The Central Idea
A power series centered at \(a\) has the form
\[ c_0+c_1(x-a)+c_2(x-a)^2+c_3(x-a)^3+\cdots. \]Taylor's idea is to choose the coefficients so that the series matches a function and all of its derivatives at \(x=a\).
Start with a General Power Series
Suppose
\[ f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n. \]At \(x=a\), every positive power of \(x-a\) vanishes.
Finding the Constant Coefficient
Substituting \(x=a\) gives
\[ f(a)=c_0. \]Therefore,
\[ \boxed{c_0=f(a)}. \]Differentiate Once
Differentiate the power series term by term:
\[ f'(x) = c_1+2c_2(x-a)+3c_3(x-a)^2+\cdots. \]At \(x=a\),
\[ f'(a)=c_1. \]Differentiate Twice
Differentiating again gives
\[ f''(x) = 2c_2+3\cdot2c_3(x-a)+\cdots. \]Therefore,
\[ f''(a)=2c_2, \]so
\[ \boxed{c_2=\frac{f''(a)}{2!}}. \]The General Coefficient Pattern
Repeating the differentiation process gives
\[ f^{(n)}(a)=n!c_n. \]Hence
\[ \boxed{ c_n=\frac{f^{(n)}(a)}{n!} }. \]The Taylor Series Formula
The Taylor series of \(f\) centered at \(a\) is
\[ \boxed{ \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n }. \]Expanded term by term:
\[ f(a) +f'(a)(x-a) +\frac{f''(a)}{2!}(x-a)^2 +\frac{f^{(3)}(a)}{3!}(x-a)^3 +\cdots. \]What Is a Maclaurin Series?
A Maclaurin series is simply a Taylor series centered at
\[ a=0. \]Therefore,
\[ \boxed{ \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n }. \]Taylor Polynomial
Stopping after degree \(N\) gives the \(N\)-th Taylor polynomial:
\[ \boxed{ T_N(x) = \sum_{n=0}^{N} \frac{f^{(n)}(a)}{n!}(x-a)^n }. \]The Polynomial Matches Derivatives
By construction,
\[ T_N(a)=f(a), \] \[ T_N'(a)=f'(a), \] \[ T_N''(a)=f''(a), \]and so on through the \(N\)-th derivative.
Each New Term Matches One More Derivative
A constant approximation matches only the function value.
A linear Taylor polynomial also matches the slope.
A quadratic Taylor polynomial additionally matches curvature information through the second derivative.
Linearization Is the First Taylor Approximation
The first-degree Taylor polynomial is
\[ T_1(x)=f(a)+f'(a)(x-a). \]This is exactly the familiar linearization formula.
Quadratic Approximation
The second-degree Taylor polynomial is
\[ T_2(x) = f(a)+f'(a)(x-a) +\frac{f''(a)}{2}(x-a)^2. \]It incorporates local curvature.
Cubic Approximation
The third-degree Taylor polynomial is
\[ T_3(x) = f(a)+f'(a)(x-a) +\frac{f''(a)}{2!}(x-a)^2 +\frac{f^{(3)}(a)}{3!}(x-a)^3. \]Why Factorials Appear
Differentiating
\[ (x-a)^n \]exactly \(n\) times produces
\[ n!. \]Dividing the coefficient by \(n!\) compensates for this multiplication.
The Exponential Function
Let
\[ f(x)=e^x. \]Every derivative is again \(e^x\):
\[ f^{(n)}(x)=e^x. \]Derivative Values at Zero for \(e^x\)
Since
\[ e^0=1, \]every derivative satisfies
\[ f^{(n)}(0)=1. \]Maclaurin Series for \(e^x\)
\[ \boxed{ e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots }. \]Finite Exponential Approximations
The first few Taylor polynomials are
\[ T_1(x)=1+x, \] \[ T_2(x)=1+x+\frac{x^2}{2}, \] \[ T_3(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}. \]Example: Approximate \(e^{0.1}\)
Using the cubic Maclaurin polynomial,
\[ e^{0.1} \approx 1+0.1+\frac{0.1^2}{2}+\frac{0.1^3}{6}. \]Therefore,
\[ e^{0.1}\approx1.105166\ldots. \]The Sine Function
Let
\[ f(x)=\sin x. \]Its derivatives cycle:
\[ \sin x,\quad \cos x,\quad -\sin x,\quad -\cos x,\quad \sin x,\ldots \]Sine Derivative Values at Zero
\[ \sin0=0, \] \[ \cos0=1, \] \[ -\sin0=0, \] \[ -\cos0=-1. \]Therefore only odd powers survive.
