Tutorials › Calculus II › Taylor Polynomials of Sums and Scalar Multiples

Home  /  Tutorials  /  Taylor Polynomials

Taylor Polynomials · Tutorial 817 of 998

Taylor Polynomials of Sums and Scalar Multiples

Derive and apply the rules for adding Taylor polynomials and multiplying them by constants at a shared center.

Beginner 9 min read

What You'll Learn

  • Define a Taylor polynomial of degree at most n at a center a
  • Derive addition and scalar multiplication rules from derivative linearity
  • Combine known Taylor formulas at zero and at a nonzero center
  • Verify combined polynomials using derivative data
  • Recognize why matching centers and degrees matters

Why Linearity Helps

Taylor polynomials are built from a function’s values and derivatives at a chosen center. When a function is a sum of simpler functions, or a constant multiple of one, it is often more efficient to combine the simpler Taylor polynomials than to calculate every derivative of the whole expression from scratch.

The rules are straightforward, but they depend on using the same center and the same degree. With those choices fixed, addition and constant multiplication act on Taylor polynomials just as they act on functions.

Definition at a Fixed Center

Let \(f\) have derivatives through order \(n\) at a real number \(a\). Its Taylor polynomial of degree at most \(n\) centered at \(a\) is

$$ T_{n,a}f(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k. $$

Here \(f^{(0)}(a)\) means \(f(a)\), and \(0!=1\). The phrase “degree at most \(n\)” matters: some of the highest coefficients may be zero, so the polynomial’s actual degree can be less than \(n\).

The center \(a\) determines where the derivatives are evaluated and which powers appear. The index \(n\) determines how many derivative terms are included. To combine Taylor polynomials by the rules in this tutorial, both must match.

Definition: For functions \(f\) and \(g\) with derivatives through order \(n\) at the same center \(a\), their Taylor polynomials \(T_{n,a}f\) and \(T_{n,a}g\) use the same powers \((x-a)^k\), for \(0\leq k\leq n\).

The Addition and Scalar Multiplication Rules

For any real constant \(c\), and functions with the required derivatives at \(a\), the rules are

$$ T_{n,a}(f+g)(x)=T_{n,a}f(x)+T_{n,a}g(x), \qquad T_{n,a}(cf)(x)=c\,T_{n,a}f(x). $$

These are exact identities between polynomials. They do not say that a Taylor polynomial is identical to its function; they say that forming the degree-\(n\) Taylor polynomial respects addition and constant multiplication.

Why the Rules Work

The reason is that derivatives are linear. Specifically, for each \(k\) from \(0\) through \(n\),

$$ (f+g)^{(k)}(a)=f^{(k)}(a)+g^{(k)}(a), \qquad (cf)^{(k)}(a)=c f^{(k)}(a). $$

Substituting these derivative identities into the definition proves both rules. The argument applies to every coefficient, including the constant term \(k=0\).

1
Write the Taylor polynomial of the sum.
Its coefficient of \((x-a)^k\) is \(\frac{(f+g)^{(k)}(a)}{k!}\).
2
Use linearity of derivatives.
Replace \((f+g)^{(k)}(a)\) by \(f^{(k)}(a)+g^{(k)}(a)\).
3
Separate the two finite sums.
The resulting terms are precisely \(T_{n,a}f(x)\) and \(T_{n,a}g(x)\).

In equations, for addition this gives

$$ \begin{aligned} T_{n,a}(f+g)(x) &=\sum_{k=0}^{n}\frac{(f+g)^{(k)}(a)}{k!}(x-a)^k\\ &=\sum_{k=0}^{n}\frac{f^{(k)}(a)+g^{(k)}(a)}{k!}(x-a)^k\\ &=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k +\sum_{k=0}^{n}\frac{g^{(k)}(a)}{k!}(x-a)^k\\ &=T_{n,a}f(x)+T_{n,a}g(x). \end{aligned} $$

For a constant multiple, the same substitution gives

$$ \begin{aligned} T_{n,a}(cf)(x) &=\sum_{k=0}^{n}\frac{(cf)^{(k)}(a)}{k!}(x-a)^k\\ &=\sum_{k=0}^{n}\frac{c f^{(k)}(a)}{k!}(x-a)^k\\ &=c\,T_{n,a}f(x). \end{aligned} $$

This proof requires the derivatives in question to exist at the common center. It does not require the functions to equal their Taylor polynomials or to have convergent infinite Taylor series.

