Taylor Polynomials & Taylor Series
Section 6.3: how to work the HW 6.3 problems
1. The idea: match every derivative at one point
Suppose \(f\) equals some power series centered at \(a\):
\[ f(x) = c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + \cdots \]
Plug in \(x = a\) and every term but the first disappears, so \(c_0 = f(a)\). Differentiate, then plug in \(x = a\) again:
\[ f'(x) = c_1 + 2c_2(x-a) + 3c_3(x-a)^2 + \cdots \quad\Longrightarrow\quad c_1 = f'(a) \]
\[ f''(x) = 2c_2 + 3\cdot 2\,c_3(x-a) + \cdots \quad\Longrightarrow\quad c_2 = \frac{f''(a)}{2} \]
Each derivative peels off one more coefficient. The \(n\)th derivative leaves \(n!\,c_n\) in front, so
\[ c_n = \frac{f^{(n)}(a)}{n!}, \qquad f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n \]
The Taylor polynomial of degree \(N\), \(P_N(x)\), is the same sum stopped at \(n = N\). With \(a = 0\) it is called a Maclaurin series.
\(P_1\) is the tangent line. \(P_2\) also matches the concavity, so it bends with the curve.
2. Taylor polynomials of degree 2 (#1–4)
Method: make a table with columns \(n\), \(f^{(n)}(x)\), \(f^{(n)}(a)\), and \(\dfrac{f^{(n)}(a)}{n!}\). The last column gives the coefficients.
Example (polynomial). \(f(x) = 3 - 2x + x^2\) at \(a = 2\).
| \(n\) | \(f^{(n)}(x)\) | \(f^{(n)}(2)\) | \(\frac{f^{(n)}(2)}{n!}\) |
|---|---|---|---|
| 0 | \(3 - 2x + x^2\) | \(3\) | \(3\) |
| 1 | \(-2 + 2x\) | \(2\) | \(2\) |
| 2 | \(2\) | \(2\) | \(1\) |
\[ P_2(x) = 3 + 2(x-2) + (x-2)^2 \]
Check: expanding gives \(3 + 2x - 4 + x^2 - 4x + 4 = x^2 - 2x + 3\). ✓ For a quadratic, \(P_2\) is the function, just rewritten around \(x = 2\).
Example (trig, chain rule). \(f(x) = \sin(2x)\) at \(a = \frac\pi4\).
| \(n\) | \(f^{(n)}(x)\) | \(f^{(n)}(\frac\pi4)\) | \(\frac{f^{(n)}(\pi/4)}{n!}\) |
|---|---|---|---|
| 0 | \(\sin(2x)\) | \(\sin\frac\pi2 = 1\) | \(1\) |
| 1 | \(2\cos(2x)\) | \(2\cos\frac\pi2 = 0\) | \(0\) |
| 2 | \(-4\sin(2x)\) | \(-4\) | \(-2\) |
\[ P_2(x) = 1 - 2\left(x - \frac\pi4\right)^2 \]
Each derivative pulls out another factor of \(2\). Forgetting it is the most common mistake on #2.
Example (logs). \(f(x) = \ln(x+2)\) at \(a = -1\).
| \(n\) | \(f^{(n)}(x)\) | \(f^{(n)}(-1)\) | \(\frac{f^{(n)}(-1)}{n!}\) |
|---|---|---|---|
| 0 | \(\ln(x+2)\) | \(\ln 1 = 0\) | \(0\) |
| 1 | \(\frac{1}{x+2}\) | \(1\) | \(1\) |
| 2 | \(-\frac{1}{(x+2)^2}\) | \(-1\) | \(-\frac12\) |
\[ P_2(x) = (x+1) - \frac12(x+1)^2 \]
Since \(a = -1\), \((x - a)\) becomes \((x + 1)\).
