Taylor Polynomials for Real Powers

Why Real Powers Have a Taylor Pattern

Taylor polynomials for powers are not limited to integer exponents. Near any positive center, the function $x^r$ is differentiable as often as needed for every real number $r$. Its successive derivatives follow a predictable pattern: each differentiation reduces the exponent by one and multiplies by the exponent that was present.

The same pattern gives Maclaurin polynomials for $(1+x)^r$. The center $x=0$ is valid for every real $r$, because $1+x$ remains positive for all $x$ sufficiently close to zero. These formulas provide a direct way to find coefficients without separately simplifying every derivative.

Falling Factorials and the Taylor Polynomial

For a real number $r$ and a nonnegative integer $k$, define the falling factorial by

r^{\underline{k}}= \begin{cases} 1, & k=0,\\ r(r-1)(r-2)\cdots(r-k+1), & k\ge 1. \end{cases}

For instance, $r^{\underline{1}}=r$, $r^{\underline{2}}=r(r-1)$, and $r^{\underline{3}}=r(r-1)(r-2)$. The exponent $r$ need not be an integer.

Definition: If $f$ has derivatives through order $n$ at $a$, its degree-$n$ Taylor polynomial centered at $a$ is $$ T_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k, $$ where $f^{(0)}=f$ and $0!=1$. Its coefficients are chosen so that $T_n^{(k)}(a)=f^{(k)}(a)$ for $0\le k\le n$.

For real powers, write $x^r=\exp(r\ln x)$ when $x>0$. This definition makes $x^r$ differentiable to every order on the positive real axis. In particular, the formula below applies at every center $a>0$. A real power $x^r$ with arbitrary real $r$ is not generally defined as a real-valued function for negative $x$, so the positive-center condition matters.

The Derivative Pattern

The first derivative is $\frac{d}{dx}x^r=rx^{r-1}$ for $x>0$. Repeating this rule gives

\frac{d^k}{dx^k}x^r = r(r-1)\cdots(r-k+1)x^{r-k} = r^{\underline{k}}x^{r-k}.

This remains valid when one of the factors is zero. For example, if $r$ is a nonnegative integer and $k>r$, then $r^{\underline{k}}=0$, as expected when differentiating a polynomial too many times.

Theorem: For $r\in\mathbb R$, $a>0$, and a nonnegative integer $n$, the degree-$n$ Taylor polynomial of $x^r$ centered at $a$ is $$ T_n(x)=\sum_{k=0}^{n}\frac{r^{\underline{k}}a^{r-k}}{k!}(x-a)^k. $$ For $r\in\mathbb R$, the degree-$n$ Maclaurin polynomial of $(1+x)^r$ is $$ T_n(x)=\sum_{k=0}^{n}\frac{r^{\underline{k}}}{k!}x^k. $$

The second formula follows from the first derivative pattern applied to $1+x$, or directly by differentiating: the $k$th derivative of $(1+x)^r$ is $r^{\underline{k}}(1+x)^{r-k}$, which at zero equals $r^{\underline{k}}$. The function is real-valued on the interval $x>-1$, which includes a neighborhood of the center zero.

Why the Formula Works

Here is a derivation of the pattern and the Taylor coefficients. The derivative formula is true for $k=0$, since $r^{\underline{0}}x^r=x^r$. Suppose it holds for some $k$. Differentiating that expression once gives

\frac{d}{dx}\left(r^{\underline{k}}x^{r-k}\right) = r^{\underline{k}}(r-k)x^{r-k-1} = r^{\underline{k+1}}x^{r-(k+1)}.

The last equality holds because the next falling-factorial product appends the factor $r-k$. Thus induction proves the derivative formula for every nonnegative integer $k$. Evaluating at $a>0$ gives $f^{(k)}(a)=r^{\underline{k}}a^{r-k}$. Substituting these derivative values into the definition of the Taylor polynomial gives the stated formula. For $(1+x)^r$, evaluating the corresponding derivative at zero replaces $(1+0)^{r-k}$ by $1$, giving its Maclaurin coefficients.

