Writing Powers Relative to the Center

Why Write Powers Relative to a Center?

A power series is organized by powers of a displacement from a chosen point. Its standard form is \(\sum_{n=0}^{\infty}c_n(x-a)^n\), where \(a\) is the center and \(c_n\) is the coefficient of the \(n\)th power. The expression \(x-a\) measures how far \(x\) is from that center, with a sign.

Polynomials can be written in this same form. For instance, the polynomial \(x^2-4x+3\), when written relative to the center \(2\), becomes \((x-2)^2-1\). This is not a different polynomial; it is the same expression organized around a different point. The choice of center determines which powers appear as the natural building blocks.

This skill is useful because the coefficients are easiest to read when the expression is written in the intended form. It also makes the role of the center concrete: set \(h=x-a\), rewrite \(x\) as \(a+h\), and express the result in powers of \(h\). The calculations below focus on this algebraic rearrangement. Rewriting an expression does not, on its own, establish any convergence property for an infinite series.

Definition: To write a polynomial relative to a center \(a\) means to express it as a finite sum of powers of \(x-a\), such as \(b_0+b_1(x-a)+\cdots+b_d(x-a)^d\). The numbers \(b_0,\ldots,b_d\) are its coefficients relative to that center.

The Basic Substitution

The key is to use the displacement from the center as a temporary variable. Set \(h=x-a\). Then \(x=a+h\). Substitute \(a+h\) wherever \(x\) appears, expand the powers, and collect like powers of \(h\). Finally replace \(h\) by \(x-a\).

Identify the center.
For a specified center \(a\), the displacement is \(x-a\).

Substitute.
Write \(x=a+h\), so the displacement \(x-a\) is simply \(h\).

Expand and collect.
Use algebra to express the polynomial as a sum of powers \(1,h,h^2,\ldots\).

Return to the original variable.
Replace every \(h\) with \(x-a\), then check by expanding back.

The sign in the substitution matters. If the center is \(a=2\), then \(h=x-2\) and \(x=h+2\). If the center is \(a=-2\), then \(h=x-(-2)=x+2\) and \(x=h-2\). In each case, the rule is \(x=a+h\); the sign follows from the value of \(a\).

Worked Rewriting and Checking

Worked Example: Rewrite a Quadratic Around the Center 2

Write \(x^2-4x+3\) in powers of \(x-2\). Let \(h=x-2\), so \(x=h+2\). Substitute this expression for \(x\):

\begin{aligned} x^2-4x+3 &=(h+2)^2-4(h+2)+3\\ &=(h^2+4h+4)-4h-8+3\\ &=h^2+(4h-4h)+(4-8+3)\\ &=h^2-1. \end{aligned}

Since \(h=x-2\), the centered form is \((x-2)^2-1\). To check, expand the answer:

$$ (x-2)^2-1=x^2-4x+4-1=x^2-4x+3. $$

This matches the original polynomial. As an additional check at the center, \(x=2\) gives \(2^2-4(2)+3=4-8+3=-1\), and the centered form gives \((2-2)^2-1=0-1=-1\).

The constant term in a centered form is the value of the polynomial at its center. In the example, the constant term is \(-1\), which agrees with the value at \(x=2\). This observation follows directly for polynomials: when \(x=a\), every positive power of \(x-a\) is zero, leaving only the constant term. It is an algebraic fact about the finite expression, not a convergence claim about infinite series.

Worked Example: Rewrite a Cubic Around the Center \(-1\)

Write \(x^3+2x^2-x+4\) in powers of \(x+1\). Because the center is \(-1\), set \(h=x-(-1)=x+1\). Thus \(x=h-1\). Substitute:

\begin{aligned} x^3+2x^2-x+4 &=(h-1)^3+2(h-1)^2-(h-1)+4\\ &=(h^3-3h^2+3h-1)+2(h^2-2h+1)-h+1+4\\ &=h^3-3h^2+3h-1+2h^2-4h+2-h+1+4\\ &=h^3+(-3+2)h^2+(3-4-1)h+(-1+2+1+4)\\ &=h^3-h^2-2h+6. \end{aligned}

Replace \(h\) by \(x+1\). The result is \((x+1)^3-(x+1)^2-2(x+1)+6\). Check by expanding the centered expression:

\begin{aligned} (x+1)^3-(x+1)^2-2(x+1)+6 &=(x^3+3x^2+3x+1)-(x^2+2x+1)-2x-2+6\\ &=x^3+(3-1)x^2+(3-2-2)x+(1-1-2+6)\\ &=x^3+2x^2-x+4. \end{aligned}

The coefficients of the powers \(1\), \(x+1\), \((x+1)^2\), and \((x+1)^3\) are, respectively, \(6\), \(-2\), \(-1\), and \(1\). Their order is determined by the exponent of the displacement.

