Free math worksheets

- Begin with the term for \(n=0\). - Increase the exponent by \(1\) for each successive term.

1. Substitute \(n=0,1,2,3,4\) into the term \(x^n\). 2. The first five terms are \(1+x+x^2+x^3+x^4\).

\(1+x+x^2+x^3+x^4\).

- Factor the constant term from the denominator first. - Stop once the denominator has the form \(1-r\).

1. Factor \(4\) from the denominator: \(4-x=4\left(1-\frac{x}{4}\right)\). 2. Therefore, \(\frac{1}{4-x}=\frac14\cdot\frac{1}{1-\frac{x}{4}}\).

\(\frac{1}{4-x}=\frac14\cdot\frac{1}{1-\frac{x}{4}}\).

- Compare how each target differs structurally from \(\frac{1}{1-x}\). - Think about which operation changes the power of the denominator and which operation creates a logarithm. - A sign change in the denominator can come from changing the input rather than differentiating or integrating.

1. Differentiating \(\frac{1}{1-x}\) gives \(\frac{1}{(1-x)^2}\), so part a) uses differentiation. 2. Integrating \(\frac{1}{1-t}\) from \(0\) to \(x\) gives \(-\ln(1-x)\), so part b) uses integration from \(0\) to \(x\). 3. Replacing \(x\) by \(-x\) changes \(\frac{1}{1-x}\) into \(\frac{1}{1+x}\), so part c) uses substitution of \(-x\).

a) Differentiate. b) Integrate from \(0\) to \(x\). c) Substitute \(-x\).

- Match the denominator to the form \(1-r\). - The sign pattern comes from powers of \(-x\). - Use the same absolute-value condition as the original geometric series.

1. Replace \(x\) by \(-x\) in \( \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n\). 2. This gives \( \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n\). 3. The first six terms are \(1-x+x^2-x^3+x^4-x^5\). 4. For \(x^8\), the coefficient is \((-1)^8=1\). 5. The substitution does not change the condition \(|x|<1\), and both endpoints diverge because the terms do not approach \(0\).

a) \( \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n\). b) \(1-x+x^2-x^3+x^4-x^5\). c) Coefficient \(1\); interval \((-1, 1)\).

- Multiply the known geometric identity by the numerator. - Decide whether to keep the original index or shift it to match the exponent. - A polynomial factor does not change where the underlying geometric series converges.

1. Start with \( \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n\) for \(|x|<1\). 2. Multiplying both sides by \(x^2\) gives \( \frac{x^2}{1-x}=\sum_{n=0}^{\infty}x^{n+2}\). 3. Reindexing with \(k=n+2\) gives \( \sum_{k=2}^{\infty}x^k\). 4. The first five nonzero terms are \(x^2+x^3+x^4+x^5+x^6\). 5. Multiplication by \(x^2\) does not change the geometric convergence condition, so the interval is \((-1, 1)\).

a) \( \frac{x^2}{1-x}=\sum_{n=0}^{\infty}x^{n+2}=\sum_{n=2}^{\infty}x^n\). b) \(x^2+x^3+x^4+x^5+x^6\). c) \((-1, 1)\).

- Factor out the constant term in the denominator. - Match the remaining expression to \(1/(1-r)\). - Translate \(|r|<1\) back into a condition on \(x\).

1. Factor the denominator: \( \frac{1}{3-x}=\frac13\cdot\frac{1}{1-\frac{x}{3}}\). 2. Apply the geometric series with ratio \(\frac{x}{3}\): \( \frac{1}{3-x}=\frac13\sum_{n=0}^{\infty}\left(\frac{x}{3}\right)^n=\sum_{n=0}^{\infty}\frac{x^n}{3^{n+1}}\). 3. The first four terms are \( \frac13+\frac{x}{9}+\frac{x^2}{27}+\frac{x^3}{81}\). 4. Convergence requires \(\left|\frac{x}{3}\right|<1\), so the interval is \((-3, 3)\).

a) \( \frac{1}{3-x}=\sum_{n=0}^{\infty}\frac{x^n}{3^{n+1}}\). b) \( \frac13+\frac{x}{9}+\frac{x^2}{27}+\frac{x^3}{81}\). c) \((-3, 3)\).

- Treat \(x^2\) as the geometric ratio. - Notice the parity of every exponent in the expansion. - Solve \(|x^2|<1\) and then test the two boundary values.