This derivation also explains why the factorial $k!$ must remain in the denominator: it is part of the Taylor-polynomial definition, not part of the repeated differentiation.

Worked Examples

Worked Example: A Fractional Power at a Positive Center

Find the degree-three Taylor polynomial of $f(x)=x^{3/2}$ centered at $a=4$. Since $a>0$, the function and all the derivatives needed here are defined at the center. Use $h=x-4$ to keep the powers of the displacement visible.

The falling-factorial values through order three are $1$, $3/2$, $(3/2)(1/2)=3/4$, and $(3/2)(1/2)(-1/2)=-3/8$. The derivatives are therefore

\begin{aligned} f(4)&=4^{3/2}=8,\\ f'(4)&=\frac32\,4^{1/2}=3,\\ f''(4)&=\frac34\,4^{-1/2}=\frac38,\\ f'''(4)&=-\frac38\,4^{-3/2}=-\frac{3}{64}. \end{aligned}

The degree-three polynomial is

\begin{aligned} T_3(x) &=f(4)+f'(4)h+\frac{f''(4)}{2!}h^2+\frac{f'''(4)}{3!}h^3\\ &=8+3h+\frac{3/8}{2}h^2+\frac{-3/64}{6}h^3\\ &=8+3(x-4)+\frac{3}{16}(x-4)^2-\frac{1}{128}(x-4)^3. \end{aligned}

Check by differentiating the polynomial: $T_3(4)=8$, $T_3'(x)=3+\frac38(x-4)-\frac{3}{128}(x-4)^2$, so $T_3'(4)=3$; $T_3''(x)=\frac38-\frac{3}{64}(x-4)$, so $T_3''(4)=3/8$; and $T_3'''(x)=-3/64$, so $T_3'''(4)=-3/64$. These are exactly the four derivative values used to construct it.

Worked Example: A Maclaurin Polynomial with a Fractional Exponent

Find the degree-three Maclaurin polynomial for $g(x)=(1+x)^{5/2}$. Its real-valued domain includes $x=0$ and an interval around it, since $1+x>0$ when $x>-1$.

The falling-factorial values are $1$, $5/2$, $(5/2)(3/2)=15/4$, and $(5/2)(3/2)(1/2)=15/8$. Hence

\begin{aligned} g(0)&=1,\\ g'(0)&=\frac52,\\ g''(0)&=\frac{15}{4},\\ g'''(0)&=\frac{15}{8}. \end{aligned}

Divide each derivative value by the factorial for its degree:

\begin{aligned} T_3(x) &=1+\frac52x+\frac{15/4}{2!}x^2+\frac{15/8}{3!}x^3\\ &=1+\frac52x+\frac{15}{8}x^2+\frac{5}{16}x^3. \end{aligned}

To check, differentiate: $T_3'(x)=\frac52+\frac{15}{4}x+\frac{15}{16}x^2$, $T_3''(x)=\frac{15}{4}+\frac{15}{8}x$, and $T_3'''(x)=\frac{15}{8}$. At zero these give $1$ for $T_3(0)$, $5/2$, $15/4$, and $15/8$ for the first three derivatives. They match the corresponding derivatives of $g$.

Worked Example: A Negative Exponent at Zero

Find the degree-three Maclaurin polynomial for $q(x)=(1+x)^{-3/2}$. The function is defined and differentiable near zero because $1+x>0$ whenever $x>-1$.

