(x)=2-x+5x^2+6x^3+\cdots\). Truncate each factor at degree \(2\), then find the product coefficients through degree \(2\).

We have \(T_2f=1+2x+3x^2\) and \(T_2g=2-x+5x^2\). The coefficients are calculated by convolution:

\begin{aligned} c_0&=1\cdot 2=2,\\ c_1&=1\cdot(-1)+2\cdot2=-1+4=3,\\ c_2&=1\cdot5+2\cdot(-1)+3\cdot2=5-2+6=9. \end{aligned}

Thus the product through degree \(2\) is \(2+3x+9x^2\). These coefficients are exact for the full product: the formula for \(c_n\) uses only factor coefficients through degree \(n\). As a direct check, multiplying the two truncated polynomials gives \(2+3x+9x^2+7x^3+15x^4\); its coefficients through degree \(2\) match the convolution calculations. The terms \(7x^3+15x^4\) are not guaranteed to match the full product.

Differentiation: The Degree Shifts

Termwise differentiation changes the power \(x^n\) to \(nx^{n-1}\) when \(n\geq 1\). Thus a coefficient of \(x^n\) in the original series becomes the coefficient of \(x^{n-1}\) in its derivative. To know the derivative through degree \(N\), we need the original series through degree \(N+1\).

f'(x)=\sum_{n=1}^{\infty}na_nx^{n-1} =\sum_{m=0}^{\infty}(m+1)a_{m+1}x^m.

The second expression relabels the index using \(m=n-1\). It makes the required input coefficients explicit: the derivative coefficient of \(x^m\) is \((m+1)a_{m+1}\).

Theorem: For a power series that can be differentiated term by term, \(D(T_Nf)=T_{N-1}(f')\) when \(N\geq1\), where \(D\) denotes differentiation. Therefore, differentiating an input truncation through degree \(N\) gives the exact derivative through degree \(N-1\). To obtain the derivative through degree \(N\), truncate the original series through degree \(N+1\).

This is different from simply truncating the derivative at degree \(N\). The coefficient of \(x^N\) in \(f'\) depends on \(a_{N+1}\), which is absent from \(T_Nf\). The one-degree increase in the required input is essential.

Worked Example: Differentiating a Truncation

Consider \(f(x)=1+2x+3x^2+4x^3+5x^4+\cdots\). First truncate at degree \(3\), then differentiate.

The degree-\(3\) truncation and its derivative are

T_3f(x)=1+2x+3x^2+4x^3, \qquad D(T_3f)=2+6x+12x^2.

The full derivative begins \(f'(x)=2+6x+12x^2+20x^3+\cdots\). The coefficients of degrees \(0,1,2\) agree, as predicted: the input was retained through degree \(3\), so the derivative is exact through degree \(2\). But the degree-\(3\) derivative coefficient is \(20\), coming from differentiating \(5x^4\). It cannot be recovered from \(T_3f\). To obtain the derivative through degree \(3\), retain \(f\) through degree \(4\); differentiating \(T_4f\) gives \(2+6x+12x^2+20x^3\), which matches through that degree.

As a check, differentiating \(5x^4\) gives \(4\cdot5x^3=20x^3\), exactly the missing term.

A Short Justification by Coefficients

The rules for multiplication and differentiation can be justified by looking directly at a fixed output degree. This avoids relying on a visual impression of which terms “look small.”

For a product coefficient of degree \(n\).
The convolution formula is \(c_n=\sum_{k=0}^{n}a_kb_{n-k}\). Every index \(k\) and \(n-k\) lies between \(0\) and \(n\). Therefore, if both factors are kept through degree \(N\), then every coefficient used in \(c_n\) is retained whenever \(n\leq N\).

For a derivative coefficient of degree \(m\).
From \(f'(x)=\sum_{m=0}^{\infty}(m+1)a_{m+1}x^m\), the coefficient of \(x^m\) depends on \(a_{m+1}\). Hence the input must be retained through degree \(m+1\) to preserve that derivative coefficient.