1. Substitute \(x^2\) for \(x\) in \( \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n\). 2. This gives \( \frac{1}{1-x^2}=\sum_{n=0}^{\infty}x^{2n}\). 3. The first five nonzero terms are \(1+x^2+x^4+x^6+x^8\). 4. Only even powers occur, so the coefficient of \(x^9\) is \(0\). 5. Convergence requires \(|x^2|<1\), which is equivalent to \(|x|<1\). Both endpoints diverge.

a) \( \frac{1}{1-x^2}=\sum_{n=0}^{\infty}x^{2n}\). b) \(1+x^2+x^4+x^6+x^8\). c) Coefficient \(0\); interval \((-1, 1)\).

- Differentiate the exponent as well as the function on the left. - Multiplying by \(x\) restores the exponent to match the index. - Read the requested coefficient directly from the general term.

1. Differentiating gives \( \frac{1}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n-1}\). 2. Multiplying by \(x\) gives \( \frac{x}{(1-x)^2}=\sum_{n=1}^{\infty}nx^n\). 3. The first five nonzero terms are \(x+2x^2+3x^3+4x^4+5x^5\). 4. The coefficient of \(x^7\) is \(7\).

a) \( \frac{x}{(1-x)^2}=\sum_{n=1}^{\infty}nx^n\). b) \(x+2x^2+3x^3+4x^4+5x^5\). c) \(7\).

- Use a new variable inside the definite integral. - Integrate each power by increasing its exponent and dividing by the new exponent. - Choose the value of \(x\) that makes \(1-x=\frac{1}{2}\).

1. Integrating the left side gives \( \int_0^x\frac{1}{1-t}\,dt=-\ln(1-x)\). 2. Integrating the series term by term gives \( -\ln(1-x)=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=\sum_{n=1}^{\infty}\frac{x^n}{n}\). 3. The first four nonzero terms are \( x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}\). 4. Since \(\ln 2=-\ln\left(1-\frac{1}{2}\right)\), the four-term approximation is \( \frac12+\frac{(1/2)^2}{2}+\frac{(1/2)^3}{3}+\frac{(1/2)^4}{4}=\frac{131}{192}\approx0.682292\).

a) \( -\ln(1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n}\). b) \( x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}\). c) \(\ln 2\approx\frac{131}{192}\approx0.682292\).

- For \(c_n\), include every pair of exponents whose sum is \(n\). - Multiply the two supplied function formulas to obtain the closed form. - Compare consecutive even and odd coefficients to identify the repeated-pair pattern.

1. The constant coefficient is \( c_0=1\cdot1=1\). 2. The coefficient of \(x\) is \( c_1=1(-1)+2(1)=1\). 3. The coefficient of \(x^2\) is \( c_2=1(1)+2(-1)+3(1)=2\). 4. The coefficient of \(x^3\) is \( c_3=1(-1)+2(1)+3(-1)+4(1)=2\). Thus the product begins \(1+x+2x^2+2x^3+\cdots\). 5. Multiplying the supplied closed forms gives \( A(x)B(x)=\frac{1}{(1-x)^2(1+x)}\). 6. The coefficients occur in equal pairs: \( 1,1,2,2,3,3,\ldots\). Therefore, for every integer \(m\geq0\), \( c_{2m}=c_{2m+1}=m+1\).

a) \( A(x)B(x)=1+x+2x^2+2x^3+\cdots\). b) \( A(x)B(x)=\frac{1}{(1-x)^2(1+x)}\). c) \( c_{2m}=c_{2m+1}=m+1\) for \(m\geq0\).

- Compare the factor \(3^n\) with the odd powers in the arctangent series. - Choose an input whose square is \(\frac13\). - Use the special-angle value of arctangent after matching the series.

1. The arctangent series is \( \arctan x=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}\). 2. Substitute \(x=\frac{1}{\sqrt3}\): \( \arctan\left(\frac{1}{\sqrt3}\right)=\frac{1}{\sqrt3}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}=\frac{S}{\sqrt3}\). 3. Since \(\arctan\left(\frac{1}{\sqrt3}\right)=\frac{\pi}{6}\), \( \frac{S}{\sqrt3}=\frac{\pi}{6}\). 4. Therefore, \( S=\frac{\pi\sqrt3}{6}\).

\( S=\frac{\pi\sqrt3}{6}\).

