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Consider the equation \( {y}^{2} + {4y} + {8x} = 4 \) . Put this equation into standard form and graph the parabola. Find the vertex, focus, and directrix. | Solution. We need a perfect square (in this case, using \( y \) ) on the left-hand side of the equation and factor out the coefficient of the non-squared variable (in this case, the \( x \) ) on the other.\n\n\[ {y}^{2} + {4y} + {8x} = 4 \]\n\n\[ {y}^{2} + {4y} = - {8x} + 4 \]\n\n\[ {y}^{2} + {4y} + 4 = - {8x} + 4 + 4\... | Yes |
A satellite dish is to be constructed in the shape of a paraboloid of revolution. If the receiver placed at the focus is located 2 ft above the vertex of the dish, and the dish is to be 12 feet wide, how deep will the dish be? | One way to approach this problem is to determine the equation of the parabola suggested to us by this data. For simplicity, we’ll assume the vertex is \( \left( {0,0}\right) \) and the parabola opens upwards. Our standard form for such a parabola is \( {x}^{2} = {4py} \) . Since the focus is 2 units above the vertex, w... | Yes |
Graph \( \frac{{\left( x + 1\right) }^{2}}{9} + \frac{{\left( y - 2\right) }^{2}}{25} = 1 \) . Find the center, the lines which contain the major and minor axes, the vertices, the endpoints of the minor axis, and the foci. | Solution. We see that this equation is in the standard form of Equation 7.4. Here \( x - h \) is \( x + 1 \) so \( h = - 1 \), and \( y - k \) is \( y - 2 \) so \( k = 2 \) . Hence, our ellipse is centered at \( \left( {-1,2}\right) \) . We see that \( {a}^{2} = 9 \) so \( a = 3 \), and \( {b}^{2} = {25} \) so \( b = 5... | Yes |
Find the equation of the ellipse with foci \( \\left( {2,1}\\right) \) and \( \\left( {4,1}\\right) \) and vertex \( \\left( {0,1}\\right) \) . | Solution. Plotting the data given to us, we have\n\n\n\nFrom this sketch, we know that the major axis is horizontal, meaning \( a > b \) . Since the center is the midpoint of the foci, we know it is \( \\left( {3,1}\... | Yes |
Graph \( {x}^{2} + 4{y}^{2} - {2x} + {24y} + {33} = 0 \) . Find the center, the lines which contain the major and minor axes, the vertices, the endpoints of the minor axis, and the foci. | Solution. Since we have a sum of squares and the squared terms have unequal coefficients, it's a good bet we have an ellipse on our hands. \( {}^{2} \) We need to complete both squares, and then divide, if necessary, to get the right-hand side equal to 1 . \n\n\[ \n{x}^{2} + 4{y}^{2} - {2x} + {24y} + {33} = 0 \n\] \n\n... | Yes |
Find the equation of the ellipse whose vertices are \( \left( {\pm 5,0}\right) \) with eccentricity \( e = \frac{1}{4} \) . | Solution. As before, we plot the data given to us\n\n\n\nFrom this sketch, we know that the major axis is horizontal, meaning \( a > b \) . With the vertices located at \( \left( {\pm 5,0}\right) \), we get \( a = 5 ... | Yes |
Example 7.4.5. Jamie and Jason want to exchange secrets (terrible secrets) from across a crowded whispering gallery. Recall that a whispering gallery is a room which, in cross section, is half of an ellipse. If the room is 40 feet high at the center and 100 feet wide at the floor, how far from the outer wall should eac... | Solution. Graphing the data yields\n\n\n\n100 units wide\n\nIt’s most convenient to imagine this ellipse centered at \( \\left( {0,0}\\right) \) . Since the ellipse is 100 units wide and 40 units tall, we get \( a = ... | Yes |
Graph the equation \( \frac{{\left( x - 2\right) }^{2}}{4} - \frac{{y}^{2}}{25} = 1 \) . Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes. | Solution. We first see that this equation is given to us in the standard form of Equation 7.6. Here \( x - h \) is \( x - 2 \) so \( h = 2 \), and \( y - k \) is \( y \) so \( k = 0 \) . Hence, our hyperbola is centered at \( \left( {2,0}\right) \) . We see that \( {a}^{2} = 4 \) so \( a = 2 \), and \( {b}^{2} = {25} \... | Yes |
Find the equation of the hyperbola with asymptotes \( y = \pm {2x} \) and vertices \( \left( {\pm 5,0}\right) \). | Solution. Plotting the data given to us, we have\n\n\n\nThis graph not only tells us that the branches of the hyperbola open to the left and to the right, it also tells us that the center is \( \left( {0,0}\right) \)... | Yes |
Consider the equation \( 9{y}^{2} - {x}^{2} - {6x} = {10} \). Put this equation in to standard form and graph. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci, and the equations of the asymptotes. | Solution. We need only complete the square on \( x \) :\n\n\[ 9{y}^{2} - {x}^{2} - {6x} = {10} \]\n\n\[ 9{y}^{2} - 1\left( {{x}^{2} + {6x}}\right) = {10} \]\n\n\[ 9{y}^{2} - \left( {{x}^{2} + {6x} + 9}\right) = {10} - 1\left( 9\right) \]\n\n\[ 9{y}^{2} - {\left( x + 3\right) }^{2} = 1 \]\n\n\[ \frac{{y}^{2}}{\frac{1}{9... | Yes |
Jeff is stationed 10 miles due west of Carl in an otherwise empty forest in an attempt to locate an elusive Sasquatch. At the stroke of midnight, Jeff records a Sasquatch call 9 seconds earlier than Carl. If the speed of sound that night is 760 miles per hour, determine a hyperbolic path along which Sasquatch must be l... | Since Jeff hears Sasquatch sooner, it is closer to Jeff than it is to Carl. Since the speed of sound is 760 miles per hour, we can determine how much closer Sasquatch is to Jeff by multiplying\n\n\[ \n{760}\frac{\text{miles}}{\text{hour}} \times \frac{1\text{ hour}}{{3600}\text{ seconds}} \times 9\text{ seconds} = {1.9... | Yes |
Example 7.5.5. By a stroke of luck, Kai was also camping in the woods during the events of the previous example. He was located 6 miles due north of Jeff and heard the Sasquatch call 18 seconds after Jeff did. Use this added information to locate Sasquatch. | Solution. Kai and Jeff are now the foci of a second hyperbola where the fixed distance \( d \) can be determined as before\n\n\[ \n{760}\frac{\\text{miles}}{\\text{hour}} \\times \\frac{1\\text{ hour}}{{3600}\\text{ seconds}} \\times {18}\\text{ seconds} = {3.8}\\text{ miles} \n\]\n\nSince Jeff was positioned at \( \\l... | Yes |