Conclusion for a prescribed cutoff.
To compute a product through degree \(N\), retain each factor through degree \(N\). To compute a derivative through degree \(N\), retain the original series through degree \(N+1\). These bounds follow from the indices in the exact coefficient formulas.

Products Followed by Differentiation

Combining the two rules is useful when the desired result is the derivative of a product. To find \((fg)'\) through degree \(N\), first we need \(fg\) through degree \(N+1\), because differentiation lowers the product degree by one. For that product, each factor must be available through degree \(N+1\).

Takeaway: To determine \((fg)'\) through degree \(N\), retain both \(f\) and \(g\) through degree \(N+1\), multiply, and then differentiate. Retaining them only through degree \(N\) generally gives the derivative only through degree \(N-1\).

Worked Example: Differentiating a Product to a Prescribed Degree

Let \(f(x)=1+x+2x^2+3x^3+\cdots\) and \(g(x)=2-x+x^2+4x^3+\cdots\). Find \((fg)'\) through degree \(2\).

A derivative through degree \(2\) requires the product through degree \(3\), so both factors must be retained through degree \(3\). Using the convolution formula:

\begin{aligned} c_0&=1\cdot2=2,\\ c_1&=1\cdot(-1)+1\cdot2=-1+2=1,\\ c_2&=1\cdot1+1\cdot(-1)+2\cdot2=1-1+4=4,\\ c_3&=1\cdot4+1\cdot1+2\cdot(-1)+3\cdot2=4+1-2+6=9. \end{aligned}

Thus \(fg\) begins \(2+x+4x^2+9x^3+\cdots\), and differentiating these terms gives

(fg)'(x)=1+8x+27x^2+\cdots.

The answer is exact through degree \(2\), because the product coefficients needed through degree \(3\) were computed from factor coefficients through degree \(3\). As a check, differentiating \(x+4x^2+9x^3\) gives \(1+8x+27x^2\), the stated result.

A Common Truncation Mistake

A common error is to use the same cutoff for the input and the derivative output. For the series in the differentiation example, one might claim that retaining terms through \(x^3\) gives the derivative through \(x^3\). The calculation shows why this is false:

D(1+2x+3x^2+4x^3)=2+6x+12x^2, \qquad [x^3]f'=4a_4=20.

Here \([x^3]f'\) denotes the coefficient of \(x^3\) in \(f'\). The truncated input contains no \(a_4\), so the coefficient \(20\) cannot appear. The correction is to keep one additional input term: \(a_4x^4\) differentiates to \(4a_4x^3\).

For products, the corresponding mistake is to assume that every term in the product of two degree-\(N\) truncations is exact. Only coefficients through degree \(N\) are guaranteed. For example, to get the product coefficient of degree \(N+1\), the convolution may use a factor coefficient of degree \(N+1\), which was discarded.

Practical Coefficient Checklist

Before truncating, specify the highest degree of the output that must be correct. Then trace backward through the operation:

For a product through degree \(N\), keep each factor through degree \(N\).
For a derivative through degree \(N\), keep the original series through degree \(N+1\).
For a product derivative through degree \(N\), keep each factor through degree \(N+1\).
Terms above the guaranteed range may be present in a polynomial calculation, but should not be treated as exact without more input coefficients.

These statements concern coefficient agreement, not an estimate of the numerical error at a particular \(x\). The key discipline is to distinguish a polynomial’s computed terms from the degrees that the available series data actually determine.

Check Your Understanding

For each question, identify the highest input degree needed or the range of exact output coefficients.

If both factors are known through degree \(5\), through what degree is their product guaranteed to be exact?
If a power series is retained through degree \(6\), through what degree is its derivative guaranteed to be exact?
What original coefficient determines the coefficient of \(x^4\) in \(f'(x)\)?
To compute \((fg)'\) through degree \(3\), through what degree must each factor be retained?
Why can the coefficient of degree \(N+1\) in a product fail to be exact when both factors were truncated at degree \(N\)?