Chapter 6 Solutions

1. Choose a center c and a fixed sequence of coefficients $a_0,a_1,\dots$.
2. A power series has the form ${\sum }_{n=0}^{\infty }{a}_n(x-c{)}^n$. Each term is a constant coefficient times a nonnegative integer power of the distance from the center.
3. Convergence is assessed at each x; the center always converges to $a_0$.

1. Choose a center c and a fixed sequence of coefficients $a_0,a_1,\dots$.
2. A power series has the form ${\sum }_{n=0}^{\infty }{a}_n(x-c{)}^n$. Each term is a constant coefficient times a nonnegative integer power of the distance from the center.
3. Convergence is assessed at each x; the center always converges to $a_0$.

1. Write $f\left(x\right)={\sum }_{i=0}^{\infty }{a}_i(x-c{)}^i$ and $g\left(x\right)={\sum }_{j=0}^{\infty }{b}_j(x-c{)}^j$. Work where both series converge absolutely, in particular inside their common radius.
2. Set $t=x-c$. The product of the i-th and j-th terms is $a_i{b}_j{t}^{i+j}$. A term has total degree n exactly when $i+j=n$.
3. For a fixed n, the possible indices are $i=0,1,\dots ,n$, with $j=n-i$. Therefore the coefficient of tⁿ is ${\sum }_{i=0}^n{a}_i{b}_{n-i}$.
4. The absolute double sum is . Absolute convergence permits grouping terms by total degree. This proves the stated product identity.

Both power series converge for and diverge at x=±1 because their terms do not tend to zero. Perform the following operations inside .

(a)

1. Add coefficients of matching powers: $f\left(x\right)+g\left(x\right)={\sum }_{n=0}^{\infty }(n^2+2n+1)x^n={\sum }_{n=0}^{\infty }(n+1{)}^2{x}^n$.
2. Differentiate the geometric series to obtain ${\sum }_{n=0}^{\infty }{\mathrm{nx}}^n=x/(1-x{)}^2$ and $f\left(x\right)=x(1+x)/(1-x{)}^3$. Also $g\left(x\right)=2x/(1-x{)}^2+1/(1-x)=(1+x)/(1-x{)}^2$.
3. Combining these rational functions gives $f\left(x\right)+g\left(x\right)=(1+x)/(1-x{)}^3$.

(b)

1. Subtract coefficients of matching powers: $f\left(x\right)-g\left(x\right)={\sum }_{n=0}^{\infty }(n^2-2n-1)x^n$.
2. Using the rational forms from part (a), the common-denominator numerator is $x(1+x)-(1+x)(1-x)=2x^2+x-1$.
3. Thus $f\left(x\right)-g\left(x\right)=(2x^2+x-1)/(1-x{)}^3$.

(c)

1. A power series can be differentiated term by term inside its radius. The n=0 term is zero, so $f\prime \left(x\right)={\sum }_{n=1}^{\infty }{n}^3{x}^{n-1}$.
2. Differentiating $x(1+x)/(1-x{)}^3$ by the product and chain rules gives $f\prime \left(x\right)=(1+4x+x^2)/(1-x{)}^4$.

(d)

1. Integrate each term and include an arbitrary constant: $\int g\left(x\right)\mathrm{dx}=C+{\sum }_{n=0}^{\infty }(2n+1)x^{n+1}/(n+1)$.
2. Rewrite $g\left(x\right)=2/(1-x{)}^2-1/(1-x)$. Choose $2x/(1-x)$ as an antiderivative of the first term; its derivative is $2/(1-x{)}^2$. The second term integrates to $\ln (1-x)$. Both chosen antiderivatives are zero at x=0, matching the integrated power series.
3. Therefore an antiderivative is $2x/(1-x)+\ln (1-x)+C$. On , $1-x>0$, so the logarithm is real.

(a)

1. Set $h=x-3$, so the polynomial is $(3+h{)}^5-(3+h{)}^3+1$.
2. Use the binomial theorem and collect like powers. The result is $217+378h+261h^2+89h^3+15h^4+h^5$. This finite Taylor series is valid for every real x. The center is 3, so it is not Maclaurin.

(b)

1. Set $h=x-1$. Then $(x-1{)}^2/x=h^2/(1+h)$.
2. Expand $1/(1+h)={\sum }_{n=0}^{\infty }(-1{)}^n{h}^n$ for . Thus the answer is ${\sum }_{n=0}^{\infty }(-1{)}^n(x-1{)}^{n+2}$. The terms do not vanish at either h endpoint. Its center is 1, so it is not Maclaurin.