- Substitute the inner expression into the standard cosine series before truncating. - Track what happens to the even powers of \(\sqrt{x}\). - For the numerical approximation, choose the value of \(x\) whose square root is \(0.5\).

1. The cosine series is \(\cos u=1-\frac{u^2}{2!}+\frac{u^4}{4!}-\frac{u^6}{6!}+\cdots\). 2. Substitute \(u=\sqrt{x}\). Each even power of \(\sqrt{x}\) becomes an integer power of \(x\): \(\cos(\sqrt{x})=1-\frac{x}{2!}+\frac{x^2}{4!}-\frac{x^3}{6!}+\cdots\). 3. Therefore, \(P_3(x)=1-\frac{x}{2}+\frac{x^2}{24}-\frac{x^3}{720}\). 4. To approximate \(\cos(0.5)\), use \(x=(0.5)^2=0.25\): \(P_3(0.25)=1-\frac{0.25}{2}+\frac{0.25^2}{24}-\frac{0.25^3}{720}=\frac{40439}{46080}\approx0.877582465\).

a) \(P_3(x)=1-\frac{x}{2}+\frac{x^2}{24}-\frac{x^3}{720}\). b) \(\cos(0.5)\approx0.877582465\).

- Apply the power rule to each term. - Find the factor that produces each term from the preceding term. - Use the infinite geometric-series formula and the condition \(|x|<1\).

1. Differentiating term by term gives \(1-x+x^2-x^3+\cdots\). 2. This is a geometric series with first term \(a=1\) and common ratio \(r=-x\). 3. Since \(|x|<1\), the series converges and its sum is \(\frac{a}{1-r}=\frac{1}{1-(-x)}=\frac{1}{1+x}\). 4. This agrees with \(f'(x)=\frac{1}{1+x}\).

1. \(1-x+x^2-x^3+\cdots\) 2. It is geometric with \(a=1\) and \(r=-x\), so its sum is \(\frac{1}{1+x}\).

- Each differentiation lowers the power and introduces a factor from the exponent. - Multiply by a suitable power of \(x\) to restore the exponent to \(n\). - Substitute the requested exponent into the coefficient formula.

1. Differentiating twice gives \( \frac{2}{(1-x)^3}=\sum_{n=2}^{\infty}n(n-1)x^{n-2}\). 2. Multiplying by \(\frac{x^2}{2}\) gives \( \frac{x^2}{(1-x)^3}=\sum_{n=2}^{\infty}\frac{n(n-1)}{2}x^n\). 3. The first four nonzero terms are \(x^2+3x^3+6x^4+10x^5\). 4. For \(n=8\), the coefficient is \( \frac{8\cdot7}{2}=28\).

\( \frac{x^2}{(1-x)^3}=\sum_{n=2}^{\infty}\frac{n(n-1)}{2}x^n\). First terms: \(x^2+3x^3+6x^4+10x^5\). Coefficient of \(x^8\): \(28\).

- Decompose the rational function before expanding. - Keep the two geometric coefficient formulas separate until the same powers are aligned. - The final interval must satisfy both convergence conditions.

1. Partial fractions give \( \frac{1}{(1-x)(1-2x)}=-\frac{1}{1-x}+\frac{2}{1-2x}\). 2. Expanding each geometric series gives \( -\sum_{n=0}^{\infty}x^n+2\sum_{n=0}^{\infty}(2x)^n\). 3. Combining coefficients gives \( \sum_{n=0}^{\infty}(2^{n+1}-1)x^n\). 4. The first four terms are \(1+3x+7x^2+15x^3\). 5. Both geometric series must converge, so \(|x|<1\) and \(|2x|<1\). The stricter condition is \(|x|<\frac{1}{2}\).

a) Coefficient of \(x^n\): \(2^{n+1}-1\). b) \(1+3x+7x^2+15x^3\). c) \( \left(-\frac12, \frac12\right)\).

- Multiplying by \(x^2\) aligns the exponent with the denominator index. - Remove the \(k=1\) term from the logarithm series. - Use the original series, not the quotient formula, to read the value at \(0\).