Example (approximation). \(f(x) = \sqrt x\) at \(a = 4\). Here \(f' = \frac{1}{2\sqrt x}\) and \(f'' = -\frac{1}{4x^{3/2}}\), so \(f(4) = 2\), \(f'(4) = \frac14\), and \(f''(4) = -\frac{1}{32}\):
\[ P_2(x) = 2 + \frac14(x-4) - \frac{1}{64}(x-4)^2 \]
Then \(\sqrt{4.1} \approx P_2(4.1) = 2 + 0.025 - 0.00015625 = 2.02484375\). The true value is \(2.0248457\ldots\), so the error is about \(2\times10^{-6}\).
Exact forms like ln(5) + (x-1) - (1/2)(x-1)^2 or e^2 + 2e^2(x-1) + ... are safest. WeBWorK shows decimals in its answer key, but it accepts the exact expressions.
Practice: \(P_2\) for \(e^{-x}\) at \(a = 2\); \(\cos(2x)\) at \(a = \frac\pi2\); \(\ln(3x)\) at \(a = 1\).
3. Recentering a polynomial (#5)
A polynomial of degree \(d\) has \(f^{(n)} = 0\) for \(n > d\). So its Taylor series is finite and equals the polynomial everywhere (\(R = \infty\)).
The fast way is the binomial theorem. Write \(x = (x - a) + a\) and expand.
Example. \(x^4\) centered at \(a = 2\): write \(x = (x-2) + 2\),
\[ x^4 = \big((x-2) + 2\big)^4 = \sum_{k=0}^4\binom4k 2^{4-k}(x-2)^k = 16 + 32(x-2) + 24(x-2)^2 + 8(x-2)^3 + (x-2)^4 \]
Check with derivatives: \(f'(2) = 4\cdot 2^3 = 32\) ✓, and \(\frac{f''(2)}{2!} = \frac{12\cdot 2^2}{2} = 24\) ✓.
Negative center. For \(a = -2\), write \(x = (x+2) - 2\), so the powers of the constant are \((-2)^{4-k}\):
\[ x^4 = \sum_{k=0}^4\binom4k(-2)^{4-k}(x+2)^k \]
To pick the right sigma expression, plug in the center. At \(x = a\), only the \(k = 0\) term survives, and it must equal \(f(a)\). Then check that the \((x - a)\) factor has the right sign.
4. Full Taylor series: find the derivative pattern (#7–9)
For a full series you need \(f^{(n)}(a)\) for every \(n\). Compute a few derivatives until the pattern repeats, then write the general term.
Example: \(e^x\) at \(a = 2\). Every derivative is \(e^x\), so \(f^{(n)}(2) = e^2\) for all \(n\):
\[ e^x = \sum_{n=0}^\infty \frac{e^2}{n!}(x-2)^n, \qquad R = \infty \]
(Ratio test: \(\frac{|x-2|}{n+1}\to 0\) for every \(x\).)
Example: \(\cos x\) at \(a = 0\). The derivatives cycle with period 4:
| \(n\) | 0 | 1 | 2 | 3 | 4 | \(\cdots\) |
|---|---|---|---|---|---|---|
| \(f^{(n)}(x)\) | \(\cos x\) | \(-\sin x\) | \(-\cos x\) | \(\sin x\) | \(\cos x\) | |
| \(f^{(n)}(0)\) | \(1\) | \(0\) | \(-1\) | \(0\) | \(1\) |
Only the even powers survive, with alternating signs:
\[ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}, \qquad R = \infty \]
Rewriting the even powers as \(x^{2n}\) (or the odd ones as \(x^{2n+1}\)) with factorial \((2n)!\) (or \((2n+1)!\)) lets \(n = 0, 1, 2, \ldots\) count the nonzero terms. The sign alternates, so it’s \((-1)^n\). The same reasoning gives \(\sin x\) at \(a = 0\) (#7), using the odd powers.
Example: \(\sin x\) at \(a = \frac\pi2\) (shifting the center). At \(\frac\pi2\) the derivatives \(\sin, \cos, -\sin, -\cos, \ldots\) evaluate to \(1, 0, -1, 0, 1, \ldots\). That’s the same pattern as \(\cos x\) at \(0\), so
\[ \sin x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}\left(x - \frac\pi2\right)^{2n} \]
Why? Let \(u = x - \frac\pi2\). Then \(\sin x = \sin\left(u + \frac\pi2\right) = \cos u\), and we already know the series for \(\cos u\). An identity that turns the shifted function into a known one saves computing any derivatives.