Solve the following systems of equations. Check your answer algebraically and graphically.\n\n1. \( \\begin{cases} {2x} - y & = 1 \\ y & = 3 \\end{cases} \) | Our first system is nearly solved for us. The second equation tells us that \( y = 3 \) . To find the corresponding value of \( x \), we substitute this value for \( y \) into the the first equation to obtain \( {2x} - 3 = 1 \), so that \( x = 2 \) . Our solution to the system is \( \\left( {2,3}\\right) \) . To check ... | Yes |
Find the quadratic function passing through the points \( \left( {-1,3}\right) ,\left( {2,4}\right) ,\left( {5, - 2}\right) \) . | Solution. According to Definition 2.5, a quadratic function has the form \( f\left( x\right) = a{x}^{2} + {bx} + c \) where \( a \neq 0 \) . Our goal is to find \( a, b \) and \( c \) so that the three given points are on the graph of \( f \) . If \( \left( {-1,3}\right) \) is on the graph of \( f \), then \( f\left( {... | Yes |
Theorem 8.4. Properties of Scalar Multiplication\n\n- Associative Property: For every \( m \times n \) matrix \( A \) and scalars \( k \) and \( r,\left( {kr}\right) A = k\left( {rA}\right) \) . | As with the other results in this section, Theorem 8.4 can be proved using the definitions of scalar multiplication and matrix addition. For example, to prove that \( k\left( {A + B}\right) = {kA} + {kB} \) for a scalar \( k \) and \( m \times n \) matrices \( A \) and \( B \), we start by adding \( A \) and \( B \), t... | No |
2. If a point \( P \) is on the hyperbola \( {x}^{2} - {y}^{2} = 4 \), show that the point \( {RP} \) is on the curve \( y = \frac{2}{x} \) . | For a generic point \( P\left( {x, y}\right) \) on the hyperbola \( {x}^{2} - {y}^{2} = 4 \), we have\n\n\[ \n{RP} = \left\lbrack \begin{array}{rr} \frac{\sqrt{2}}{2} & - \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{array}\right\rbrack \left\lbrack \begin{array}{l} x \\ y \end{array}\right\rbrack ... | Yes |
1. If \( A \) is invertible then \( {A}^{-1} \) is unique. | To establish the first property, we assume that \( A \) is invertible and suppose the matrices \( B \) and \( C \) act as inverses for \( A \) . That is, \( {BA} = {AB} = {I}_{n} \) and \( {CA} = {AC} = {I}_{n} \) . We need to show that \( B \) and \( C \) are, in fact, the same matrix. To see this, we note that \( B =... | Yes |
Consider the circuit diagram below. \( {}^{7} \) We have two batteries with source voltages \( V{B}_{1} \) and \( V{B}_{2} \), measured in volts \( V \), along with six resistors with resistances \( {R}_{1} \) through \( {R}_{6} \), measured in kiloohms, \( {k\Omega } \) . Using Ohm’s Law and Kirchhoff’s Voltage Law, w... | The system of linear equations associated with this circuit is\n\n\[ \begin{cases} \left( {{R}_{1} + {R}_{3}}\right) {i}_{1} - {R}_{3}{i}_{2} - {R}_{1}{i}_{4} & = V{B}_{1} \\ - {R}_{3}{i}_{1} + \left( {{R}_{2} + {R}_{3} + {R}_{4}}\right) {i}_{2} - {R}_{4}{i}_{3} - {R}_{2}{i}_{4} & = 0 \\ - {R}_{4}{i}_{2} + \left( {{R}_... | Yes |
Theorem 8.7. Properties of the Determinant: Let \( A = {\left\lbrack {a}_{ij}\right\rbrack }_{n \times n} \) . - We may find the determinant by expanding along any row. That is, for any \( 1 \leq k \leq n \) , \[ \det \left( A\right) = {a}_{k1}{C}_{k1} + {a}_{k2}{C}_{k2} + \ldots + {a}_{kn}{C}_{kn} \] | Unfortunately, while we can easily demonstrate the results in Theorem 8.7, the proofs of most of these properties are beyond the scope of this text. We could prove these properties for generic \( 2 \times 2 \) or even \( 3 \times 3 \) matrices by brute force computation, but this manner of proof belies the elegance and... | No |
Theorem 8.8. Cramer’s Rule: Suppose \( {AX} = B \) is the matrix form of a system of \( n \) linear equations in \( n \) unknowns where \( A \) is the coefficient matrix, \( X \) is the unknowns matrix, and \( B \) is the constant matrix. If \( \det \left( A\right) \neq 0 \), then the corresponding system is consistent... | \[ {x}_{j} = \frac{\det \left( {A}_{j}\right) }{\det \left( A\right) } \] where \( {A}_{j} \) is the matrix \( A \) whose \( j \) th column has been replaced by the constants in \( B \). | Yes |
Solve \( \\begin{cases} 2{x}_{1} - 3{x}_{2} & = 4 \\\\ 5{x}_{1} + {x}_{2} & = - 2 \\end{cases} \) for \( {x}_{1} \) and \( {x}_{2} \) | Writing this system in matrix form, we find\n\n\[ A = \\left\\lbrack \\begin{array}{rr} 2 & - 3 \\\\ 5 & 1 \\end{array}\\right\\rbrack \\;X = \\left\\lbrack \\begin{array}{l} {x}_{1} \\\\ {x}_{2} \\end{array}\\right\\rbrack \\;B = \\left\\lbrack \\begin{array}{r} 4 \\\\ - 2 \\end{array}\\right\\rbrack \]\n\nTo find the... | Yes |
Theorem 8.10. Suppose \( R\left( x\right) = \frac{N\left( x\right) }{D\left( x\right) } \) is a rational function where the degree of \( N\left( x\right) \) less than the degree of \( D\left( x\right) \) and \( N\left( x\right) \) and \( D\left( x\right) \) have no common factors. \( {}^{a} \n\n- If \( \alpha \) is a r... | The proof of Theorem 8.10 is best left to a course in Abstract Algebra. Notice that the theorem provides for the general case, so we need to use subscripts, \( {A}_{1},{A}_{2} \), etc., to denote different unknown coefficients as opposed to the usual convention of \( A, B \), etc.. The stress on multiplicities is to he... | Yes |
Theorem 8.11. Suppose\n\n\[ \n{a}_{n}{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{2}{x}^{2} + {a}_{1}x + {a}_{0} = {b}_{m}{x}^{m} + {m}_{m - 1}{x}^{m - 1} + \cdots + {b}_{2}{x}^{2} + {b}_{1}x + {b}_{0} \n\]\n\nfor all \( x \) in an open interval \( I \) . Then \( n = m \) and \( {a}_{i} = {b}_{i} \) for all \( i = ... | Believe it or not, the proof of Theorem 8.11 is a consequence of Theorem 3.14. Define \( p\left( x\right) \) to be the difference of the left hand side of the equation in Theorem 8.11 and the right hand side. Then \( p\left( x\right) = 0 \) for all \( x \) in the open interval \( I \) . If \( p\left( x\right) \) were a... | Yes |