1. Multiplying by \(x^2\) gives \( x^2F(x)=\sum_{n=0}^{\infty}\frac{x^{n+2}}{n+2}=\sum_{k=2}^{\infty}\frac{x^k}{k}\). 2. Since \( -\ln(1-x)=\sum_{k=1}^{\infty}\frac{x^k}{k}\), \( x^2F(x)=-\ln(1-x)-x\). 3. Thus \( F(x)=\frac{-\ln(1-x)-x}{x^2}\) for \(x\neq0\). 4. The series itself gives \(F(0)=\frac{1}{2}\). 5. The first four terms are \( \frac12+\frac{x}{3}+\frac{x^2}{4}+\frac{x^3}{5}\). 6. At \(x=1\), the series diverges like the harmonic series. At \(x=-1\), it converges conditionally. The interval is \([-1, 1)\).

a) \( F(x)=\frac{-\ln(1-x)-x}{x^2}\) for \(x\neq0\). b) \(F(0)=\frac{1}{2}\). c) \( \frac12+\frac{x}{3}+\frac{x^2}{4}+\frac{x^3}{5}\), with interval \([-1, 1)\).

- Make the geometric ratio equal to \(-x^2\). - Use the derivative of arctangent to choose the correct antiderivative. - Recheck the two boundary values after integrating.

1. Substitute \(-x^2\) into the geometric series: \( \frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^nx^{2n}\) for \(|x|<1\). 2. Since \(\arctan x=\int_0^x\frac{1}{1+t^2}\,dt\), term-by-term integration gives \( \arctan x=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}\). 3. At \(x=1\) and \(x=-1\), the resulting numerical series are alternating harmonic-type series, so both converge. The interval is \([-1, 1]\).

a) \( \frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^nx^{2n}\). b) \( \arctan x=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}\). c) \([-1, 1]\).

- For a fixed total exponent \(n\), pair exponent \(k\) from one series with exponent \(n-k\) from the other. - The second series contributes coefficient \(1\) to every pairing. - After finding the coefficient formula, examine what those coefficients approach as \(n\) grows. - For points at or beyond a proposed radius, check whether the individual series terms can approach \(0\).

1. For two power series \(\sum a_kx^k\) and \(\sum b_jx^j\), the coefficient of \(x^n\) in their Cauchy product is \(c_n=\sum_{k=0}^{n}a_kb_{n-k}\). Here, \(a_k=\frac{1}{k!}\) and \(b_{n-k}=1\), so \(c_n=\sum_{k=0}^{n}\frac{1}{k!}\). 2. Therefore, \(\frac{e^x}{1-x}=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}\frac{1}{k!}\right)x^n\) for \(|x|<1\), where both original series converge. 3. The terms through \(x^3\) are \(1+2x+\frac52x^2+\frac83x^3\). 4. The coefficient of \(x^4\) is \(1+1+\frac12+\frac16+\frac{1}{24}=\frac{65}{24}\). 5. To show the interval is exactly \((-1, 1)\), note that \(c_n=\sum_{k=0}^{n}\frac{1}{k!}\to e\). Thus, if \(|x|\geq1\), the terms \(c_nx^n\) do not approach \(0\): for \(|x|=1\), their magnitudes approach \(e\), and for \(|x|>1\), their magnitudes grow. Hence the series diverges for \(|x|\geq1\), so its interval of convergence is \((-1, 1)\).

a) \(c_n=\sum_{k=0}^{n}\frac{1}{k!}\). b) \(1+2x+\frac52x^2+\frac83x^3\). c) Coefficient \(\frac{65}{24}\); interval \((-1, 1)\).

- Decompose the rational function into two simpler fractions. - Factor each denominator to expose a geometric ratio. - The nearest restriction to the center determines the radius, but both boundary values still require checks.

1. Partial fractions give \( \frac{1}{(2-x)(x+4)}=\frac16\left(\frac{1}{2-x}+\frac{1}{x+4}\right)\). 2. Expand each factor: \( \frac{1}{2-x}=\frac12\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^n\) for \(|x|<2\), and \( \frac{1}{x+4}=\frac14\sum_{n=0}^{\infty}\left(-\frac{x}{4}\right)^n\) for \(|x|<4\). 3. Therefore, \( f(x)=\sum_{n=0}^{\infty}\left(\frac{1}{12\cdot2^n}+\frac{(-1)^n}{24\cdot4^n}\right)x^n\). 4. The first three terms are \( \frac18+\frac{x}{32}+\frac{3x^2}{128}\). 5. Both expansions are valid only where their intervals overlap, so \(|x|<2\). At \(x=2\), the function has a singularity; at \(x=-2\), the first geometric component has ratio \(-1\), so the terms do not approach \(0\). Thus \(R=2\) and the interval is \((-2, 2)\).

a) \( f(x)=\sum_{n=0}^{\infty}\left(\frac{1}{12\cdot2^n}+\frac{(-1)^n}{24\cdot4^n}\right)x^n\). b) \( \frac18+\frac{x}{32}+\frac{3x^2}{128}\). c) Radius \(R=2\); interval \((-2, 2)\).

