Finding Taylor Polynomial Approximations of Functions
Introduced| English | Meaning |
|---|---|
| Taylor polynomial/ˈteɪlə ˌpɒlɪˈnəʊmɪəl/ | A polynomial approximation built from a function's value and successive derivatives at a chosen point. |
| derivative/dɪˈrɪvətɪv/ | The instantaneous rate of change of a function with respect to its input. |
| tangent line/ˈtændʒənt laɪn/ | A straight line through a point on a curve with the same gradient as the curve at that point. |
| coefficient/ˌkəʊɪˈfɪʃənt/ | A numerical or constant factor multiplying a term in an expression. |
| factorial/fækˈtɔːrɪəl/ | The product of the positive integers up to a given positive integer; zero factorial is defined as one. |
| Maclaurin polynomial/məˈklɔːrɪn ˌpɒlɪˈnəʊmɪəl/ | A Taylor polynomial centred at 0. |
Estimate a growth factor with a polynomial
- A growth model needs $e^{0.1}$. The short calculation $1+0.1+0.1^2/2=1.105$ gives an estimate using simple arithmetic.
- A Taylor polynomial 泰勒多项式 provides these terms by matching the function and its derivatives 导数 at a chosen center.
Match local information at the center
- At $x=a$, $P_n(a)=f(a)$ and $P_n^{(k)}(a)=f^{(k)}(a)$ for $1\le k\le n$, when these derivatives exist.
- Matching value and slope gives a tangent line 切线. Higher-order information adds terms that describe local bending and other changes.
Every derivative used to build a Taylor polynomial is evaluated at the ____ a.
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Derivative values at a become constant coefficients.
A Taylor polynomial through order 2 at a matches which values there?
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It matches derivatives through order 2, including order 0.
Turn derivative values into coefficients
- The coefficient 系数 of $(x-a)^k$ is $f^{(k)}(a)/k!$. Divide by the factorial 阶乘, not just by $k$.
- For example, at $a=1$, suppose the derivative values through order $3$ are $2,-3,8,12$. The coefficients are $2,-3,4,2$.
At a=1, $f''(1)=8$. What is the coefficient of $(x-1)^2$?
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Divide 8 by 2!=2 to get 4.
At a=1, $f^{(3)}(1)=12$. What is the coefficient of $(x-1)^3$?
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Divide 12 by 3!=6.
Keep the sum finite
- A Taylor polynomial through order $n$ includes $k=0,1,\ldots,n$. Its actual degree is at most $n$; a coefficient can be zero.
- Since $f^{(0)}(a)=f(a)$ and $0!=1$, the first term is the constant $f(a)$.
Evaluate derivatives at $a$ before forming the coefficients. Keep $(x-a)$ in every positive-power term. Replacing it with $x$ changes the center unless $a=0$.
Which is the kth Taylor term at center a?
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Evaluate the derivative at a, divide by k!, and use the shifted power.
What is $0!$?
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This makes the constant term f(a).
The kth Taylor coefficient is the kth derivative value divided by k!.
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The denominator is the factorial of the derivative order.
Which indices are included in $P_3$?
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Order 3 includes four possible powers, but some coefficients may be zero.
Maclaurin means center zero
- A Maclaurin polynomial 麦克劳林多项式 is a Taylor polynomial with $a=0$. Its powers are $x^k$.
- Every derivative of $e^x$ at $0$ equals $1$, so its polynomial through order $2$ is $P_2(x)=1+x+x^2/2!$.
For $e^x$ at center $0$, divide the derivative values $1,1,1$ by $0!,1!,2!$:
The coefficient of $x^2$ is $1/2$, not $1$.
A Maclaurin polynomial has center...
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It is the Taylor construction with a=0.
What is the Maclaurin polynomial for $e^x$ through order 2?
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The derivative values at 0 are 1,1,1; divide by 0!,1!,2!.
Evaluate the polynomial at the target point
- At $x=0.1$, $P_2(0.1)=1+0.1+0.01/2=1.105$. Write $e^{0.1}\approx1.105$, since this is an approximation.
- A numerical comparison gives $e^{0.1}\approx1.105170918$, so the error is about $0.000170918$. A guaranteed error bound needs a separate argument.
Function and Taylor polynomial
Build a finite polynomial from derivative values at the selected center. The default eˣ example uses order 2 about 0 and evaluates at 0.1, giving 1.105. Compare order with degree and inspect approximation errors.