The falling-factorial values are

\begin{aligned} (-3/2)^{\underline{0}}&=1,\\ (-3/2)^{\underline{1}}&=-\frac32,\\ (-3/2)^{\underline{2}}&=\left(-\frac32\right)\left(-\frac52\right)=\frac{15}{4},\\ (-3/2)^{\underline{3}}&=\left(-\frac32\right)\left(-\frac52\right)\left(-\frac72\right)=-\frac{105}{8}. \end{aligned}

The coefficients are these values divided by $k!$. Thus

\begin{aligned} T_3(x) &=1-\frac32x+\frac{15/4}{2!}x^2+\frac{-105/8}{3!}x^3\\ &=1-\frac32x+\frac{15}{8}x^2-\frac{35}{16}x^3. \end{aligned}

Check the derivative data: $T_3(0)=1$; $T_3'(x)=-\frac32+\frac{15}{4}x-\frac{105}{16}x^2$, so $T_3'(0)=-3/2$; $T_3''(x)=\frac{15}{4}-\frac{105}{8}x$, so $T_3''(0)=15/4$; and $T_3'''(x)=-105/8$, so $T_3'''(0)=-105/8$. These agree with the derivatives of $q$ at zero.

Using the Formula at a Positive Center

At a general positive center $a$, the coefficient of $(x-a)^k$ contains $a^{r-k}$:

\text{coefficient of }(x-a)^k = \frac{r^{\underline{k}}a^{r-k}}{k!}.

This factor matters. The Maclaurin formula for $(1+x)^r$ has coefficients $r^{\underline{k}}/k!$ because the base $1+x$ equals $1$ at its center. It is not the general coefficient formula for $x^r$ centered at an arbitrary positive $a$. One can also write the positive-center polynomial using $h=x-a$, which makes each term's degree and coefficient especially clear.

A useful practical sequence is to identify the center and degree, list the falling-factorial factors through that degree, evaluate the required powers at the center, and then divide the $k$th derivative value by $k!$. Checking the derivatives of the resulting polynomial at the center catches arithmetic errors in the coefficients.

A Common Factorial Error

A frequent mistake is to use the $k$th derivative value itself as the coefficient of $(x-a)^k$. For example, for $g(x)=(1+x)^{5/2}$, the second derivative at zero is $15/4$. Writing the quadratic term as $(15/4)x^2$ is wrong: differentiating that term twice gives $15/2$, not the required $15/4$.

\frac{15}{4}x^2 \quad\Longrightarrow\quad \frac{d^2}{dx^2}\left(\frac{15}{4}x^2\right)=\frac{15}{2}.

The correction is to divide by $2!$:

\frac{g''(0)}{2!}x^2 = \frac{15/4}{2}x^2 = \frac{15}{8}x^2, \qquad \frac{d^2}{dx^2}\left(\frac{15}{8}x^2\right)=\frac{15}{4}.

The factorial is what converts derivative data into polynomial coefficients. Also keep the domain condition in view: the general real power $x^r$ uses a positive center, while $(1+x)^r$ at zero is valid because its base is positive in a neighborhood of zero.

What These Polynomials Tell Us

A Taylor polynomial is designed to match a function's value and a specified number of derivatives at one center. For a degree-$n$ polynomial, the matching conditions are $T_n^{(k)}(a)=f^{(k)}(a)$ for every $k$ from zero through $n$. This is the precise reason the falling-factorial formulas produce the desired coefficients.

The polynomial is a local model, not automatically an identity for the function. A noninteger power such as $(1+x)^{5/2}$ is generally not equal to its degree-three polynomial at every nearby $x$. The formula here constructs the Taylor polynomial from derivative data; by itself, it does not claim how large the approximation error is or where an infinite Taylor series converges.

Check Your Understanding

Use the falling-factorial pattern and the Taylor-polynomial definition to answer these questions.

Write $r^{\underline{4}}$ as a product, and state the value of $r^{\underline{0}}$.
What is the degree-two Taylor polynomial of $x^{1/2}$ centered at $a=1$?
Find the coefficient of $x^3$ in the degree-three Maclaurin polynomial for $(1+x)^{1/3}$.
Why is zero a valid center for $(1+x)^r$ for every real $r$, but not generally a valid center for $x^r$ when $r$ is noninteger?
For a Taylor polynomial centered at $a$, what derivative condition can you use to check its coefficient of $(x-a)^k$?