- Split the logarithm of a quotient into a difference. - Compare what happens to odd and even powers when the two logarithm series are subtracted. - Match exponent \(7\) to the general odd exponent \(2n+1\).

1. Write \( H(x)=\frac12[\ln(1+x)-\ln(1-x)]\). 2. In the difference, the even-power terms cancel and the odd-power terms double. 3. Therefore, \( H(x)=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}=x+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\cdots\). 4. The coefficient of \(x^7\) is \(\frac{1}{7}\).

a) \( H(x)=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}\). b) \( x+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}\). c) \( \frac17\).

- Use the supplied factorization to create a geometric denominator. - Multiply the geometric series by both terms of \(1-x\). - Organize the result by consecutive powers before testing endpoints. - At an endpoint, ordinary convergence is determined by the original partial sums, not by an unproved regrouping.

1. The identity gives \( \frac{1}{1+x+x^2}=\frac{1-x}{1-x^3}\). 2. For \(|x|<1\), \( \frac{1}{1-x^3}=\sum_{k=0}^{\infty}x^{3k}\). Multiplying by \(1-x\) gives \( \frac{1}{1+x+x^2}=\sum_{k=0}^{\infty}(x^{3k}-x^{3k+1})\). 3. The first eight terms by consecutive powers are \( 1-x+0x^2+x^3-x^4+0x^5+x^6-x^7\). 4. The geometric derivation requires \(|x|<1\). At \(x=1\), the ordinary power-series terms have repeating coefficients \(1,-1,0\), so the terms do not approach \(0\). At \(x=-1\), the terms also fail to approach \(0\). Thus both endpoints diverge, and the interval is \((-1, 1)\). 5. Substituting \(x=1\) into the grouped formula changes the sequence of partial sums by pairing terms at a point where the original series diverges. The identity and grouping were justified only for \(|x|<1\), so the zero block values do not assign an ordinary endpoint sum.

a) \( \frac{1}{1+x+x^2}=\sum_{k=0}^{\infty}(x^{3k}-x^{3k+1})\) for \(|x|<1\). b) \( 1-x+0x^2+x^3-x^4+0x^5+x^6-x^7\). c) \((-1, 1)\). d) The grouped identity is valid only inside \(|x|<1\). At \(x=1\), the original consecutive terms do not approach \(0\), so the power series diverges.

- Use the derivative of inverse sine to identify the needed antiderivative. - Increase each even exponent by \(1\) and divide by the new exponent. - Preserve the original convergence condition during term-by-term integration.

1. Since \( \arcsin x=\int_0^x\frac{1}{\sqrt{1-t^2}}\,dt\), integrate each displayed term. 2. This gives \( \arcsin x=x+\frac{x^3}{6}+\frac{3x^5}{40}+\frac{5x^7}{112}+\cdots\). 3. The coefficient of \(x^7\) is \(\frac{5}{112}\). 4. The binomial series is valid for \(|x^2|<1\), so the open interval is \((-1, 1)\).

a) \( x+\frac{x^3}{6}+\frac{3x^5}{40}+\frac{5x^7}{112}\). b) \( \frac{5}{112}\). c) \((-1, 1)\) before endpoint checks.

- Rewrite every occurrence of \(x\) using the displacement \(x-1\). - Factor the denominator so the remaining expression has the form \(1-r\). - The geometric series converges when the absolute value of its ratio is less than \(1\).

1. Write the denominator in terms of the displacement from the center: \( 4-x=3-(x-1)=3\left(1-\frac{x-1}{3}\right)\). 2. Apply the geometric-series identity: \( \frac{1}{4-x}=\frac13\sum_{n=0}^{\infty}\left(\frac{x-1}{3}\right)^n=\sum_{n=0}^{\infty}\frac{(x-1)^n}{3^{n+1}}\). 3. The first four nonzero terms are \( \frac13+\frac{x-1}{9}+\frac{(x-1)^2}{27}+\frac{(x-1)^3}{81}\). 4. The coefficient of \((x-1)^5\) is \(\frac{1}{3^6}=\frac{1}{729}\). 5. Convergence requires \(|x-1|<3\), so the interval is \((-2, 4)\). At both endpoints the geometric ratio is \(-1\) or \(1\), so the series diverges.

a) \( \sum_{n=0}^{\infty}\frac{(x-1)^n}{3^{n+1}}\). b) \( \frac13+\frac{x-1}{9}+\frac{(x-1)^2}{27}+\frac{(x-1)^3}{81}\). c) Coefficient \( \frac{1}{729}\); interval \((-2, 4)\).