Maclaurin Series for \(\sin x\)
\[ \boxed{ \sin x = x-\frac{x^3}{3!} +\frac{x^5}{5!} -\frac{x^7}{7!} +\cdots }. \]In sigma notation,
\[ \sin x = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!}. \]The Cosine Function
Let
\[ f(x)=\cos x. \]Its derivative cycle is
\[ \cos x,\quad -\sin x,\quad -\cos x,\quad \sin x,\quad \cos x,\ldots \]Cosine Derivative Values at Zero
\[ \cos0=1, \] \[ -\sin0=0, \] \[ -\cos0=-1, \] \[ \sin0=0. \]Therefore only even powers survive.
Maclaurin Series for \(\cos x\)
\[ \boxed{ \cos x = 1-\frac{x^2}{2!} +\frac{x^4}{4!} -\frac{x^6}{6!} +\cdots }. \]In sigma notation,
\[ \cos x = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n}}{(2n)!}. \]Odd and Even Structure
The Maclaurin series mirrors symmetry:
- \(\sin x\) is odd, so its series contains only odd powers;
- \(\cos x\) is even, so its series contains only even powers.
Generating Series by Integration
Recall the geometric identity
\[ \frac1{1+x} = 1-x+x^2-x^3+\cdots, \qquad |x|<1. \]Integrating term by term gives a logarithmic series.
Maclaurin Series for \(\ln(1+x)\)
\[ \boxed{ \ln(1+x) = x-\frac{x^2}{2} +\frac{x^3}{3} -\frac{x^4}{4} +\cdots }. \]In sigma notation,
\[ \ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^n}{n}. \]Convergence for the Logarithmic Series
The geometric derivation initially gives
\[ |x|<1. \]Endpoint testing extends the interval to
\[ \boxed{(-1,1]}. \]At \(x=1\)
The logarithmic series becomes
\[ 1-\frac12+\frac13-\frac14+\cdots, \]which converges.
Its sum is
\[ \ln2. \]At \(x=-1\)
The logarithmic series becomes
\[ -1-\frac12-\frac13-\frac14-\cdots, \]which diverges.
Maclaurin Series for \(1/(1-x)\)
\[ \boxed{ \frac1{1-x} = 1+x+x^2+x^3+\cdots }. \]Although this series is geometric rather than derived through repeated differentiation here, it is also the Maclaurin series of \(1/(1-x)\).
Maclaurin Series for \(1/(1+x)\)
\[ \boxed{ \frac1{1+x} = 1-x+x^2-x^3+\cdots }. \]The Arctangent Series
Starting from
\[ \frac1{1+x^2} = 1-x^2+x^4-x^6+\cdots \]and integrating term by term gives
\[ \boxed{ \arctan x = x-\frac{x^3}{3} +\frac{x^5}{5} -\frac{x^7}{7} +\cdots }. \]A Famous Special Value
Since
\[ \arctan1=\frac{\pi}{4}, \]the arctangent series gives
\[ \frac{\pi}{4} = 1-\frac13+\frac15-\frac17+\cdots. \]Taylor Series About a Nonzero Center
Taylor series need not be centered at zero.
For center \(a\),
\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \]when the series actually represents \(f\).
Example: Taylor Series for \(e^x\) About \(a=1\)
Every derivative of \(e^x\) is \(e^x\), so
\[ f^{(n)}(1)=e. \]Therefore,
\[ \boxed{ e^x = e\sum_{n=0}^{\infty} \frac{(x-1)^n}{n!} }. \]The First Terms About \(a=1\)
\[ e^x = e \left[ 1+(x-1)+\frac{(x-1)^2}{2!} +\frac{(x-1)^3}{3!} +\cdots \right]. \]Choosing a Useful Center
A center near the point where you want to approximate a function often produces a more efficient local approximation.
This is one reason Taylor series are written about arbitrary centers.