Resolve the following rational functions into partial fractions.\n\n1. \( R\left( x\right) = \frac{x + 5}{2{x}^{2} - x - 1} \) | We begin by factoring the denominator to find \( 2{x}^{2} - x - 1 = \left( {{2x} + 1}\right) \left( {x - 1}\right) \) . We get \( x = - \frac{1}{2} \) and \( x = 1 \) are both zeros of multiplicity one and thus we know\n\n\[ \frac{x + 5}{2{x}^{2} - x - 1} = \frac{x + 5}{\left( {{2x} + 1}\right) \left( {x - 1}\right) } ... | Yes |
1. \( \\begin{cases} {x}^{2} + {y}^{2} & = 4 \\ 4{x}^{2} + 9{y}^{2} & = {36} \\end{cases} \) | Since both equations contain \( {x}^{2} \) and \( {y}^{2} \) only, we can eliminate one of the variables as we did in Section 8.1.\n\n\[ \n\\left\\{ \\begin{matrix} \\left( {E1}\\right) & {x}^{2} + {y}^{2} & = & 4 \\ \\left( {E2}\\right) & 4{x}^{2} + 9{y}^{2} & = & {36} \\end{matrix}\\right. \\;\\xrightarrow[{-{4E1} + ... | No |
Solve the following systems of equations. Verify your answers algebraically and graphically, as appropriate.\n\n1. \( \\left\\{ \\begin{array}{l} {x}^{2} + {2xy} - {16} = 0 \\\\ {y}^{2} + {2xy} - {16} = 0 \\end{array}\\right. \) | At first glance, it doesn't appear as though elimination will do us any good since it's clear that we cannot completely eliminate one of the variables. The alternative, solving one of the equations for one variable and substituting it into the other, is full of unpleasantness. Returning to elimination, we note that it ... | Yes |
Sketch the solution to the following nonlinear inequalities in the plane.\n\n1. \( {y}^{2} - 4 \leq x < y + 2 \) | The inequality \( {y}^{2} - 4 \leq x < y + 2 \) is a compound inequality. It translates as \( {y}^{2} - 4 \leq x \) and \( x < y + 2 \) . As usual, we solve each inequality and take the set theoretic intersection to determine the region which satisfies both inequalities. To solve \( {y}^{2} - 4 \leq x \), we write\n\n-... | Yes |
Write the first four terms of the following sequences.\n\n1. \( {a}_{n} = \frac{{5}^{n - 1}}{{3}^{n}}, n \geq 1 \) 2. \( {b}_{k} = \frac{{\left( -1\right) }^{k}}{{2k} + 1}, k \geq 0 \)\n\n3. \( \{ {2n} - 1{\} }_{n = 1}^{\infty } \) 4. \( {\left\{ \frac{1 + {\left( -1\right) }^{i}}{i}\right\} }_{i = 2}^{\infty } \)\n\n5... | ## Solution.\n\n1. Since we are given \( n \geq 1 \), the first four terms of the sequence are \( {a}_{1},{a}_{2},{a}_{3} \) and \( {a}_{4} \) . Since the notation \( {a}_{1} \) means the same thing as \( a\left( 1\right) \), we obtain our first term by replacing every occurrence of \( n \) in the formula for \( {a}_{n... | Yes |
Determine if the following sequences are arithmetic, geometric or neither. If arithmetic, find the common difference \( d \) ; if geometric, find the common ratio \( r \) .\n\n1. \( {a}_{n} = \frac{{5}^{n - 1}}{{3}^{n}}, n \geq 1 \) | From Example 9.1.1, we know that the first four terms of this sequence are \( \frac{1}{3},\frac{5}{9},\frac{25}{27} \) and \( \frac{125}{81} \) . To see if this is an arithmetic sequence, we look at the successive differences of terms. We find that \( {a}_{2} - {a}_{1} = \frac{5}{9} - \frac{1}{3} = \frac{2}{9} \) and \... | Yes |
Find an explicit formula for the \( {n}^{\text{th }} \) term of the following sequences.\n\n1. \( {0.9},{0.09},{0.009},{0.0009},\ldots \) | Although this sequence may seem strange, the reader can verify it is actually a geometric sequence with common ratio \( r = {0.1} = \frac{1}{10} \) . With \( a = {0.9} = \frac{9}{10} \), we get \( {a}_{n} = \frac{9}{10}{\left( \frac{1}{10}\right) }^{n - 1} \) for \( n \geq 0 \) . Simplifying, we get \( {a}_{n} = \frac{... | No |
1. If monthly payments of \$50 are made, find the value of the annuity in 30 years. | We have \( r = {0.06} \) and \( n = {12} \) so that \( i = \frac{r}{n} = \frac{0.06}{12} = {0.005} \) . With \( P = {50} \) and \( t = {30} \) ,\n\n\[ A = \frac{{50}\left( {{\left( 1 + {0.005}\right) }^{\left( {12}\right) \left( {30}\right) } - 1}\right) }{0.005} \approx {50225.75} \]\n\nOur final answer is \$50,225.75... | Yes |
Theorem 9.2. Geometric Series: Given the sequence \( {a}_{k} = a{r}^{k - 1} \) for \( k \geq 1 \), where \( \left| r\right| < 1 \), \[ a + {ar} + a{r}^{2} + \ldots = \mathop{\sum }\limits_{{k = 1}}^{\infty }a{r}^{k - 1} = \frac{a}{1 - r} \] If \( \left| r\right| \geq 1 \), the sum \( a + {ar} + a{r}^{2} + \ldots \) is ... | The justification of the result in Theorem 9.2 comes from taking the formula in Equation 9.2 for the sum of the first \( n \) terms of a geometric sequence and examining the formula as \( n \rightarrow \infty \) . Assuming \( \left| r\right| < 1 \) means \( - 1 < r < 1 \), so \( {r}^{n} \rightarrow 0 \) as \( n \righta... | Yes |
1. The sum formula for arithmetic sequences: \( \mathop{\sum }\limits_{{j = 1}}^{n}\left( {a + \left( {j - 1}\right) d}\right) = \frac{n}{2}\left( {{2a} + \left( {n - 1}\right) d}\right) \) . | We set \( P\left( n\right) \) to be the equation we are asked to prove. For \( n = 1 \), we compare both sides of the equation given in \( P\left( n\right) \)\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{1}\left( {a + \left( {j - 1}\right) d}\right) \overset{?}{ = }\frac{1}{2}\left( {{2a} + \left( {1 - 1}\right) d}\right) \]... | Yes |
Theorem 9.3. For natural numbers \( n \) and \( j \) with \( n \geq j \) , \n\n\[ \n\left( \begin{matrix} n \\ j - 1 \end{matrix}\right) + \left( \begin{array}{l} n \\ j \end{array}\right) = \left( \begin{matrix} n + 1 \\ j \end{matrix}\right) \n\] | \[ \n\left( \begin{matrix} n \\ j - 1 \end{matrix}\right) + \left( \begin{array}{l} n \\ j \end{array}\right) = \frac{n!}{\left( {j - 1}\right) !\left( {n - \left( {j - 1}\right) }\right) !} + \frac{n!}{j!\left( {n - j}\right) !} \n\] \n\n\[ \n= \frac{n!}{\left( {j - 1}\right) !\left( {n - j + 1}\right) !} + \frac{n!}{... | Yes |
Theorem 9.4. Binomial Theorem: For nonzero real numbers \( a \) and \( b \) ,\n\n\[{\left( a + b\right) }^{n} = \mathop{\sum }\limits_{{j = 0}}^{n}\left( \begin{array}{l} n \\ j \end{array}\right) {a}^{n - j}{b}^{j}\]\n\nfor all natural numbers \( n \) . | To prove the Binomial Theorem, we let \( P\left( n\right) \) be the expansion formula given in the statement of the theorem and we note that \( P\left( 1\right) \) is true since\n\n\[{\left( a + b\right) }^{1}\overset{?}{ = }\mathop{\sum }\limits_{{j = 0}}^{1}\left( \begin{array}{l} 1 \\ j \end{array}\right) {a}^{1 - j... | Yes |