(c)

1. Integrate $1/(1-x)={\sum }_{n=0}^{\infty }{x}^n$ from zero to x and change the sign.
2. This gives $\ln (1-x)=-{\sum }_{n=1}^{\infty }{x}^n/n$. It converges for : at −1 the alternating harmonic series converges, while at 1 the harmonic series diverges. Its center is 0, so it is Maclaurin.

(d)

1. Set $h=x-1$ and use $\cos (1+h)=\cos \left(1\right)\cos \left(h\right)-\sin \left(1\right)\sin \left(h\right)$.
2. Substitute the ordinary cosine and sine series: $\cos \left(1\right){\sum }_{n=0}^{\infty }(-1{)}^n{h}^{2n}/(2n)!-\sin (1\left){\sum }_{n=0}^{\infty }\right(-1{)}^n{h}^{2n+1}/(2n+1)!$. It converges for all real h. The center is 1, so it is not Maclaurin.

(e)

1. Set $h=x-5$. Since $e^x=e^5{e}^h$, substitute the exponential series.
2. The result is $e^5{\sum }_{n=0}^{\infty }(x-5{)}^n/n!$, valid for all real x. The center is 5, so it is not Maclaurin.

(f)

1. Expand $1/(1+x^2)={\sum }_{n=0}^{\infty }(-1{)}^n{x}^{2n}$ inside and integrate from 0 to x.
2. This gives $\arctan x={\sum }_{n=0}^{\infty }(-1{)}^n{x}^{2n+1}/(2n+1)$. Both endpoints converge by the alternating series test, so the interval is $-1\le x\le 1$. The center is 0, so it is Maclaurin.

(g)

1. Start with $e^x={\sum }_{n=0}^{\infty }{x}^n/n!$ and multiply each term by x.
2. The result is ${\mathrm{xe}}^x={\sum }_{n=0}^{\infty }{x}^{n+1}/n!$, valid for every real x. Its center is 0, so it is Maclaurin.

1. For $a_n=x^n$, the geometric ratio is x, so convergence requires .
2. At x=1 the terms are 1; at x=−1 they alternate ±1. Neither endpoint has terms tending to zero.

1. The series is geometric with ratio $(x-2)/3$.
2. Convergence requires , or . At each endpoint the ratio is ±1, so the terms fail the zero-limit test.

1. The ratio test gives a limit of $\left|x\right|$, so the radius is 1.
2. At x=1, the harmonic series diverges. At x=−1, the alternating harmonic series converges. Outside radius 1 the terms do not approach zero.

1. The ratio of consecutive magnitudes tends to $\left|x\right|$, giving radius 1.
2. At both endpoints the magnitudes are $1/n^2$, whose p-series converges. Both endpoints are included.

1. The ratio of consecutive magnitudes is $\left|x\right|/(n+1)$.
2. This tends to zero for every fixed x. The ratio test therefore gives convergence for every real x.

1. For nonzero x, the ratio of consecutive term magnitudes is $(n+1)\left|x\right|$, which tends to infinity. The terms eventually grow and the series diverges.
2. At x=0, all positive-degree terms vanish and the constant term is 1.

1. The geometric series $1/(1-x)={\sum }_{n=0}^{\infty }{x}^n$ holds for .
2. Differentiate term by term within the radius to obtain $1/(1-x{)}^2={\sum }_{n=1}^{\infty }{\mathrm{nx}}^{n-1}$.
3. The differentiated series still has radius 1, and its terms fail to approach zero at either endpoint.

1. Integrate the geometric series from 0 to x inside .
2. This gives $-\ln (1-x)={\sum }_{n=0}^{\infty }{x}^{n+1}/(n+1)$.
3. The resulting series converges at x=−1 by the alternating test but diverges at x=1. Thus its interval is .

1. For $f\left(x\right)=e^x$, every derivative at x=1 equals e.
2. Insert those values in $P_3\left(x\right)={\sum }_{n=0}^3{f}^{\left(n\right)}\left(1\right)(x-1{)}^n/n!$.

1. For $f\left(x\right)=\ln x$, the first four derivative values at 2 are $f\left(2\right)=\ln 2$, $f\prime \left(2\right)=1/2$, $f″\left(2\right)=-1/4$, and $f‴\left(2\right)=1/4$.
2. Divide the second and third derivative values by 2! and 3!, respectively, in the Taylor formula.

