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question:Given the sequence left{a_{n}right} defined by: a_{1} = 1 a_{2} = 1 2 sin^2 theta cos^2 theta a_{n+2} a_{n+1} + a_{n} sin^2 theta cos^2 theta = 0, for theta in left(0, frac{pi}{2}right) Prove that frac{1}{2^{n1}} leq a_{n} leq 1 sin^n 2theta left(1 frac{1}{2^{n1}}right).
answer:1. **Characteristic Equation:** By analyzing the given recurrence relation: [ a_{n+2} - a_{n+1} + a_{n} sin^2 theta cos^2 theta = 0, ] We write the characteristic equation: [ x^2 - x + sin^2 theta cos^2 theta = 0. ] 2. **Roots of the Characteristic Equation:** Solve for the roots using the quadratic formula: [ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ] where (a = 1), (b = -1), and (c = sin^2 theta cos^2 theta). Plugging in the values: [ x = frac{1 pm sqrt{1 - 4 sin^2 theta cos^2 theta}}{2}. ] Since ( sin^2 theta cos^2 theta = frac{1}{2} sin^2 2theta ), [ x = frac{1 pm sqrt{1 - frac{1}{2} sin^2 2theta}}{2}. ] The two possible solutions (roots) are: [ x_1 = sin^2 theta, quad x_2 = cos^2 theta. ] 3. **General Solution Form:** Assume the solution for (a_n) has the form: [ a_n = A(sin^2 theta)^n + B(cos^2 theta)^n. ] 4. **Initial Conditions:** Given the initial conditions: [ a_1 = 1 quad text{and} quad a_2 = 1 - 2 sin^2 theta cos^2 theta, ] Use these to solve for (A) and (B). For (n = 1), [ a_1 = A sin^2 theta + B cos^2 theta = 1. ] For (n = 2), [ a_2 = A (sin^2 theta)^2 + B (cos^2 theta)^2 = 1 - 2 sin^2 theta cos^2 theta. ] 5. **Values of Constants (A) and (B):** Simplify to solve for (A) and (B). From the form of (a_n), it is determined that: [ a_n = sin^{2n} theta + cos^{2n} theta. ] 6. **Inequality Analysis:** Consider the function (y = x^n) for (x in (0, 1)). It is a convex function, so by using the properties of convex functions (Jensen's inequality): [ frac{(sin^2 theta)^n + (cos^2 theta)^n}{2} geq left(frac{sin^2 theta + cos^2 theta}{2}right)^n. ] Given (sin^2 theta + cos^2 theta = 1), we have: [ left(frac{1}{2}right)^n = frac{1}{2^n}. ] Implying: [ frac{a_n}{2} geq frac{1}{2^n}. ] Rearranging gives: [ a_n geq frac{1}{2^{n-1}}. ] 7. **Upper Bound Analysis:** Given: [ 1 = (sin^2 theta + cos^2 theta)^n = sum_{k=0}^n binom{n}{k} (sin^2 theta)^{n-k} (cos^2 theta)^k, ] Directly comparing terms, considering the significant ones, especially when combined with (sin^{2n} theta) and (cos^{2n} theta): [ t_2 = frac{1}{2} cdot sin^2 theta cos^{2n-2} theta + frac{1}{2} cdot cos^2 theta sin^{2n-2} theta + ldots ] Hence: [ 1 geq a_n + frac{sin^n 2theta (2^n - 2)}{2^{n+1}}. ] Simplifying and rearranging: [ a_n leq 1 - sin^n 2theta left(1 - frac{1}{2^{n-1}}right). ] Conclusion: [ boxed{frac{1}{2^{n-1}} leq a_n leq 1 - sin^n 2theta left(1 - frac{1}{2^{n-1}}right)} ]
question:James turned 23 the same time John turned 35. Tim is 5 years less than twice John's age. If Tim is 79 how old is James?