Worked Example: Rewrite a Quadratic Around the Center \(-2\)

Write \(2x^2+3x+1\) in powers of \(x+2\). Set \(h=x+2\), so \(x=h-2\). Substitute and combine all terms with the same power of \(h\):

\begin{aligned} 2x^2+3x+1 &=2(h-2)^2+3(h-2)+1\\ &=2(h^2-4h+4)+3h-6+1\\ &=2h^2-8h+8+3h-6+1\\ &=2h^2+(-8+3)h+(8-6+1)\\ &=2h^2-5h+3. \end{aligned}

Replacing \(h\) by \(x+2\) gives \(2(x+2)^2-5(x+2)+3\). To check by expansion:

\begin{aligned} 2(x+2)^2-5(x+2)+3 &=2(x^2+4x+4)-5x-10+3\\ &=2x^2+8x+8-5x-10+3\\ &=2x^2+(8-5)x+(8-10+3)\\ &=2x^2+3x+1. \end{aligned}

At the center \(x=-2\), the original expression gives \(2(4)+3(-2)+1=8-6+1=3\). The centered expression gives \(2(0)^2-5(0)+3=3\), as required.

Why the Binomial Theorem Gives the Coefficients

The substitution method works systematically because each original power \(x^j\) can be expanded using \(x=a+(x-a)\). The binomial theorem states that for a nonnegative integer \(j\), \((u+v)^j=\sum_{k=0}^{j}\binom{j}{k}u^{j-k}v^k\), where \(\binom{j}{k}\) is the binomial coefficient. Take \(u=a\) and \(v=x-a\). Then

x^j=(a+(x-a))^j =\sum_{k=0}^{j}\binom{j}{k}a^{j-k}(x-a)^k.

Now suppose \(P(x)=\sum_{j=0}^{d}p_jx^j\) is a polynomial of degree at most \(d\). Replace each \(x^j\) by the binomial expansion above. The contribution to the power \((x-a)^k\) from the term \(p_jx^j\) is \(p_j\binom{j}{k}a^{j-k}\), and it occurs only when \(j\geq k\). Collecting those contributions gives the formula below.

Result: For \(P(x)=\sum_{j=0}^{d}p_jx^j\), its expression in powers of \(x-a\) is \(P(x)=\sum_{k=0}^{d}b_k(x-a)^k\), where \(b_k=\sum_{j=k}^{d}p_j\binom{j}{k}a^{j-k}\). This follows by expanding each original power with the binomial theorem and collecting equal powers of \(x-a\).

This derivation also explains why a fixed center determines a definite list of coefficients. After substituting \(h=x-a\), the expression is an ordinary polynomial in \(h\). The coefficients of a polynomial in a variable are unique, so there cannot be two different coefficient lists for the same polynomial and the same center.

Common Sign and Centering Mistakes

A frequent error is to confuse the center \(a\) with the expression \(x+a\). The centered power is \(x-a\), not \(x+a\), unless the center itself is negative. For center \(2\), using \(x+2\) would mean centering at \(-2\), not at \(2\).

Another common mistake is to get the constant term wrong after completing a square. For example, the incorrect calculation

$$ x^2-4x+3=(x-2)^2+3 $$

fails because expanding the right-hand side gives \(x^2-4x+4+3=x^2-4x+7\), not the original polynomial. The correction is to compare constants: \((x-2)^2=x^2-4x+4\), so a constant of \(3\) requires subtracting \(1\), giving \(x^2-4x+3=(x-2)^2-1\). Expanding the proposed answer is a reliable way to catch this error.

Takeaway: For center \(a\), always set \(h=x-a\) and therefore \(x=a+h\). Expand, collect equal powers, and check by substituting \(h=x-a\) and expanding back.

How Centered Powers Reveal Series Structure

In a power series \(\sum_{n=0}^{\infty}c_n(x-a)^n\), the index \(n\) labels the power of the displacement from \(a\), and \(c_n\) is the coefficient attached to that power. Writing a polynomial in powers of \(x-a\) uses exactly the same organization, except that only finitely many coefficients are nonzero. For example, the cubic centered at \(-1\) above has coefficients \(c_0=6\), \(c_1=-2\), \(c_2=-1\), and \(c_3=1\).

Choosing a center changes the way coefficients are displayed, not the polynomial itself. A polynomial can be expressed in powers of \(x\), or in powers of \(x-a\) for any specified real number \(a\). The centered form makes explicit which constant, linear, quadratic, and higher-order displacement terms are present relative to that point. This is the structural feature shared with power series.

Keep the algebraic claim separate from claims about infinite sums. A finite polynomial can always be expanded and regrouped by the substitution and binomial theorem. An infinite series raises additional questions that depend on its terms and on convergence; those conclusions do not follow merely from changing the appearance of its powers.

Check Your Understanding

For each question, use the substitution \(h=x-a\) and check any proposed centered expression by expanding it back.

Rewrite \(x^2-6x+8\) in powers of \(x-3\).
Rewrite \(x^2+4x+1\) in powers of \(x+2\).
For center \(a=4\), what expression should replace \(x\) in terms of \(h=x-a\)?
Why does the constant coefficient in a polynomial written in powers of \(x-a\) equal its value at \(x=a\)?
Does rewriting a polynomial in powers of \(x-a\) by itself establish convergence of an infinite power series? Explain briefly.