Example: Approximate Near \(x=1\)
To approximate \(e^{1.02}\), a Taylor series centered at \(1\) uses the small quantity
\[ x-1=0.02. \]Small powers of \(0.02\) decrease rapidly.
Series Equality Is Not Automatic
A function may possess derivatives of every order and still require proof that its Taylor series actually converges to the function.
The formal coefficient construction alone does not guarantee equality.
The Taylor Series and the Function Are Distinct Objects
The expression
\[ \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \]is the Taylor series generated by \(f\).
Whether that series equals \(f(x)\) away from the center is a convergence-and-error question.
The Remainder
If
\[ T_N(x) \]is the degree-\(N\) Taylor polynomial, define the remainder by
\[ R_N(x)=f(x)-T_N(x). \]The next lesson develops systematic methods for estimating this quantity.
Function Representation Requires the Remainder to Vanish
For the Taylor series to represent \(f(x)\), we need
\[ \lim_{N\to\infty}R_N(x)=0. \]Then the polynomial approximations converge to the function.
Taylor Polynomials as Local Models
Near \(x=a\), the Taylor polynomial reproduces increasingly detailed local derivative information.
In many applications, a relatively low-degree polynomial already gives a useful approximation.
Why Polynomials Are Computationally Convenient
Polynomials are easy to:
- evaluate;
- differentiate;
- integrate;
- combine algebraically;
- use in numerical algorithms.
Taylor series exploit these advantages for more complicated functions.
Example: Quadratic Approximation to \(\cos x\)
The Maclaurin series begins
\[ \cos x = 1-\frac{x^2}{2!} +\frac{x^4}{4!} -\cdots. \]The quadratic approximation is
\[ \boxed{ \cos x\approx1-\frac{x^2}{2} }. \]Example: Cubic Approximation to \(\sin x\)
\[ \boxed{ \sin x \approx x-\frac{x^3}{6} }. \]This approximation is especially effective for \(x\) near zero.
Example: Quadratic Approximation to \(e^x\)
\[ \boxed{ e^x \approx 1+x+\frac{x^2}{2} }. \]Using Known Series Instead of Recomputing Derivatives
Once a standard Taylor series is known, substitution, multiplication, differentiation, and integration can often generate new expansions faster than repeated derivative calculation.
Example: Substitute \(x^2\) into \(e^x\)
Starting from
\[ e^u = \sum_{n=0}^{\infty}\frac{u^n}{n!}, \]set \(u=x^2\):
\[ \boxed{ e^{x^2} = \sum_{n=0}^{\infty}\frac{x^{2n}}{n!} }. \]Example: Substitute \(2x\) into \(\sin x\)
\[ \sin(2x) = 2x-\frac{(2x)^3}{3!} +\frac{(2x)^5}{5!} -\cdots. \]Example: Multiply a Known Series
Since
\[ e^x = 1+x+\frac{x^2}{2!}+\cdots, \]multiplying by \(x^2\) gives
\[ x^2e^x = x^2+x^3+\frac{x^4}{2!}+\cdots. \]Example: Differentiate a Known Series
Starting from
\[ \sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots, \]term-by-term differentiation gives
\[ \cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots. \]Example: Integrate a Known Series
Starting from
\[ \frac1{1+x} = 1-x+x^2-x^3+\cdots, \]integration gives
\[ \ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots. \]Coefficient Extraction
If
\[ f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n, \]then
\[ \boxed{ c_n=\frac{f^{(n)}(a)}{n!} }. \]Taylor coefficients encode derivative information.
Derivatives Can Be Recovered from Coefficients
The same identity can be read in reverse:
\[ \boxed{ f^{(n)}(a)=n!c_n }. \]Thus a known power series makes derivative values at its center immediately available.
Example: Read a Derivative from a Series
Suppose
\[ f(x)=3+2x-5x^2+7x^3+\cdots. \]Since the coefficient of \(x^3\) is \(7\),
\[ \frac{f^{(3)}(0)}{3!}=7. \]Therefore,
\[ f^{(3)}(0)=42. \]Common Error: Forgetting the Factorial
The coefficient of \((x-a)^n\) is not simply \(f^{(n)}(a)\).