- Begin with the sine series and divide every term by the variable. - Integrating from \(0\) makes the constant of integration automatic. - Use the coefficient of \(x^7\) to recover the seventh derivative at the center.

1. Divide the sine series by \(t\): \( \frac{\sin t}{t}=\sum_{n=0}^{\infty}\frac{(-1)^nt^{2n}}{(2n+1)!}\). 2. Integrate term by term from \(0\) to \(x\): \( F(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)(2n+1)!}\). 3. The first four nonzero terms are \( x-\frac{x^3}{18}+\frac{x^5}{600}-\frac{x^7}{35280}\). 4. The coefficient of \(x^7\) is \(-\frac{1}{35280}\), so \( F^{(7)}(0)=7!\left(-\frac{1}{35280}\right)=-\frac17\).

a) \( F(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)(2n+1)!}\). b) \( x-\frac{x^3}{18}+\frac{x^5}{600}-\frac{x^7}{35280}\). c) \( F^{(7)}(0)=-\frac17\).

- View the denominator as \(1-(-x^4)\). - Integrate each power before substituting the upper limit. - Keep the first three values of \(n\): \(0\), \(1\), and \(2\).

1. Substitute \(-x^4\) into the geometric series: \( \frac{1}{1+x^4}=\sum_{n=0}^{\infty}(-1)^nx^{4n}\) for \(|x|<1\). 2. Integrating term by term from \(0\) to \(\frac12\) gives \( I=\sum_{n=0}^{\infty}\frac{(-1)^n\left(\frac12\right)^{4n+1}}{4n+1}\). 3. Using the first three nonzero terms, \( I\approx\frac12-\frac{\left(\frac12\right)^5}{5}+\frac{\left(\frac12\right)^9}{9}\). 4. Therefore, \( I\approx\frac12-\frac{1}{160}+\frac{1}{4608}=\frac{11381}{23040}\approx0.493967014\).

a) \( \frac{1}{1+x^4}=\sum_{n=0}^{\infty}(-1)^nx^{4n}\), for \(|x|<1\). b) \( I\approx\frac{11381}{23040}\approx0.493967014\).

- Decompose the coefficient before working with the series. - Relate both resulting sums to the logarithm series, paying attention to the shifted denominator. - Test the two endpoints using the original positive coefficient formula.

1. Split the coefficient: \( F(x)=\sum_{n=1}^{\infty}\frac{x^n}{n}-\sum_{n=1}^{\infty}\frac{x^n}{n+1}\). 2. The first sum is \(-\ln(1-x)\). For the second sum, \( \sum_{n=1}^{\infty}\frac{x^n}{n+1}=\frac{-\ln(1-x)-x}{x}\) for \(x\neq0\). 3. Therefore, \( F(x)=1+\left(\frac1x-1\right)\ln(1-x)\) for \(x\neq0\). The original series gives the removable value \(F(0)=0\). 4. The first four nonzero terms are \( \frac{x}{2}+\frac{x^2}{6}+\frac{x^3}{12}+\frac{x^4}{20}\). 5. The radius is \(1\). At both \(x=1\) and \(x=-1\), the absolute-value series is \( \sum_{n=1}^{\infty}\frac{1}{n(n+1)}\), which converges. Thus the interval is \([-1, 1]\). 6. At \(x=\frac12\), \( F\left(\frac12\right)=1+\ln\left(\frac12\right)=1-\ln 2\).

a) \( F(x)=1+\left(\frac1x-1\right)\ln(1-x)\) for \(x\neq0\), with \(F(0)=0\). b) \( \frac{x}{2}+\frac{x^2}{6}+\frac{x^3}{12}+\frac{x^4}{20}+\cdots\); interval \([-1, 1]\). c) \( F\left(\frac12\right)=1-\ln 2\).

- Use the finite geometric-sum formula rather than estimating the tail term by term. - Substitute \(x=0.4\) only after deriving the general remainder. - Check one value of \(N\) that fails and the next value that succeeds.