Example 9.4.2. Use the Binomial Theorem to find the following.\n\n1. \( {\left( x - 2\right) }^{4} \) 2. \( {2.1}^{3} \)\n\n3. The term containing \( {x}^{3} \) in the expansion \( {\left( 2x + y\right) }^{5} \) | Solution.\n\n1. Since \( {\left( x - 2\right) }^{4} = {\left( x + \left( -2\right) \right) }^{4} \), we identify \( a = x, b = - 2 \) and \( n = 4 \) and obtain\n\n\[ \n{\left( x - 2\right) }^{4} = \mathop{\sum }\limits_{{j = 0}}^{4}\left( \begin{array}{l} 4 \\ j \end{array}\right) {x}^{4 - j}{\left( -2\right) }^{j} \n... | No |
The equation \( y = \arccos \left( {-\frac{1}{2}}\right) = {\cos }^{-1}\left( {-\frac{1}{2}}\right) \) means that\n\n\[ \cos \left( y\right) = - \frac{1}{2}\text{ and }0 \leq y \leq \pi . \] | That is, we are trying to find the arc \( y \) whose cosine is \( - \frac{1}{2} \) and \( 0 \leq y \leq \pi \) . Using our knowledge of cosine values for common arcs, we notice that \( \cos \left( \frac{\pi }{3}\right) = \frac{1}{2} \) . So we conclude that the reference angle \( \widehat{y} \) for \( y \) is \( \wideh... | Yes |
For the equation \( \cos \left( x\right) = {0.7} \) | we first use the result about the inverse cosine function on page 150, which states that for \( t \) in the closed interval \( \left\lbrack {0,\pi }\right\rbrack \) ,\n\n\[{\cos }^{-1}\left( {\cos \left( t\right) }\right) = t\]\n\nSo we \ | No |
We will determine the exact value of \( \cos \left( {\arcsin \left( {-\frac{2}{7}}\right) }\right) \) . | We let\\theta = \arcsin \left( {-\frac{2}{7}}\right) . We then know that\n\n\\sin \left( \\theta \\right) = - \\frac{2}{7}\\text{ and } - \\frac{\\pi }{2} \\leq \\theta \\leq \\frac{\\pi }{2}\n\nWe note that since \\sin \left( \\theta \\right) < 0, we actually know that - \\frac{\\pi }{2} \\leq \\theta \\leq 0 .\n\nSo ... | Yes |
Example 3.8 Suppose that one of the acute angles of a right triangle has a measure of \( {35}^{ \circ } \) and that the side adjacent to this angle is 8 inches long. Determine the other acute angle of the right triangle and the lengths of the other two sides. | Solution. The first thing we do is draw a picture of the triangle. (The picture does not have to be perfect but it should reasonably reflect the given information.) In making the diagram, we should also label the unknown parts of the triangle. One wav to do this is shown in the diagram.\n\n![835f3845-8e2a-4ca2-8cfb-515... | Yes |
Consider the equation with the equation \( \cos \left( {x - \frac{\pi }{2}}\right) = \sin \left( {x + \frac{\pi }{2}}\right) \) that we encountered in our Beginning Activity. | Although you can check that \( \cos \left( {x - \frac{\pi }{2}}\right) \) and \( \sin \left( {x + \frac{\pi }{2}}\right) \) are equal at some values, \( \frac{\pi }{4} \) for example, they are not equal at all values \( - \cos \left( {0 - \frac{\pi }{2}}\right) = 0 \) but \( \sin \left( {0 + \frac{\pi }{2}}\right) = 1 ... | Yes |
Consider the equation\n\n\\[ \n2\\sin \\left( x\\right) = 1\\text{.}\n\\]\n\nWe want to find all values of \\( x \\) that satisfy this equation. | Notice that this equation looks a lot like the linear equation \\( {2y} = 1 \\), with \\( \\sin \\left( x\\right) \\) in place of \\( y \\) . So this trigonometric equation is of linear type and we say that it is linear in \\( \\sin \\left( x\\right) \\) . We know how to solve \\( {2y} = 1 \\), we simply divide both si... | Yes |
Consider the trigonometric equation\n\n\\[ \n{\\cos }^{2}\\left( x\\right) - {\\sin }^{2}\\left( x\\right) = 1.\n\\]\n\n(7) | This equation is complicated by the fact that there are two different trigonometric functions involved. In this case we use the Pythagorean Identity\n\n\\[ \n{\\sin }^{2}\\left( x\\right) + {\\cos }^{2}\\left( x\\right) = 1\n\\]\n\nby solving for \\( {\\cos }^{2}\\left( x\\right) \\) to obtain\n\n\\[ \n{\\cos }^{2}\\le... | Yes |
Let us return to our problem of finding \( \cos \left( \frac{\pi }{12}\right) \) . Since we know \( \frac{\pi }{12} = \frac{\pi }{3} - \frac{\pi }{4} \) , we can use the Cosine Difference Identity with \( A = \frac{\pi }{3} \) and \( B = \frac{\pi }{4} \) to obtain \( \cos \left( \frac{\pi }{12}\right) = \cos \left( {\... | \[ \cos \left( \frac{\pi }{12}\right) = \cos \left( {\frac{\pi }{3} - \frac{\pi }{4}}\right) \] \[ = \cos \left( \frac{\pi }{3}\right) \cos \left( \frac{\pi }{4}\right) + \sin \left( \frac{\pi }{3}\right) \sin \left( \frac{\pi }{4}\right) \] \[ = \left( \frac{1}{2}\right) \left( \frac{\sqrt{2}}{2}\right) + \left( \frac... | Yes |
Consider the equation\n\n\\[ \cos \\left( \\theta \\right) \\cos \\left( \\frac{\\pi }{5}\\right) - \\sin \\left( \\theta \\right) \\sin \\left( \\frac{\\pi }{5}\\right) = \\frac{\\sqrt{3}}{2}. \\]\n\nOn the surface this equation looks quite complicated, but we can apply an identity to simplify it to the point where it... | Notice that left side of this equation has the form \\( \\cos \\left( A\\right) \\cos \\left( B\\right) - \\sin \\left( A\\right) \\sin \\left( B\\right) \\) with \\( A = \\theta \\) and \\( B = \\frac{\\pi }{5} \\). We can use the Cosine Sum Identity \\( \\cos \\left( {A + B}\\right) = \\cos \\left( A\\right) \\cos \\... | Yes |
Consider the equation\n\n\\[ \n2\\cos \\left( {2\\theta }\\right) - 1 = 0 \n\\] | This is an equation that is linear in \\( \\cos \\left( {2\\theta }\\right) \\), so we can apply the same ideas as we did earlier to this equation. We solve for \\( \\cos \\left( {2\\theta }\\right) \\) to see that\n\n\\[ \n\\cos \\left( {2\\theta }\\right) = \\frac{1}{2} \n\\]\n\nWe know the angles at which the cosine... | Yes |
Let us return to the problem stated at the beginning of this section to solve the equation\n\n\\[ \n\\sin \\left( {3x}\\right) + \\sin \\left( x\\right) = 0.\n\\] | Using the Sum-to-Product\n\n\\[ \n\\sin \\left( A\\right) + \\sin \\left( B\\right) = 2\\sin \\left( \\frac{A + B}{2}\\right) \\cos \\left( \\frac{A - B}{2}\\right)\n\\]\n\nwith \\( A = x \\) and \\( B = {3x} \\) we can rewrite the equation as follows:\n\n\\[ \n\\sin \\left( {3x}\\right) + \\sin \\left( x\\right) = 0\n... | Yes |