It is
\[ \frac{f^{(n)}(a)}{n!}. \]Common Error: Confusing Taylor and Maclaurin
Every Maclaurin series is a Taylor series.
A Maclaurin series specifically uses center
\[ a=0. \]Common Error: Replacing \(x-a\) with \(x\)
For a Taylor series centered at \(a\ne0\), the powers must be
\[ (x-a)^n. \]Using \(x^n\) changes the center.
Common Error: Using the Wrong Derivative Value
Taylor coefficients use derivatives evaluated at the center:
\[ f^{(n)}(a), \]not the general derivative expression \(f^{(n)}(x)\).
Common Error: Dropping Zero Coefficients Carelessly
For functions such as sine and cosine, many derivative values at zero vanish.
Recognizing the zero pattern is useful, but the surviving powers must still be indexed correctly.
Common Error: Assuming the Series Always Equals the Function
Constructing the Taylor series does not by itself prove
\[ f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n. \]Equality requires convergence of the remainder to zero.
Common Error: Ignoring the Interval of Convergence
A Taylor or Maclaurin series is still a power series.
Its representation may be valid only within a particular convergence interval.
Common Error: Using Degrees Instead of Radians
The standard power series for
\[ \sin x \quad\text{and}\quad \cos x \]assume that \(x\) is measured in radians.
A Reliable Derivative-Based Procedure
- Choose the center \(a\).
- Compute \(f(a)\).
- Compute several derivatives.
- Evaluate each derivative at \(a\).
- Look for a repeating pattern.
- Use \(f^{(n)}(a)/n!\) as the coefficient.
- Attach the power \((x-a)^n\).
- Write several explicit terms before compressing into sigma notation.
A Reliable Known-Series Procedure
- Start from a standard power series.
- Identify the required substitution or algebraic transformation.
- Apply the transformation carefully to every term.
- Preserve the convergence condition.
- Recheck the first several terms against the target function.
Worked Example: Build a Maclaurin Polynomial
Find the degree-\(4\) Maclaurin polynomial for
\[ f(x)=e^x. \]Since all derivatives at zero equal \(1\),
\[ \boxed{ T_4(x) = 1+x+\frac{x^2}{2!} +\frac{x^3}{3!} +\frac{x^4}{4!} }. \]Worked Example: Degree-\(5\) Polynomial for \(\sin x\)
The Maclaurin expansion begins
\[ \sin x = x-\frac{x^3}{3!} +\frac{x^5}{5!} -\cdots. \]Therefore,
\[ \boxed{ T_5(x) = x-\frac{x^3}{6} +\frac{x^5}{120} }. \]Worked Example: Degree-\(4\) Polynomial for \(\cos x\)
\[ \boxed{ T_4(x) = 1-\frac{x^2}{2} +\frac{x^4}{24} }. \]Worked Example: Taylor Polynomial About \(a=1\)
Find the degree-\(2\) Taylor polynomial for \(e^x\) centered at \(1\).
Since
\[ f(1)=f'(1)=f''(1)=e, \]we obtain
\[ \boxed{ T_2(x) = e+e(x-1)+\frac{e}{2}(x-1)^2 }. \]Worked Example: Read Derivatives from Coefficients
Suppose
\[ f(x)=1-3x+4x^2-2x^3+\cdots. \]Then
\[ f(0)=1, \] \[ f'(0)=-3, \] \[ f''(0)=2!\cdot4=8, \] \[ f^{(3)}(0)=3!\cdot(-2)=-12. \]Worked Example: Build from a Known Series
Find a series for
\[ \frac1{1-x^2}. \]Start from
\[ \frac1{1-u} = \sum_{n=0}^{\infty}u^n. \]Set \(u=x^2\):
\[ \boxed{ \frac1{1-x^2} = \sum_{n=0}^{\infty}x^{2n} = 1+x^2+x^4+x^6+\cdots } \]for \(|x|<1\).
Practice 1: General Taylor Formula
Write the first four terms of the Taylor series for a function \(f\) centered at \(a\) in terms of \(f(a)\), \(f'(a)\), \(f''(a)\), and \(f^{(3)}(a)\).
