SUMS AND PRODUCTS OF CONTINUED FRACTIONS by CiA).
... k>2, it is trivial that Theorem 1 becomes false if the lower bound 2 for the partial quotients is replaced by any larger integer. Even the equality 5(&)+S(&) = [0, 2&-1] is easily seen to be false for k>2, as consideration of the effect of the removal of the first middle interval in the subdivision ...
... k>2, it is trivial that Theorem 1 becomes false if the lower bound 2 for the partial quotients is replaced by any larger integer. Even the equality 5(&)+S(&) = [0, 2&-1] is easily seen to be false for k>2, as consideration of the effect of the removal of the first middle interval in the subdivision ...
Solving Inequalities
... Substituting, we gave (–5)2 + (–5) > 12? That is true. So the interval from negative infinity to – 4 works. Let’s try zero. (0)2 + 0 > 12? Is not true, so the interval between – 4 and 3 does not work. And for the third interval, let’s try +5. (5)2 + 5 >12? That is true. So the interval from 5 to pos ...
... Substituting, we gave (–5)2 + (–5) > 12? That is true. So the interval from negative infinity to – 4 works. Let’s try zero. (0)2 + 0 > 12? Is not true, so the interval between – 4 and 3 does not work. And for the third interval, let’s try +5. (5)2 + 5 >12? That is true. So the interval from 5 to pos ...
3.1 Extrema on an Interval
... Interval To find the extrema of a continuous function on a closed interval [a,b], use the ...
... Interval To find the extrema of a continuous function on a closed interval [a,b], use the ...
PC-P.1
... teacher page and bring them to class with you if you would like – choosing a format that works best for note taking. Extended Periods: we will proceed with the next days notes, then add some mindstretching activities! ...
... teacher page and bring them to class with you if you would like – choosing a format that works best for note taking. Extended Periods: we will proceed with the next days notes, then add some mindstretching activities! ...
Section 1.7 Inequalities
... Let’s get x by itself. Remember that if we do something to one side of this compound inequality, that we must do it to all three sides: −5 − 1 ≤ 2x + 1 − 1 < 9 − 1 −6 ≤ 2x < 8 −6 2x 8 ...
... Let’s get x by itself. Remember that if we do something to one side of this compound inequality, that we must do it to all three sides: −5 − 1 ≤ 2x + 1 − 1 < 9 − 1 −6 ≤ 2x < 8 −6 2x 8 ...
Math 403 SOLUTIONS FOR HWK #4 RATIONAL, ALGEBRAIC
... Let U be an open set in (R, ρ). An open interval I ⊂ U is called a composite interval in U if there is no other interval J ⊂ U such that I ⊂ J. 1) Prove that any two composite intervals are disjoint. 2) Let x ∈ U . Consider the union I of all intervals Ix such that x ∈ Ix and Ix ⊂ U . Prove that x ∈ ...
... Let U be an open set in (R, ρ). An open interval I ⊂ U is called a composite interval in U if there is no other interval J ⊂ U such that I ⊂ J. 1) Prove that any two composite intervals are disjoint. 2) Let x ∈ U . Consider the union I of all intervals Ix such that x ∈ Ix and Ix ⊂ U . Prove that x ∈ ...
Existence of a Unique Solution
... Let set y1 , y2 ,......., yn of n linearly independent solutions of the homogeneous linear nthorder differential equation on an interval I is said to be a fundamental set of solutions on the interval. There exists a fundamental set of solutions for the homogeneous linear nth-order DE on an interval ...
... Let set y1 , y2 ,......., yn of n linearly independent solutions of the homogeneous linear nthorder differential equation on an interval I is said to be a fundamental set of solutions on the interval. There exists a fundamental set of solutions for the homogeneous linear nth-order DE on an interval ...
Increasing and Decreasing Functions
... decreasing, use the following steps: (1) Locate the critical number of f in (a, b), and use these numbers to determine test intervals (2) Determine the sign of f’(x) at one test value in each of the intervals (3) If f’(x) is positive, the function is increasing in that interval. If f’(x) is ne ...
... decreasing, use the following steps: (1) Locate the critical number of f in (a, b), and use these numbers to determine test intervals (2) Determine the sign of f’(x) at one test value in each of the intervals (3) If f’(x) is positive, the function is increasing in that interval. If f’(x) is ne ...
Interval Notation
... In this lesson you solved inequalities and found that the solution can usually be represented as a piece or "interval" of the number line. You learned to graph this solution set on the number line by shading that piece and placing a circle, open or filled, at an endpoint of that interval. Another wa ...
... In this lesson you solved inequalities and found that the solution can usually be represented as a piece or "interval" of the number line. You learned to graph this solution set on the number line by shading that piece and placing a circle, open or filled, at an endpoint of that interval. Another wa ...
Section 5.1
... f c does not exist. Note: The sign of the derivative of f does not reveal the direction of curvature of the graph of f. For this we need the 2nd derivative: B.) Concavity – describes the direction of curvature of the graph of f. The graph is concave up on intervals where the tangent lines have ...