Worked Example: Combining Maclaurin Polynomials

Worked Example: Combining Maclaurin Polynomials

Find the degree-four Maclaurin polynomial for \(F(x)=e^x+3\sin x\), using the known degree-four Maclaurin polynomials for the exponential and sine functions.

The center is \(a=0\). The known polynomials, truncated at degree four, are

$$ T_{4,0}e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}, \qquad T_{4,0}\sin x=x-\frac{x^3}{6}. $$

Apply addition and scalar multiplication, then collect like powers:

$$ \begin{aligned} T_{4,0}F(x) &=T_{4,0}e^x+3T_{4,0}\sin x\\ &=\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\right) +3\left(x-\frac{x^3}{6}\right)\\ &=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24} +3x-\frac{x^3}{2}\\ &=1+4x+\frac{x^2}{2}-\frac{x^3}{3}+\frac{x^4}{24}. \end{aligned} $$

Check the result from the derivatives of \(F\). At zero, the values of \(F,F',F'',F'''\), and \(F^{(4)}\) are respectively \(1,4,1,-2,1\). Dividing these by \(0!,1!,2!,3!,4!\) gives the coefficients \(1,4,\frac12,-\frac13,\frac1{24}\), exactly as in the polynomial above.

Worked Example: Adding at a Nonzero Center

Worked Example: Adding at a Nonzero Center

Find the degree-three Taylor polynomial of \(F(x)=\ln x+\frac{2}{x}\) centered at \(a=1\). Both functions are defined near \(1\), so their Taylor polynomials can be combined there.

Write \(h=x-1\) to make the center visible. The known degree-three polynomials at \(1\) are

$$ T_{3,1}\ln x=h-\frac{h^2}{2}+\frac{h^3}{3}, \qquad T_{3,1}\left(\frac1x\right)=1-h+h^2-h^3. $$

Multiply the second polynomial by \(2\), then add term by term:

$$ \begin{aligned} T_{3,1}F(x) &=\left(h-\frac{h^2}{2}+\frac{h^3}{3}\right) +2(1-h+h^2-h^3)\\ &=h-\frac{h^2}{2}+\frac{h^3}{3} +2-2h+2h^2-2h^3\\ &=2-h+\frac{3}{2}h^2-\frac{5}{3}h^3\\ &=2-(x-1)+\frac{3}{2}(x-1)^2-\frac{5}{3}(x-1)^3. \end{aligned} $$

For a check, evaluate the derivatives of \(F\) at \(1\):

$$ \begin{aligned} F(1)&=2,\\ F'(x)&=\frac1x-\frac{2}{x^2}, & F'(1)&=-1,\\ F''(x)&=-\frac1{x^2}+\frac{4}{x^3}, & F''(1)&=3,\\ F'''(x)&=\frac2{x^3}-\frac{12}{x^4}, & F'''(1)&=-10. \end{aligned} $$

The Taylor coefficients are \(2\), \(-1\), \(\frac{3}{2}\), and \(-\frac{10}{6}=-\frac{5}{3}\). These agree with the combined polynomial.

Worked Example: Scalar Multiples and Cancellation

Worked Example: Scalar Multiples and Cancellation

Find the degree-three Taylor polynomial of \(G(x)=4\cos x-2\sin x\) centered at \(a=\frac{\pi}{2}\).