Practice 2: Maclaurin Polynomial for \(e^x\)
Find the degree-\(3\) Maclaurin polynomial for
\[ e^x. \]Practice 3: Maclaurin Polynomial for \(\sin x\)
Find the degree-\(7\) Maclaurin polynomial for
\[ \sin x. \]Practice 4: Maclaurin Polynomial for \(\cos x\)
Find the degree-\(6\) Maclaurin polynomial for
\[ \cos x. \]Practice 5: Read a Derivative
Suppose
\[ f(x)=2+5x-3x^2+4x^3+\cdots. \]Find \(f^{(3)}(0)\).
Practice 6: Shifted Center
Find the degree-\(2\) Taylor polynomial for
\[ e^x \]centered at \(a=2\).
Practice 7: Use a Known Geometric Series
Express
\[ \frac1{1+3x} \]as a Maclaurin series and state the basic convergence condition.
Practice 8: Substitute into \(e^x\)
Use the Maclaurin series for \(e^x\) to write the first four nonzero terms of
\[ e^{-x^2}. \]Practice 9: Integrate a Known Series
Starting from
\[ \frac1{1+x} = 1-x+x^2-x^3+\cdots, \]write the first four nonzero terms of the Maclaurin series for
\[ \ln(1+x). \]Practice 10: Explain Representation
Explain why matching every derivative at the center does not, by itself, prove that a Taylor series equals the original function everywhere.
Practice Answers
Practice 1
\[ \boxed{ f(a) +f'(a)(x-a) +\frac{f''(a)}{2!}(x-a)^2 +\frac{f^{(3)}(a)}{3!}(x-a)^3 }. \]Practice 2
\[ \boxed{ T_3(x) = 1+x+\frac{x^2}{2} +\frac{x^3}{6} }. \]Practice 3
\[ \boxed{ T_7(x) = x-\frac{x^3}{3!} +\frac{x^5}{5!} -\frac{x^7}{7!} }. \]Practice 4
\[ \boxed{ T_6(x) = 1-\frac{x^2}{2!} +\frac{x^4}{4!} -\frac{x^6}{6!} }. \]Practice 5
The coefficient of \(x^3\) is \(4\), so
\[ \frac{f^{(3)}(0)}{3!}=4. \]Therefore,
\[ \boxed{ f^{(3)}(0)=24 }. \]Practice 6
Every derivative of \(e^x\) evaluated at \(2\) equals \(e^2\). Thus
\[ \boxed{ T_2(x) = e^2+e^2(x-2) +\frac{e^2}{2}(x-2)^2 }. \]Practice 7
Write
\[ \frac1{1+3x} = \frac1{1-(-3x)}. \]Therefore,
\[ \boxed{ \frac1{1+3x} = \sum_{n=0}^{\infty}(-3x)^n = 1-3x+9x^2-27x^3+\cdots } \]for
\[ \boxed{|x|<\frac13}. \]Practice 8
Substitute \(-x^2\) into the exponential series:
\[ \boxed{ e^{-x^2} = 1-x^2+\frac{x^4}{2!} -\frac{x^6}{3!} +\cdots }. \]Practice 9
\[ \boxed{ \ln(1+x) = x-\frac{x^2}{2} +\frac{x^3}{3} -\frac{x^4}{4} +\cdots }. \]Practice 10
The Taylor coefficients force agreement of the function and its derivatives at the center. To conclude equality away from the center, the Taylor remainder
\[ R_N(x)=f(x)-T_N(x) \]must approach zero as \(N\to\infty\). Without that additional convergence result, the formal Taylor series and the original function need not be identical away from the center.
What to Remember
The Taylor series of \(f\) about \(a\) is
\[ \boxed{ \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n }. \]The Maclaurin series is the special case \(a=0\):
\[ \boxed{ \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n }. \]- the coefficients encode derivative values;
- finite truncations are Taylor polynomials;
- linearization is the first-degree Taylor polynomial;
- standard series can generate new series by substitution, differentiation, integration, and algebra;
- matching derivatives does not alone guarantee that the infinite series equals the function.
Where We Go Next
Taylor polynomials provide increasingly detailed approximations, but an approximation is useful only when we understand its accuracy.
The next lesson, Error and Remainder Estimates, develops tools for bounding \[ R_N(x)=f(x)-T_N(x) \] and deciding how many terms are needed to reach a specified tolerance.