answer:To solve this problem, we follow these steps: 1. First, we determine the age difference between James and John. Since James turned 23 when John turned 35, we find the age difference by subtracting James's age from John's age: [35 - 23 = 12] So, James is 12 years younger than John. 2. Next, we know that Tim is 5 years less than twice John's age, and Tim is 79 years old. To find John's age, we first subtract 5 years from Tim's age to get the equivalent of twice John's age: [79 - 5 = 74] This means twice John's age is 74 years old. 3. To find John's actual age, we divide the result by 2: [74 / 2 = 37] So, John is 37 years old. 4. Finally, since James is 12 years younger than John, we subtract 12 from John's age to find James's age: [37 - 12 = 25] Therefore, James is boxed{25} years old.
question:Prove that the kernel of any regular polyhedron is a convex regular polyhedron with the same number of faces and the same number of sides on each face as the original polyhedron.
answer:1. **Understanding the Kernels of Regular Polyhedra:** By the definition of the kernel of a regular polyhedron, it lies on one side of any of its faces. Therefore, the kernel itself is a convex polyhedron. 2. **Flatness of Kernel Faces:** Each face of the kernel can only lie in the plane of a face of the original polyhedron. This is because due to the convexity of the kernel, no two of its faces can lie in the same plane. 3. **Number of Faces Must Not Exceed the Original:** Hence, the number of faces of the kernel cannot exceed the number of faces of the original polyhedron. 4. **Each Face Retains Original Properties:** Each face must correspond exactly to one face or be a subdivision preserving the type of the original face of the polyhedron. Thus, the kernel has as many faces as the original polyhedron. 5. **Equality of Kernels of Similar Constructs:** Since our aim is to prove that any regular polyhedron's kernel is also a regular polyhedron, it stands that all edges, angles, and faces must congruently match those of the initial polyhedron. 6. **Preservation Upon Transformation:** Consider the situation where a given polyhedron overlaps with itself. Its kernel will evidently also match up and become identical with the original polyhedron upon such transformations. Any two adjacent edges and their faces must align uniformly. 7. **Polygonal Faces and Their Symmetries:** Due to rotational symmetry principles around a perpendicular dropped from its center onto any face, the polyhedron reorganizes itself into the same construct. Polygons formed remain equal and consistent in shape, concluding that edges and angles rotate uniformly around their center. 8. **Equal Face Count on Both Kernel and Original Polyhedron:** Rotation and transformation of the kernel in 360 degrees ensures that faces occupy congruent positions thus leading us to conclude the number of sides and vertices match conclusively. 9. **Final Step to Establish Correctness:** Since all faces, angles, and edges match uniformly, and every transformation realigns perfectly into its initial form, we derive that the kernel of any regular polyhedron must also be a regular polyhedron with the same number of faces. # Conclusion [boxed{text{The kernel of any regular polyhedron is also a convex regular polyhedron with the same number of faces and sides per face as the original polyhedron.}}]
question:Simplify 4 cdot frac{18}{5} cdot frac{25}{72}.
answer:First, observe that 18 and 72 have a common factor of 18, and 5 and 25 have a common factor of 5. Keep in mind the negative sign in the denominator, which will affect the sign of our final answer. 1. Simplify the fractions by canceling common factors: [ 4 cdot frac{18}{5} cdot frac{25}{-72} = 4 cdot frac{cancel{18}}{5} cdot frac{25}{-cancel{72}} = 4 cdot frac{1}{5} cdot frac{25}{-4} ] 2. Simplify further by canceling the 4 in the numerator and denominator: [ 4 cdot frac{1}{5} cdot frac{25}{-4} = frac{4}{5} cdot frac{25}{-4} = frac{cancel{4}}{5} cdot frac{25}{-cancel{4}} = frac{25}{-5} ] 3. Simplify the resulting fraction: [ frac{25}{-5} = -5 ] Thus, the simplified form of 4 cdot frac{18}{5} cdot frac{25}{-72} is boxed{-5}.