Calculate $P_2(0.1)$ for $P_2(x)=1+x+x^2/2$.
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1+0.1+0.01/2=1.105.
The calculation $P_2(0.1)=1.105$ proves $e^{0.1}=1.105$ exactly.
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The polynomial value is exact for P₂; its use for the exponential is an approximation.
Use a center where the values are easy
- For $f(x)=\ln x$, choose $a=1$. The values are $f(1)=0$, $f'(1)=1$, $f''(1)=-1$ and $f'''(1)=2$.
- With $h=x-1$, this gives $P_3(x)=h-h^2/2+h^3/3$. At $x=1.1$, use $h=0.1$, giving $\ln(1.1)\approx0.0953333\ldots$.
For ln x at center 1, $f^{(3)}(1)=2$. What is the coefficient of $(x-1)^3$?
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Divide 2 by 3!=6 to get 1/3.
To approximate ln(1.1) using center 1, what value of h=x−1 is used?
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1.1−1=0.1.
A point left of the center gives a negative change
- The earlier derivative table gives $P_3(x)=2-3(x-1)+4(x-1)^2+2(x-1)^3$. At $x=0.9$, $h=-0.1$.
- So $P_3(0.9)=2+0.3+0.04-0.002=2.338$. Even powers make a negative $h$ positive; odd powers keep its sign.
The target is x=0.9 and the center is 1. What is x−a?
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Use the signed change, not its absolute distance, when evaluating powers.
Evaluate $2-3h+4h^2+2h^3$ at h=−0.1.
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2+0.3+0.04−0.002=2.338.
Zero coefficients separate order from degree
- For cosine about $0$, the constant, linear, quadratic and cubic coefficients are $1,0,-1/2,0$. Thus $P_2=P_3=1-x^2/2$.
- The polynomial has degree $2$, and it matches derivatives through order $3$. Three nonzero sine terms, by contrast, reach degree $5$.
A Taylor polynomial through order 3 must have degree 3.
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For cosine at 0, the cubic coefficient is 0 and P₃ has degree 2.
What is the degree of the cosine polynomial $P_3=1-x^2/2$?
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The highest nonzero power is 2, despite the Taylor order 3.
Check accuracy instead of assuming it
- Matching derivatives is local information. More Taylor terms often help, but they do not guarantee a smaller error at every chosen point.
- For $e^x$ at $x=-2$, $P_0=1$ has error about $0.8647$, while $P_1=1+x=-1$ has error about $1.1353$. The higher-order approximation is worse here.
Which statement is guaranteed by the Taylor construction?
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The polynomial agrees with the function at its center; universal monotonic error claims do not follow.
For $P_1(x)=1+x$, what is $P_1(-2)$?
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1−2=−1.
Construct a fresh polynomial
- Use a new table at center $-1$ to build and evaluate a cubic approximation. Keep the shifted input and each factorial.
- Then construct a fourth-order cosine approximation about $\pi$ and a third-order Maclaurin approximation for $e^{2x}$.
At a=−1, a function has $f(-1)=4$, $f'(-1)=2$, $f''(-1)=-6$, $f^{(3)}(-1)=24$. Build its Taylor polynomial through order 3 and use it to estimate f(−0.8).
Use h=x+1 and divide each derivative by its order factorial.
Construct the Taylor polynomial for cos x through order 4 about a=π. State its actual degree and use it to estimate cos(π+0.2).
At π the derivative values through order 4 are−1,0,1,0,−1.
Find the Maclaurin polynomial through order 3 for $e^{2x}$, then estimate $e^{0.2}$ using x=0.1.
The kth derivative at 0 equals 2ᵏ, not 1.
Verify the construction
- Check the value and derivatives at the center. State the Taylor order, actual degree and target point separately.
- Use an approximation sign for $f(x)\approx P_n(x)$. Equality at the center is guaranteed; accuracy at another point needs checking.
For a Taylor polynomial through order $n$, use $P_n(x)=\sum_{k=0}^{n}f^{(k)}(a)(x-a)^k/k!$. Maclaurin means $a=0$. Derivative values determine the coefficients, zero coefficients can lower the degree, and local matching is not a universal accuracy guarantee.
Review: if $f^{(4)}(a)=48$, what is the coefficient of $(x-a)^4$?
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48/4!=48/24=2.