1. The finite geometric-sum formula gives \( S_N(x)=\frac{1-x^{N+1}}{1-x}\). 2. Therefore, \( R_N(x)=\frac{1}{1-x}-\frac{1-x^{N+1}}{1-x}=\frac{x^{N+1}}{1-x}\). 3. At \(x=0.4\), \( |R_N(0.4)|=\frac{0.4^{N+1}}{0.6}\). 4. For \(N=7\), \( |R_7(0.4)|=\frac{0.4^8}{0.6}\approx0.00109227\), which is too large. 5. For \(N=8\), \( |R_8(0.4)|=\frac{0.4^9}{0.6}\approx0.000436907\), which is below \(0.001\). Thus the least value is \(N=8\). 6. The corresponding approximation is \( S_8(0.4)=\frac{1-0.4^9}{0.6}=1.66622976\).

a) \( R_N(x)=\frac{x^{N+1}}{1-x}\). b) Least value: \(N=8\). c) \( \frac{1}{1-0.4}\approx S_8(0.4)=1.66622976\), with exact error \( \frac{0.4^9}{0.6}\approx0.000436907\).

- Use the vertical guide in each panel and compare vertical distances from the partial-sum curves to \(f\); do not begin by reconstructing their formulas. - The magnified panel is there to distinguish the two approximations that are nearly indistinguishable on the broader view near \(0.5\). - Rewrite the denominator so the ratio in a geometric series is visible. - Compare the two marked inputs with the convergence condition of that series.

1. Panels a) and b) show that at \(x=0.5\), \(P_{10}\) is closest to \(f\), then \(P_6\), then \(P_2\). All three partial sums lie below \(f(0.5)\). 2. Panel c) shows that at \(x=1.2\), \(P_2\) is closest to \(f\), then \(P_6\), then \(P_{10}\). All three lie below \(f(1.2)\), and increasing the degree makes the approximation worse at this input. 3. Since \(\frac{1}{1+x^2}=\frac{1}{1-(-x^2)}\), the geometric-series identity gives \(\frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^n x^{2n}\) when \(|-x^2|<1\), equivalently \(|x|<1\). 4. Thus \(x=0.5\) is inside the convergence interval, where successive partial sums approach \(f\), while \(x=1.2\) is outside it, where these partial sums need not approach the function even though \(f(1.2)\) itself is defined.

a) At \(x=0.5\): \(P_{10}\), then \(P_6\), then \(P_2\) from most to least accurate; all three lie below \(f(0.5)\). b) At \(x=1.2\): \(P_2\), then \(P_6\), then \(P_{10}\) from most to least accurate; increasing degree makes the approximation worse there. c) \(\frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^n x^{2n}\) for \(|x|<1\). The graph improves at \(0.5\) because that input is inside the interval of convergence and deteriorates at \(1.2\) because that input is outside it.

- Track what replacing \(x\) by \(-x\) does to both the function and the powers in the series. - After differentiating a power series, check the exponent and index before multiplying by \(x\). - In the second route, compare the sign of the transformed function with the target function before matching coefficients. - The radius is unchanged by termwise differentiation, but endpoint behavior still has to be checked for the transformed series.

1. Replacing \(x\) by \(-x\) gives \(\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^n x^n\). Differentiating, \(-\frac{1}{(1+x)^2}=\sum_{n=1}^{\infty}n(-1)^n x^{n-1}\). Multiplying by \(-x\) gives \(\frac{x}{(1+x)^2}=\sum_{n=1}^{\infty}n(-1)^{n+1}x^n\). 2. Differentiating the original geometric series first gives \(\frac{1}{(1-x)^2}=\sum_{n=1}^{\infty}n x^{n-1}\). Multiplying by \(x\), \(\frac{x}{(1-x)^2}=\sum_{n=1}^{\infty}n x^n\). Replacing \(x\) by \(-x\) gives \(-\frac{x}{(1+x)^2}=\sum_{n=1}^{\infty}n(-1)^n x^n\). Multiplying by \(-1\) again yields \(\frac{x}{(1+x)^2}=\sum_{n=1}^{\infty}n(-1)^{n+1}x^n\). 3. Both routes therefore give the same coefficient of \(x^n\), namely \(n(-1)^{n+1}\). The first five nonzero terms are \(x-2x^2+3x^3-4x^4+5x^5\). Differentiation and the substitution \(x\mapsto -x\) preserve the radius \(1\). At \(x=1\), the terms \(n(-1)^{n+1}\) do not approach \(0\), and at \(x=-1\), the terms equal \(-n\), which also do not approach \(0\). Thus both endpoints diverge and the interval of convergence is \((-1,1)\).

a) \(\frac{x}{(1+x)^2}=\sum_{n=1}^{\infty}n(-1)^{n+1}x^n\). b) The second route gives the same series. c) Coefficient of \(x^n\): \(n(-1)^{n+1}\); first terms \(x-2x^2+3x^3-4x^4+5x^5\); interval \((-1,1)\).