We will find the solutions to the equation\n\n\[ \n{x}^{4} = - 8 + 8\sqrt{3}i \n\] | Note that we can write the right hand side of this equation in trigonometric form as\n\n\[ \n- 8 + 8\sqrt{3}i = {16}\left( {\cos \left( \frac{2\pi }{3}\right) + i\sin \left( \frac{2\pi }{3}\right) }\right) .\n\]\n\nThe fourth roots of \( - 8 + 8\sqrt{3}i \) are then\n\n\[ \n{x}_{0} = \sqrt[4]{16}\left\lbrack {\cos \lef... | Yes |
Example 5.19 (Converting from Rectangular to Polar Coordinates)\n\nTo determine polar coordinates for the\n\nthe point \( \\left( {-2,2}\\right) \) in rectangular coordi-\nnates, we first draw a picture and note that\n\n\\[ \nr^2 = \\left( -2\\right)^2 + 2^2 = 8.\n\\] | Since it is usually easier to work with a positive value for \( r \), we will use \( r = \\sqrt{8} \).\n\nWe also see that \( \\tan \\left( \\theta \\right) = \\frac{3}{-3} = - 1 \). We can use many different values for \( \\theta \) but to keep it easy, we use \( \\theta \) as shown in the diagram. For the reference a... | Yes |
Problem 1. As you saw when you filled in the details of our development of the Quadratic Formula 4 the substitution \( x = y - \frac{b}{2a} \) was crucial because it turned\n\n\[ \n{x}^{2} + \frac{b}{a}x + \frac{c}{a} = 0 \n\]\n\ninto\n\n\[ \n{y}^{2} = k \n\] | where \( k \) depends only on \( a, b \), and \( c \) . In the sixteenth century a similar technique was used by Ludovico Ferrari (1522-1565) to reduce the general cubic equation\n\n\[ \na{x}^{3} + b{x}^{2} + {cx} + d = 0 \n\]\n\n(6)\n\ninto the so-called \ | No |
Suppose that \( {r}_{1} \) and \( {r}_{2} \) are solutions of \( a{x}^{2} + {bx} + c = 0 \) . Suppose further that \( {r}_{1} \geq {r}_{2} \) . Then\n\n\[ a{x}^{2} + {bx} + c = a\left( {x - {r}_{1}}\right) \left( {x - {r}_{2}}\right) \]\n\n\[ = a\left\lbrack {{x}^{2} - \left( {{r}_{1} + {r}_{2}}\right) x + {\left( {r}_... | Equations 7 and 8 can be solved simultaneously to yield\n\n\[ {r}_{1} = \frac{-b + \sqrt{{b}^{2} - {4ac}}}{2a} \]\n\n\[ {r}_{2} = \frac{-b - \sqrt{{b}^{2} - {4ac}}}{2a}. \] | Yes |
Problem 3. Let \( p \) be a prime number and \( a, b \) positive integers such that \( p \mid \left( {a \cdot b}\right) \) . Show that \( p \mid a \) or \( p \mid b \) . | [Hint: If \( p \mid a \) then we are done. If not then notice that \( p \) is a prime factor of \( a \cdot b \) . What does the Fundamental Theorem of Arithmetic say about the prime factors of \( a \cdot b \) compared to the prime factors of \( a \) and \( b \) ?] | No |
Let \( a, b, c, d \in \mathbb{N} \) and find a rational number between \( a/b \) and \( c/d \) . | Consider again the rational numbers \( a/b \) and \( c/d \) . If we think of these as lengths we can ask,\ | No |
Theorem 2. Let \( a, b, c \), and \( d \) be integers. There is a number \( \alpha \in \mathbb{Q} \) such that \( {M\alpha } = a/b \) and \( {N\alpha } = c/d \) where \( M \) and \( N \) are also integers. | Proof: To prove this theorem we will display \( \alpha, M \) and \( N \) . It is your responsibility to confirm that these actually work. Here they are: \( \alpha = 1/{bd}, M = {ad} \) , and \( N = {cb} \) . | No |
Problem 8. Confirm that \( \alpha, M \), and \( N \) as given in the proof of Theorem 2 satisfy the requirements of the theorem. | It should be clear that it is necessary for \( a, b, c \), and \( d \) to be integers for everything to work out. Otherwise \( M \) and \( N \) will not also be integers as required. | No |
Between any two distinct real numbers there is a rational number. | Sketch of Proof: We will outline the proof of part (a) of Theorem 3 and indicate how it can be used to prove part b.\n\nLet \( \alpha \) and \( \beta \) be real numbers with \( \alpha > \beta \) . There are two cases.\n\nCase 1: \( \alpha - \beta > 1 \) . In this case there is at least one integer between \( \alpha \) ... | No |
Show that the equations \( x = \frac{t - \sin t}{{4q}{c}^{2}}, y = \frac{1 - \cos t}{{4q}{c}^{2}} \) satisfy equation 2.4. | Bernoulli recognized this solution to be an inverted cycloid, the curve traced by a fixed point on a circle as the circle rolls along a horizontal surface. | No |
Solve the following Initial Value problem 4 Find \( y\left( x\right) \) given that \( \frac{\mathrm{d}y}{\mathrm{\;d}x} = y, y\left( 0\right) = 1 \) . | Assuming the solution can be expressed as a power series we have\n\n\[ y = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{x}^{n} = {a}_{0} + {a}_{1}x + {a}_{2}{x}^{2} + \cdots . \]\n\nDifferentiating gives us\n\n\[ \frac{\mathrm{d}y}{\mathrm{\;d}x} = {a}_{1} + 2{a}_{2}x + 3{a}_{3}{x}^{2} + 4{a}_{4}{x}^{3} + \ldots \]... | Yes |
Property 2. \( E\left( {x + y}\right) = E\left( x\right) E\left( y\right) \) . | To see this we multiply the two series together, so we have\n\n\[ E\left( x\right) E\left( y\right) = \left( {\mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{n!}{x}^{n}}\right) \left( {\mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{n!}{y}^{n}}\right) \]\n\n\[ = \left( {\frac{{x}^{0}}{0!} + \frac{{x}^{1}}{1!} + \fr... | Yes |
Problem 20. Prove Property 6. | In light of Property 6, we see that for any rational number \( r, E\left( r\right) = {e}^{r} \) . Not only does this give us the series representation \( {e}^{r} = \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{n!}{r}^{n} \) for any rational number \( r \), but it gives us a way to define \( {e}^{x} \) for irrationa... | Yes |
Use the geometric series, \( \frac{1}{1 - x} = 1 + x + {x}^{2} + {x}^{3} + \cdots = \mathop{\sum }\limits_{{n = 0}}^{\infty }{x}^{n} \), to obtain a series for \( \frac{1}{1 + {x}^{2}} \) and use this to obtain the series | \[ \arctan x = x - \frac{1}{3}{x}^{3} + \frac{1}{5}{x}^{5} - \cdots = \mathop{\sum }\limits_{{n = 0}}^{\infty }{\left( -1\right) }^{n}\frac{1}{{2n} + 1}{x}^{{2n} + 1}. \] | No |