1. Set $h=x-1$, so $x^3=(1+h{)}^3$.
2. The binomial theorem gives $1+3h+3h^2+h^3$. There are no higher terms.

1. Write $x=2+h$. Then $1/x=(1/2)/(1+h/2)$.
2. Use the geometric series with ratio $-h/2$. The answer is ${\sum }_{n=0}^{\infty }(-1{)}^n(x-2{)}^n/2^{n+1}$.
3. The series converges when . At either endpoint its terms fail to tend to zero.

1. Set $h=x-\pi /2$. The angle-addition identity gives $\sin (\pi /2+h)=\cos h$.
2. Substitute the cosine Maclaurin series to get ${\sum }_{n=0}^{\infty }(-1{)}^n(x-\pi /2{)}^{2n}/\left(2n\right)!$. It converges for every real x.

1. In a Taylor series centered at 3, the coefficient of $(x-3{)}^n$ equals $f^{\left(n\right)}\left(3\right)/n!$.
2. The linear coefficient gives $f\prime \left(3\right)=-2$. The quadratic coefficient 4 gives $f″\left(3\right)=2!·4=8$.

1. The cubic Taylor polynomial for $e^x$ at zero is $1+x+x^2/2+x^3/6$.
2. On $0\le x\le 0.1$, its fourth derivative is $e^x\le e^{0.1}$. Taylor’s remainder bound is $\left|R_3\right(0.1\left)\right|\le e^{0.1}(0.1{)}^4/4!$.
3. Evaluating the bound gives approximately $0.000004605$, while the polynomial value is $1.105166667$.

1. A Taylor series uses derivative values at its center c.
2. Choosing $c=0$ gives ${\sum }_{n=0}^{\infty }{f}^{\left(n\right)}\left(0\right)x^n/n!$, which is called the Maclaurin series.

1. Start from $e^u={\sum }_{n=0}^{\infty }{u}^n/n!$.
2. Substitute $u=2x$ to obtain $e^{2x}={\sum }_{n=0}^{\infty }{2}^n{x}^n/n!$. The exponential series converges for every real u, hence every real x.

1. Use $\cos u={\sum }_{n=0}^{\infty }(-1{)}^n{u}^{2n}/(2n)!$.
2. Set $u=3x$, giving $1-9x^2/2!+81x^4/4!-729x^6/6!+\dots$. The series converges for all real x.

1. Start with $\ln (1+u)={\sum }_{n=1}^{\infty }(-1{)}^{n+1}{u}^n/n$ for .
2. Substitute $u=2x$. This yields ${\sum }_{n=1}^{\infty }(-1{)}^{n+1}{2}^n{x}^n/n$.
3. At $x=1/2$ the alternating harmonic series converges; at $x=-1/2$ the magnitude harmonic series diverges.

1. Substitute $u=x^2$ into $\sin u=u-u^3/3!+u^5/5!-\dots$.
2. The first three nonzero terms are $x^2-x^6/6+x^{10}/120$. The full series has infinite radius.

1. Use $e^x=1+x+x^2/2+x^3/6+\dots$ and $\sin x=x-x^3/6+\dots$.
2. Collect terms through degree 3: the x coefficient is 1, the x² coefficient is 1, and the x³ coefficient is $1/2-1/6=1/3$.

1. The geometric formula $1/(1-u)={\sum }_{n=0}^{\infty }{u}^n$ applies when .
2. Set $u=-x^2$, obtaining $1/(1+x^2)={\sum }_{n=0}^{\infty }(-1{)}^n{x}^{2n}$.
3. At x=±1 the terms alternate ±1 and fail to tend to zero. The interval is $(-1,1)$.

1. Multiplying $e^x=\sum x^n/n!$ by x² gives $x^2{e}^x=\sum x^{n+2}/n!$.
2. The x⁵ term requires $n+2=5$, hence n=3. Its coefficient is $1/3!=1/6$.

1. The Maclaurin expansion of $1/(1-x)$ is the geometric series $\sum x^n$, whose radius is 1.
2. The function itself exists for every $x\ne 1$, but the series diverges when $\left|x\right|\ge 1$. A function’s domain can therefore exceed the convergence interval of one Taylor expansion.

1. For $f\left(x\right)=\mathrm{sqrt}(1+x)$, $f\left(0\right)=1$, $f\prime \left(0\right)=1/2$, and $f″\left(0\right)=-1/4$.
2. Divide the second derivative by 2! in the Taylor formula. This gives the quadratic Maclaurin polynomial $1+x/2-x^2/8$.