- Use the fact that every term contributes positively to the known value at \(x=\frac12\). - Rewrite powers of \(\frac14\) using powers of \(\frac12\). - Identify exactly where the argument would fail if positive and negative coefficients could cancel. - For part d), identify the exact step that uses each term being nonnegative.

1. Since \( F\left(\frac12\right)=\sum_{n=0}^{\infty}\frac{a_n}{2^n}=3\) and every term is nonnegative, each individual term satisfies \( \frac{a_n}{2^n}\leq3\). Thus \(a_n\leq3\cdot2^n\). 2. The tail at \(x=\frac14\) is \( F\left(\frac14\right)-P_N\left(\frac14\right)=\sum_{n=N+1}^{\infty}\frac{a_n}{4^n}\). Rewrite each term as \( \frac{a_n}{4^n}=\frac{a_n}{2^n}\left(\frac12\right)^n\). 3. For \(n\geq N+1\), \( \left(\frac12\right)^n\leq\left(\frac12\right)^{N+1}\). Therefore, \( F\left(\frac14\right)-P_N\left(\frac14\right)\leq\left(\frac12\right)^{N+1}\sum_{n=N+1}^{\infty}\frac{a_n}{2^n}\leq\frac{3}{2^{N+1}}\). 4. The condition \(\frac{3}{2^{N+1}}<0.001\) requires \(2^{N+1}>3000\). Since \(2^{11}=2048\) and \(2^{12}=4096\), the least value is \(N=11\). 5. If coefficients could have opposite signs, cancellation at \(x=\frac12\) could hide large individual terms and large tails. Nonnegativity prevents that cancellation.

a) \(a_n\leq3\cdot2^n\). b) \( 0\leq F\left(\frac14\right)-P_N\left(\frac14\right)\leq\frac{3}{2^{N+1}}\). c) \(N=11\). d) Nonnegativity prevents cancellation, allowing the known total to control every coefficient and the tail.

- Rewrite the integrand so a geometric ratio involving \(t^2\) is visible. - Termwise integration changes both the powers and the coefficients; then test the two boundary inputs separately. - For the omitted positive tail, compare every denominator from the first omitted term onward with one fixed denominator. - After that comparison, the remaining powers have a common ratio that can be summed geometrically.

1. For \(|t|<1\), \(\frac{1}{1-t^2}=\sum_{n=0}^{\infty}t^{2n}\). Integrating term by term from \(0\) to \(x\) gives \(F(x)=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}\). 2. The series has radius \(1\). At \(x=1\), it becomes \(\sum_{n=0}^{\infty}\frac{1}{2n+1}\), which diverges. At \(x=-1\), it becomes \(-\sum_{n=0}^{\infty}\frac{1}{2n+1}\), which also diverges. Thus the interval of convergence is \((-1, 1)\). 3. Keeping terms through \(x^5\) gives \(F(0.3)\approx0.3+\frac{0.3^3}{3}+\frac{0.3^5}{5}=0.309486\). 4. The omitted tail is \(R=\sum_{n=3}^{\infty}\frac{0.3^{2n+1}}{2n+1}\). For \(n\geq3\), \(\frac{1}{2n+1}\leq\frac17\), so \(0<R\leq\frac17\sum_{n=3}^{\infty}0.3^{2n+1}\). The remaining powers form a geometric series with first term \(0.3^7\) and ratio \(0.3^2\). Therefore, \(0<R\leq\frac{0.3^7}{7(1-0.3^2)}\approx3.433281\times10^{-5}<3.434\times10^{-5}<3.44\times10^{-5}\).

a) \(F(x)=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}\). b) \((-1, 1)\). c) \(F(0.3)\approx0.309486\). d) \(0<R\leq\frac{0.3^7}{7(1-0.3^2)}<3.434\times10^{-5}\).