Practice: \(e^x\) at \(a = -2\); \(\cos x\) at \(a = \pi\) (hint: \(\cos(u + \pi) = -\cos u\)); \(\sin x\) at \(a = 0\).
5. Shortcuts: reuse series you already know (#6, 10)
If \(f(x) = \sum c_n(x-a)^n\) on an interval around \(a\), then the \(c_n\) are the Taylor coefficients \(\frac{f^{(n)}(a)}{n!}\). So any legitimate method gives the Taylor series: substitution, algebra, differentiation, integration.
Example: \(\frac1x\) centered at \(a = 2\). Rewrite in terms of \((x - 2)\), then use the geometric series:
\[ \frac1x = \frac{1}{2 + (x-2)} = \frac12\cdot\frac{1}{1 + \frac{x-2}{2}} = \frac12\sum_{n=0}^\infty\left(-\frac{x-2}{2}\right)^n = \sum_{n=0}^\infty\frac{(-1)^n}{2^{n+1}}(x-2)^n \]
This is valid for \(|x - 2| < 2\), i.e. \((0, 4)\). That makes sense: the series can’t reach past \(x = 0\), where \(\frac1x\) blows up.
(From derivatives, \(f^{(n)}(x) = (-1)^n n!\, x^{-n-1}\), so \(\frac{f^{(n)}(2)}{n!} = \frac{(-1)^n}{2^{n+1}}\). Same answer, more work.)
Example: \(\frac{1}{(1+x)^2}\) centered at \(0\). Recognize a derivative: \(\frac{d}{dx}\frac{1}{1+x} = -\frac{1}{(1+x)^2}\). Start from \(\frac{1}{1+x} = \sum(-1)^n x^n\):
\[ \frac{1}{(1+x)^2} = -\frac{d}{dx}\sum_{n=0}^\infty (-1)^n x^n = -\sum_{n=1}^\infty (-1)^n n\,x^{n-1} = \sum_{n=0}^\infty (-1)^n(n+1)\,x^n \]
(The last step reindexes \(n \to n+1\), which turns \(-(-1)^{n+1}\) into \((-1)^n\).)
Higher powers. Differentiating \(\frac{1}{1-x}\) twice gives \(\frac{2}{(1-x)^3} = \sum_{n=2}^\infty n(n-1)x^{n-2}\). Watch the sign when the factor is written \((x - 1)\) instead of \((1 - x)\): \((x-1)^3 = -(1-x)^3\).
6. Standard series to know
| \(f(x)\) | Taylor series at \(0\) | Converges for |
|---|---|---|
| \(\dfrac{1}{1-x}\) | \(\displaystyle\sum_{n=0}^\infty x^n\) | \((-1, 1)\) |
| \(e^x\) | \(\displaystyle\sum_{n=0}^\infty \frac{x^n}{n!}\) | all \(x\) |
| \(\sin x\) | \(\displaystyle\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}\) | all \(x\) |
| \(\cos x\) | \(\displaystyle\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}\) | all \(x\) |
| \(\ln(1+x)\) | \(\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n+1} x^{n}}{n}\) | \((-1, 1]\) |
| \(\arctan x\) | \(\displaystyle\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}\) | \([-1, 1]\) |
Summary checklist
| Problem type | HW 6.3 | Key move |
|---|---|---|
| Degree-2 Taylor polynomial | #1–4 | table of \(f^{(n)}(a)\); divide by \(n!\); chain-rule constants |
| Polynomial at a new center | #5 | binomial theorem with \(x = (x-a) + a\) |
| Full series by derivatives | #7–9 | find the repeating pattern; index only the nonzero terms |
| Shifted trig | #8 | use an identity to reduce to a known series |
| Full series by shortcuts | #6, 10 | rewrite in \((x-a)\) + geometric series; differentiate known series |