Set \(h=x-\frac{\pi}{2}\). The degree-three Taylor polynomials at this center follow from the sine and cosine formulas:

$$ T_{3,\pi/2}\cos x=-h+\frac{h^3}{6}, \qquad T_{3,\pi/2}\sin x=1-\frac{h^2}{2}. $$

Now scale each polynomial and combine:

$$ \begin{aligned} T_{3,\pi/2}G(x) &=4\left(-h+\frac{h^3}{6}\right) -2\left(1-\frac{h^2}{2}\right)\\ &=-4h+\frac{4h^3}{6}-2+h^2\\ &=-2-4h+h^2+\frac{2}{3}h^3\\ &=-2-4\left(x-\frac{\pi}{2}\right) +\left(x-\frac{\pi}{2}\right)^2 +\frac{2}{3}\left(x-\frac{\pi}{2}\right)^3. \end{aligned} $$

To check, \(G(\frac{\pi}{2})=-2\), \(G'(\frac{\pi}{2})=-4\), \(G''(\frac{\pi}{2})=2\), and \(G'''(\frac{\pi}{2})=4\). Dividing by \(0!,1!,2!,3!\) gives \(-2,-4,1,\frac23\), matching the displayed coefficients. The zero coefficient of \(h^2\) in the cosine polynomial and the zero coefficient of \(h\) in the sine polynomial are retained implicitly; no extra terms are needed.

Conditions and Practical Use

For the rules to apply directly, the functions must be expanded about the same center and to the same order. If a formula is known to a higher degree than needed, discard terms above the target degree before combining. If it is known only to a lower degree, it does not supply the missing higher-order coefficients.

The addition rule also extends to any finite sum by repeated application. For real constants \(c_1,\ldots,c_m\) and functions \(f_1,\ldots,f_m\) with derivatives through order \(n\) at \(a\),

$$ T_{n,a}\left(\sum_{j=1}^{m}c_j f_j\right)(x) =\sum_{j=1}^{m}c_jT_{n,a}f_j(x). $$

This is especially useful when an expression is a linear combination of familiar functions. One can reuse their Taylor polynomials, multiply each by its constant, and combine matching powers of \((x-a)\). Coefficients may cancel, as they did in the examples, so a term present in one polynomial need not appear in the final result.

A Common Center Mistake

A polynomial written in powers of \(x-a\) is still an ordinary polynomial in \(x\), so it may be tempting to add Taylor polynomials even when their centers differ. The resulting polynomial is defined, but it need not be the Taylor polynomial of the sum at either center.

Common Mistake: Adding Polynomials from Different Centers

Consider \(f(x)=x\) and \(g(x)=x^2\), and seek a degree-one Taylor polynomial for \(f+g\) centered at \(0\). A mistaken approach uses \(T_{1,0}f=x\) and \(T_{1,1}g=1+2(x-1)=2x-1\), then adds them to obtain \(3x-1\). The second polynomial is centered at \(1\), not \(0\), so this use of the addition rule is invalid.

At the required common center \(0\), the polynomials are \(T_{1,0}f=x\) and \(T_{1,0}g=0\), since \(g(0)=0\) and \(g'(0)=0\). Thus the correct result is

$$ T_{1,0}(f+g)(x)=x+0=x. $$

Checking directly, \((f+g)(0)=0\) and \((f+g)'(0)=1\), so its degree-one Taylor polynomial at zero is indeed \(0+1x=x\).

Check Your Understanding

Use a common center and degree in each problem. When you finish, verify your coefficients from derivative values where practical.

  1. State the addition rule for degree-\(n\) Taylor polynomials at a center \(a\), including the condition needed to apply it.
  2. Explain why multiplying a function by a constant multiplies each Taylor coefficient by that constant.
  3. Combine \(T_{2,0}e^x=1+x+\frac{x^2}{2}\) and \(T_{2,0}\cos x=1-\frac{x^2}{2}\) to find the degree-two Maclaurin polynomial of \(e^x+\cos x\).
  4. Why is it not valid to apply the addition rule directly to two Taylor polynomials with different centers?
  5. If a function’s Taylor polynomial has degree at most \(n\), must its coefficient of \((x-a)^n\) be nonzero? Explain.

Previous

Next

Back to Tutorials