... f c does not exist. Note: The sign of the derivative of f does not reveal the direction of curvature of the graph of f. For this we need the 2nd derivative: B.) Concavity – describes the direction of curvature of the graph of f. The graph is concave up on intervals where the tangent lines have ...
Interval Notation and Review of Inequalities
... Interval Notation is another way for denoting the solution for an inequality by describing the interval containing the numbers that satisfy that inequality. The numbers at the end of the interval are enclosed in brackets [ or ] (the boundary is a solution/ filled in circle) or parentheses ( or ) (th ...
... Interval Notation is another way for denoting the solution for an inequality by describing the interval containing the numbers that satisfy that inequality. The numbers at the end of the interval are enclosed in brackets [ or ] (the boundary is a solution/ filled in circle) or parentheses ( or ) (th ...
NESTED INTERVALS
... Sufficiency. Suppose jupuqj < for all p; q > N and any > 0. Then all the numbers u N; uN1;. . . lie in a finite interval, i.e., the set is bounded and infinite. Hence, by the Bolzano–Weierstrass theorem there is at least one limit point, say a. If a is the only limit point, we have the desired proof and ...
... Sufficiency. Suppose jupuqj < for all p; q > N and any > 0. Then all the numbers u N; uN1;. . . lie in a finite interval, i.e., the set is bounded and infinite. Hence, by the Bolzano–Weierstrass theorem there is at least one limit point, say a. If a is the only limit point, we have the desired proof and ...
xx - UTEP Math
... Theorem 3.5 – Let f be a function that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). 1. If f ' x 0 for all x in (a, b), then f is increasing on [a, b]. 2. If f ' x 0 for all x in (a, b), then f is decreasing on [a, b]. 3. If f ' x 0 for ...
... Theorem 3.5 – Let f be a function that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). 1. If f ' x 0 for all x in (a, b), then f is increasing on [a, b]. 2. If f ' x 0 for all x in (a, b), then f is decreasing on [a, b]. 3. If f ' x 0 for ...
real number
... (a, b) means a < x < b. This is an open interval because the endpoints are not included in the interval. ...
... (a, b) means a < x < b. This is an open interval because the endpoints are not included in the interval. ...
Solution
... The derivative is f 0 (x) = 3x2 + 15x which has roots x = −5, 0. These are our critical points. Since we are restricted to an interval, we check all the critical points inside that interval (both happen to be inside), as well as the endpoints of the interval: f (−7) = 27.5 f (−5) = 65.5 f (0) = 3 f ...
... The derivative is f 0 (x) = 3x2 + 15x which has roots x = −5, 0. These are our critical points. Since we are restricted to an interval, we check all the critical points inside that interval (both happen to be inside), as well as the endpoints of the interval: f (−7) = 27.5 f (−5) = 65.5 f (0) = 3 f ...
solving rational inequalities
... 1. If needed, rewrite the inequality so that 0 is on one side. This usually requires you to add or subtract term(s) to both sides of the inequality. 2. If needed, simplify the non-zero side so that there is a single fraction. This usually requires finding a common denominator. 3. Set the numerator e ...
... 1. If needed, rewrite the inequality so that 0 is on one side. This usually requires you to add or subtract term(s) to both sides of the inequality. 2. If needed, simplify the non-zero side so that there is a single fraction. This usually requires finding a common denominator. 3. Set the numerator e ...
Tree Structure for Set of Intervals
... Root nor any node on the left or right boundary path of the tree can have any intervals of the canonical interval decomposition attached to it because their node intervals are unbounded and we are representing only finite intervals. ...
... Root nor any node on the left or right boundary path of the tree can have any intervals of the canonical interval decomposition attached to it because their node intervals are unbounded and we are representing only finite intervals. ...
solving rational inequalities
... 2. Replace the inequality with = and solve for x. These numbers are called Solution Cut Points. (SCP) 3. Draw a number line and label the cut-points as follows: a. Always label Domain Cut Points with an open circle. b. Solution Cut-Points i. If the problem contains ≤ or ≥, use closed circles. ii. If ...
... 2. Replace the inequality with = and solve for x. These numbers are called Solution Cut Points. (SCP) 3. Draw a number line and label the cut-points as follows: a. Always label Domain Cut Points with an open circle. b. Solution Cut-Points i. If the problem contains ≤ or ≥, use closed circles. ii. If ...
PDF
... Theorem. If the real function f is continuous on the interval [0, ∞) and the limit lim f (x) exists as a finite number a, then f is uniformly continuous x→∞ on that interval. Proof. Let ε > 0. According to the limit condition, there is a positive number M such that ε ...
... Theorem. If the real function f is continuous on the interval [0, ∞) and the limit lim f (x) exists as a finite number a, then f is uniformly continuous x→∞ on that interval. Proof. Let ε > 0. According to the limit condition, there is a positive number M such that ε ...
A Review of Basic Function Ideas
... Function: No x’s repeat (it is okay if the y’s repeat). On a graph, you can use the Vertical line test – each vertical line can only touch one point on the line) Domain: x’s ...
... Function: No x’s repeat (it is okay if the y’s repeat). On a graph, you can use the Vertical line test – each vertical line can only touch one point on the line) Domain: x’s ...