Problem 26. Use the series \( {\left( 1 + x\right) }^{\frac{1}{2}} = \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{\mathop{\prod }\limits_{{j = 0}}^{{n - 1}}\left( {\frac{1}{2} - j}\right) }{n!}{x}^{n} \) to obtain the series | \[ \frac{\pi }{4} = {\int }_{x = 0}^{1}\sqrt{1 - {x}^{2}}\mathrm{\;d}x = \mathop{\sum }\limits_{{n = 0}}^{\infty }\left( \frac{\mathop{\prod }\limits_{{j = 0}}^{{n - 1}}\left( {\frac{1}{2} - j}\right) }{n!}\right) \left( \frac{{\left( -1\right) }^{n}}{{2n} + 1}\right) = 1 - \frac{1}{6} - \frac{1}{40} - \frac{1}{112} - ... | Yes |
Problem 28. Let \( k \) be a positive integer. Find the power series, centered at zero, for \( f\left( x\right) = {\left( 1 - x\right) }^{-k} \) | (a) Differentiating the geometric series \( \left( {k - 1}\right) \) times.\n\n(b) Applying the binomial series.\n\n(c) Compare these two results. | No |
Problem 30. Use the geometric series to obtain the series\n\n\[ \ln \left( {1 + x}\right) = x - \frac{1}{2}{x}^{2} + \frac{1}{3}{x}^{3} - \cdots \] | \[ = \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{{\left( -1\right) }^{n}}{n + 1}{x}^{n + 1}. \] | Yes |
Problem 36. Prove Theorem 4. | From Theorem 4 we see that if we do start with the function \( f\left( x\right) \) then no matter how we obtain its power series, the result will always be the same. The series\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{{f}^{\left( n\right) }\left( a\right) }{n!}{\left( x - a\right) }^{n} = f\left( a\right) +... | No |
Theorem 5. If \( {f}^{\prime },{f}^{\prime \prime },\ldots ,{f}^{\left( n + 1\right) } \) are all continuous on an interval containing a and \( x \), then\n\n\[ f\left( x\right) = f\left( a\right) + \frac{{f}^{\prime }\left( a\right) }{1!}\left( {x - a}\right) + \frac{{f}^{\prime \prime }\left( a\right) }{2!}{\left( x ... | Notice that the case when \( n = 0 \) is really a restatement of the Fundamental Theorem of Calculus. Specifically, the FTC says \( {\int }_{t = a}^{x}{f}^{\prime }\left( t\right) \mathrm{d}t = f\left( x\right) - f\left( a\right) \) which we can rewrite as\n\n\[ f\left( x\right) = f\left( a\right) + \frac{1}{0!}{\int }... | Yes |
Theorem 6. Let a be any real number. There exists a rearrangement of the series \( 1 - \frac{1}{2} + \frac{1}{3} - \cdots \) which converges to a. | Armed with this fact, we can see why Theorem 6 is true. First note that\n\n\[ - \frac{1}{2} - \frac{1}{4} - \frac{1}{6} - \cdots = - \frac{1}{2}\left( {1 + \frac{1}{2} + \frac{1}{3} + \cdots }\right) = - \infty \]\n\nand\n\n\[ 1 + \frac{1}{3} + \frac{1}{5} + \cdots \geq \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots ... | Yes |
Problem 44. Show that there is a rearrangement of \( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots \) which diverges to \( - \infty \) . | It is fun to know that we can rearrange some series to make them add up to anything you like but there is a more fundamental idea at play here. That the negative terms of the alternating Harmonic Series diverge to negative infinity and the positive terms diverge to positive infinity make the convergence of the alternat... | No |
Example 3. Use the definition of convergence to zero to prove\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{\sin n}{n} = 0 \] | Proof: Let \( \varepsilon > 0 \) . Let \( N = \frac{1}{\varepsilon } \) . If \( n > N \), then \( n > \frac{1}{\varepsilon } \) and \( \frac{1}{n} < \varepsilon \) . Thus\n\n\( \left| \frac{\sin n}{n}\right| \leq \frac{1}{n} < \varepsilon \) . Hence by definition, \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac... | Yes |
Use the definition of convergence to zero to prove\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{n + 4}{{n}^{2} + 1} = 0 \] | Proof: Let \( \varepsilon > 0 \) . Let \( N = \max \left( {4,\frac{2}{\varepsilon }}\right) \) . If \( n > N \), then \( n > 4 \) and \( n > \frac{2}{\varepsilon } \) . Thus we have \( n > 4 \) and \( \frac{2}{n} < \varepsilon \) . Therefore\n\n\[ \left| \frac{n + 4}{{n}^{2} + 1}\right| = \frac{n + 4}{{n}^{2} + 1} < \f... | Yes |
Example 5. Use the definition to prove that the sequence\n\n\[ \n{\left( 1 + {\left( -1\right) }^{n}\right) }_{n = 0}^{\infty } = \left( {2,0,2,0,2,\ldots }\right) \n\]\n\ndoes not converge to zero. | Before we provide this proof, let’s analyze what it means for a sequence \( \left( {s}_{n}\right) \) to not converge to zero. Converging to zero means that any time a distance \( \varepsilon > 0 \) is given, we must be able to respond with a number \( N \) such that \( \left| {s}_{n}\right| < \varepsilon \) for every \... | Yes |
Prove \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{n}{n + {100}} = 1 \) . | Proof: Let \( \varepsilon > 0 \) . Let \( N = \frac{100}{\varepsilon } \) . If \( n > N \), then \( n > \frac{100}{\varepsilon } \) and so \( \frac{100}{n} < \varepsilon \) .\n\nHence\n\[ \left| {\frac{n}{n + {100}} - 1}\right| = \left| \frac{n - \left( {n + {100}}\right) }{n + {100}}\right| = \frac{100}{n + {100}} < \... | Yes |
Problem 62. Let \( b > 0 \) . Use the definition to prove \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{b}^{\left( \frac{1}{n}\right) } = 1 \) . | [Hint: You will probably need to separate this into two cases: \( 0 < b < 1 \) and \( b \geq 1 \).] | No |
Theorem 7. If \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = a \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} = b \), then \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{a}_{n} + {b}_{n}}\right) = a + b \) . | SCRAPWORK: If we let \( \varepsilon > 0 \), then we want \( N \) so that if \( n > N \), then \( \left| {\left( {{a}_{n} + {b}_{n}}\right) - \left( {a + b}\right) }\right| < \varepsilon \) . We know that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = a \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty... | Yes |
Theorem 8. If \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = a \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} = b \), then \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{a}_{n}{b}_{n}}\right) = {ab} \) . | Given \( \varepsilon > 0 \), we want \( N \) so that if \( n > N \), then \( \left| {{a}_{n}{b}_{n} - {ab}}\right| < \) \( \varepsilon \) . One of the standard tricks in analysis is to \ | No |
Problem 67. Prove Lemma 2. | [Hint: We know that there exists \( N \) such that if \( n > N \), then \( \left| {{a}_{n} - a}\right| < 1 \) . Let \( B = \max \left( {\left| {a}_{1}\right| ,\left| {a}_{2}\right| ,\ldots ,\left| {a}_{\lceil N\rceil }\right| ,\left| a\right| + 1}\right) \), where \( \lceil N\rceil \) represents the smallest integer gr... | No |
Theorem 9. Suppose \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = a \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} = b \) . Also suppose \( b \neq 0 \) and \( {b}_{n} \neq 0,\forall n \) . Then \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( \frac{{a}_{n}}{{b}_{n}}\right) = \frac... | To prove this, let's look at the special case of trying to prove \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( \frac{1}{{b}_{n}}\right) = \frac{1}{b} \) . The general case will follow from this and Theorem 8 Consider \( \left| {\frac{1}{{b}_{n}} - \frac{1}{b}}\right| = \frac{\left| b - {b}_{n}\right| }{\left... | Yes |
Identify all of the theorems implicitly used to show that\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{3{n}^{3} - {100n} + 1}{5{n}^{3} + 4{n}^{2} - 7} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{n}^{3}\left( {3 - \frac{100}{{n}^{2}} + \frac{1}{{n}^{3}}}\right) }{{n}^{3}\left( {5 + \frac{4}{n} ... | Notice that this presumes that all of the individual limits exist. This will become evident as the limit is decomposed. | No |
Problem 72. Prove Theorem 10. | The Squeeze Theorem holds even if \( {r}_{n} \leq {s}_{n} \leq {t}_{n} \) holds for only sufficiently large \( n \) ; i.e., for \( n \) larger than some fixed \( {N}_{0} \) . This is true because when you find an \( {N}_{1} \) that works in the original proof, this can be modified by choosing \( N = \max \left( {{N}_{0... | Yes |
Prove \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{n + 1}{{n}^{2}} = 0 \) . | Proof: Notice that \( 0 \leq \frac{n + 1}{{n}^{2}} \leq \frac{n + n}{{n}^{2}} = \frac{2}{n} \) . Since \( \mathop{\lim }\limits_{{n \rightarrow \infty }}0 = 0 = \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{2}{n} \), then by the Squeeze Theorem, \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{n + 1}{{n}^... | Yes |
Consider the sequence, \( {\left( n\right) }_{n = 1}^{\infty } \). This clearly diverges by getting larger and larger ... Ooops! Let’s be careful. The sequence \( {\left( 1 - \frac{1}{n}\right) }_{n = 1}^{\infty } \) gets larger and larger too, but it converges. What we meant to say was that the terms of the sequence \... | To show divergence we must show that the sequence satisfies the negation of the definition of convergence. That is, we must show that for every \( r \in \mathbb{R} \) there is an \( \varepsilon > 0 \) such that for every \( N \in \mathbb{R} \), there is an \( n > N \) with \( \left| {n - r}\right| \geq \varepsilon \). ... | Yes |
Let \( \left( {s}_{n}\right) \) and \( \left( {t}_{n}\right) \) be sequences with \( {s}_{n} \leq {t}_{n},\forall n \) . Suppose \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{s}_{n} = s \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{t}_{n} = t \) . Prove \( s \leq t \) . | [Hint: Assume for contradiction, that \( s > t \) and use the definition of convergence with \( \varepsilon = \frac{s - t}{2} \) to produce an \( n \) with \( {s}_{n} > {t}_{n} \). | No |
Problem 87. Suppose \( \left( {s}_{n}\right) \) is a sequence of positive numbers such that\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( \frac{{s}_{n + 1}}{{s}_{n}}\right) = L \]\n\n(a) Prove that if \( L < 1 \), then \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{s}_{n} = 0 \) . | [Hint: Choose \( R \) with \( L < R < 1 \) . By the previous problem, \( \exists N \) such that if \( n > N \), then \( \frac{{s}_{n + 1}}{{s}_{n}} < R \) . Let \( {n}_{0} > N \) be fixed and show \( {s}_{{n}_{0} + k} < {R}^{k}{s}_{{n}_{0}} \) . Conclude that \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{s}_{{n}_{... | No |
Problem 89. Prove Theorem 11. | [Hint: You might want to use Problem 87 of Chapter 4. Also there are two cases to consider: \( a < x \) and \( x < a \) (the case \( x = a \) is trivial). You will find that this is true in general. This is why we will often indicate that \( t \) is between \( a \) and \( x \) as in the theorem. In the case \( x < a \)... | No |
Problem 92. Prove Theorem 12 for the case where \( x < a \) . | [Hint: Note that\n\n\[{\int }_{t = a}^{x}{f}^{\left( n + 1\right) }\left( t\right) {\left( x - t\right) }^{n}\mathrm{\;d}t = {\left( -1\right) }^{n + 1}{\int }_{t = x}^{a}{f}^{\left( n + 1\right) }\left( t\right) {\left( t - x\right) }^{n}\mathrm{\;d}t.\n\]\n\nUse the same argument on this integral. It will work out in... | No |
This problem investigates the Taylor series representation\n\n\[ \frac{1}{1 + x} = 1 - x + {x}^{2} - {x}^{3} + \cdots \] | (a) Use the fact that \( \frac{1 - {\left( -x\right) }^{n + 1}}{1 + x} = 1 - x + {x}^{2} - {x}^{3} + \cdots + {\left( -x\right) }^{n} \) to compute the remainder\n\n\[ \frac{1}{1 + x} - \left( {1 - x + {x}^{2} - {x}^{3} + \cdots + {\left( -x\right) }^{n}}\right) . \]\n\nSpecifically, compute this remainder when \( x = ... | No |
Theorem 13. The binomial series\n\n\[ 1 + \frac{1}{2}x + \frac{\frac{1}{2}\left( {\frac{1}{2} - 1}\right) }{2!}{x}^{2} + \frac{\frac{1}{2}\left( {\frac{1}{2} - 1}\right) \left( {\frac{1}{2} - 2}\right) }{3!}{x}^{3} + \cdots \]\n\nconverges to \( \sqrt{1 + x} \) for \( x \in \left\lbrack {0,1}\right\rbrack \) . | Proof: First note that the binomial series is, in fact, the Taylor series for the function \( f\left( x\right) = \sqrt{1 + x} \) expanded about \( a = 0 \) . If we let \( x \) be a fixed number with \( 0 \leq x \leq 1 \), then it suffices to show that the Lagrange form of the remainder converges to 0 . With this in min... | Yes |
Problem 97. Suppose \( - 1 < x \leq c \leq 0 \) and consider the function \( g\left( c\right) = \frac{c - x}{1 + c} \) . Show that on \( \left\lbrack {x,0}\right\rbrack, g \) is increasing and use this to conclude that for \( - 1 < x \leq \) \( c \leq 0 \)\n\n\[ \frac{c - x}{1 + c} \leq \left| x\right| \] | Use this fact to finish the proof that the binomial series converges to \( \sqrt{1 + x} \) for \( - 1 < x < 0 \) . | No |
Example 9. Use the definition of continuity to show that \( f\left( x\right) = x \) is continuous at any point \( a \) . | Proof: Let \( \varepsilon > 0 \) . Let \( \delta = \varepsilon \) . If \( \left| {x - a}\right| < \delta \), then\n\n\[ \left| {f\left( x\right) - f\left( a\right) }\right| = \left| {x - a}\right| < \varepsilon \]\n\nThus by the definition, \( f \) is continuous at \( a \) . | Yes |
Example 10. Use the definition of continuity to show that \( f\left( x\right) = {x}^{2} \) is continuous at \( a = 0 \) . | Proof: Let \( \varepsilon > 0 \) . Let \( \delta = \sqrt{\varepsilon } \) . If \( \left| {x - 0}\right| < \delta \), then \( \left| x\right| < \sqrt{\varepsilon } \) . Thus\n\n\[ \left| {{x}^{2} - {0}^{2}}\right| = {\left| x\right| }^{2} < {\left( \sqrt{\varepsilon }\right) }^{2} = \varepsilon . \]\n\nThus by the defin... | Yes |
Use the definition of continuity to prove that \( f\left( x\right) = \sqrt{x} \) is continuous at \( a = 1 \) . | Proof: Let \( \varepsilon > 0 \) . Let \( \delta = \varepsilon \) . If \( \left| {x - 1}\right| < \delta \), then \( \left| {x - 1}\right| < \varepsilon \), and so\n\n\[ \left| {\sqrt{x} - \sqrt{1}}\right| = \left| \frac{\left( {\sqrt{x} - 1}\right) \left( {\sqrt{x} + 1}\right) }{\sqrt{x} + 1}\right| = \frac{\left| x -... | Yes |
Example 12. Use Theorem 15 to prove that\n\n\\[ \nf\\left( x\\right) = \\left\\{ \\begin{array}{ll} \\frac{\\left| x\\right| }{x}, & \\text{ if }x \\neq 0 \\\\ 0, & \\text{ if }x = 0 \\end{array}\\right.\n\\]\n\nis not continuous at 0 . | Proof: First notice that \\( f \\) can be written as\n\n\\[ \nf\\left( x\\right) = \\left\\{ \\begin{array}{ll} 1 & \\text{ if }x > 0 \\\\ - 1 & \\text{ if }x < 0 \\\\ 0 & \\text{ if }x = 0 \\end{array}\\right.\n\\]\n\nTo show that \\( f \\) is not continuous at 0, all we need to do is create a single sequence \\( \\le... | Yes |
Use Theorem 15 to show that \( T \) is not continuous at 0, no matter what value is chosen for \( b \) . | Sketch of Proof: We've seen how we can use Theorem 15 now we need to prove Theorem 15. The forward direction is fairly straightforward. So we assume that \( f \) is continuous at \( a \) and start with a sequence \( \left( {x}_{n}\right) \) which converges to \( a \) . What is left to show is that \( \mathop{\lim }\lim... | No |
Theorem 16. Suppose \( f \) and \( g \) are both continuous at a. Then \( f + g \) and \( f \cdot g \) are continuous at a. | Proof: We could use the definition of continuity to prove Theorem 16, but Theorem 15 makes our job much easier. For example, to show that \( f + g \) is continuous, consider any sequence \( \left( {x}_{n}\right) \) which converges to \( a \) . Since \( f \) is continuous at \( a \), then by Theorem 15, \( \mathop{\lim ... | Yes |
Consider the function \( D\left( x\right) = \frac{{x}^{2} - 1}{x - 1}, x \neq 1 \) . You probably recognize this as the difference quotient used to compute the derivative of \( f\left( x\right) = {x}^{2} \) at \( x = 1 \), so we strongly suspect that \( \mathop{\lim }\limits_{{x \rightarrow 1}}\frac{{x}^{2} - 1}{x - 1}... | Proof: Let \( \varepsilon > 0 \) and let \( \delta = \varepsilon \) . If \( 0 < \left| {x - 1}\right| < \delta \), then\n\n\[ \left| {\frac{{x}^{2} - 1}{x - 1} - 2}\right| = \left| {\left( {x + 1}\right) - 2}\right| = \left| {x - 1}\right| < \delta = \varepsilon . \] | Yes |
(a) \( \mathop{\lim }\limits_{{x \rightarrow 1}}\frac{{x}^{3} - 1}{x - 1} = 3 \) | Hint:\n\n\[ \left| {\frac{{x}^{3} - 1}{x - 1} - 3}\right| = \left| {{x}^{2} + x + 1 - 3}\right| \]\n\n\[ \leq \left| {{x}^{2} - 1}\right| + \left| {x - 1}\right| \]\n\n\[ = \left| {{\left( x - 1 + 1\right) }^{2} - 1}\right| + \left| {x - 1}\right| \]\n\n\[ = \left| {{\left( x - 1\right) }^{2} + 2\left( {x - 1}\right) }... | No |
Theorem 18. Suppose \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = L \) and \( \mathop{\lim }\limits_{{x \rightarrow a}}g\left( x\right) = M \), then\n\n(a) \( \mathop{\lim }\limits_{{x \rightarrow a}}\left( {f\left( x\right) + g\left( x\right) }\right) = L + M \) | Proof: Let \( \left( {x}_{n}\right) \) be a sequence such that \( {x}_{n} \neq a \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = a \) . Since \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = L \) and \( \mathop{\lim }\limits_{{x \rightarrow a}}g\left( x\right) = M \) we see that \( \mat... | Yes |
Problem 123. Prove Theorem 19. | Returning to \( \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\sin x}{x} \) we’ll see that the Squeeze Theorem is just what we need. First notice that since \( D\left( x\right) = \sin x/x \) is an even function, we only need to focus on \( x > 0 \) in our inequalities. Consider the unit circle. | No |
Theorem 21. (Fermat’s Theorem) Suppose \( f \) is differentiable in some interval \( \\left( {a, b}\\right) \) containing \( c \) . If \( f\\left( c\\right) \\geq f\\left( x\\right) \) for every \( x \) in \( \\left( {a, b}\\right) \), then \( {f}^{\\prime }\\left( c\\right) = \) 0 . | Proof: Since \( {f}^{\\prime }\\left( c\\right) \) exists we know that if \( {\\left( {h}_{n}\\right) }_{n = 1}^{\\infty } \) converges to zero then the sequence \( {a}_{n} = \\frac{f\\left( {c + {h}_{n}}\\right) - f\\left( c\\right) }{{h}_{n}} \) converges to \( {f}^{\\prime }\\left( c\\right) \) . The proof consists ... | No |
Theorem 22. (The Mean Value Theorem) Suppose \( {f}^{\prime } \) exists for every \( x \in \left( {a, b}\right) \) and \( f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) . Then there is a real number \( c \in \left( {a, b}\right) \) such that \[ {f}^{\prime }\left( c\right) = \frac{f\left( b\right) - f\lef... | However, it would be difficult to prove the MVT right now. So we will first state and prove Rolle's Theorem, which can be seen as a special case of the MVT. The proof of the MVT will then follow easily. | No |
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