A sequence is a function whose domain is basically a set of positive integers. We denote these list of values with a variable and a subscript of 1, 2, 3,... all the way up to n. This shows a series of a set of numbers. 2 + (-1)^1, 2 + (-1)^2 + 2 + (-1)^3 + 2 + (-1)^4.. 1, 3, 1, 3 Sequences approaching limiting values converge, and sequences approaching a value that does not exist are said to diverge. The definition of a limit of a sequence is: lim (n approaches ∞) a(nth term) = L where L is a limit. If L exists, the sequence converges. If limit L of a sequence does not exist, it diverges. For the above example, because the sequence alternates between 1 and 3, the limit does not exist. Therefore, the above sequence diverges. Squeeze Theorem for Sequences
Monotonic Sequences and Bounded SequencesA sequence is monotonic when its terms are non-decreasing or non-increasing. To determine whether a sequence is monotonic, simply check whether the terms in the series are increasing or alternating. If it is changing between two values, it is not monotonic. A sequence is bounded above when the real number M is greater than the sequence for all numbers. M is the upper bound in this case.A sequence is bounded below when the real number M is less than the sequence for all numbers. M is the lower bound in this case. When a sequence is bounded above and below, it is bounded.
Therefore, from that, an - L < ϵ which means that {an} converges to L. It is super interesting to delve into the depths of series and their applications in the real world too. Explore on! Good luck!
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Happy holidays, everyone! Hope you're having a great start to break! Today, I am going to be elaborating more on another important concept in calculus known as the disk method.
**NOTE: the shape extends from 0 to 1 (along x-axis) so the shape ends at 1** In order to find the volume of the shaded region, we have to subtract the volume of the cone shape from the volume of the dome shape. As per the solid of revolution formula, we know that V = π∫(b to a) [R(x)]^2 dx.
π∫(1 to 0)(x^1/2)^2 dx - π∫(1 to 0)(x)^2 dx π∫(1 to 0)x dx - π∫(1 to 0) x^2 dx = π[1/2 x^2] - π[1/3 x^3] = (substitue 1 and 0) 1/2 π - 1/3 π (3-2 π)/6 = 1/6π So the volume of the purple area is 1/6 π!! With this method, you can practically find the volume of any distinct 3 dimensional shape! Very interesting, right? Good luck and happy holidays,
source: http://indusladies.com/community/threads/dot-kolams.77039/page-2 Can we use integrals to figure out the area of the insides of such complicated designs and shapes? Well, to introduce you to these designs, called 'kolams', a typical design drawn during the season of lights and happiness, I have posted a picture above for you to get the idea. So, lets say we had a simple kolam to deal with at first, such as this one: source: http://www.a1tamilnadu.com/singlePost.asp?cid=1199#.Wgj22bA-dPM Lets say the problem was to find the area inside of this complex shape. I want to try to use integrals to find the area of this shape. So, therefore, lets use a small, tiny arbitrary curve from this whole design. In other words, take a magnifying glass and find a curvature in which we can find a small approximation of the area inside that small curve. So, lets pick a small curve from this kolam and draw a x-y axis behind it to integrate it and approximate the area. This curve looks a lot like a natural log curve. We can use the left rectangular approximation method to find the approximate area. if we integrate ln(x) from x = 1 to x = 4, we get 2.5451.... ∫ [at interval 1,4] ln(x) d(x) = 2.5451.. However, lets try using rectangles to approximate the areas. So, for the first rectangle (from x=1 to x=2), we can solve the area, A, by getting the y value at x = 1 and multiplying it by one to get the area of that particular rectangle. Therefore, the area of the first rectangle would be ln(1) which is the y value, times 1, which yields 0. The next rectangle (from x=2 to x=3) is ln(2) times again 1 (which represents the width) which yields 0.6. The next rectangle (from x=3 to 4) is ln(3) times 1 = 1.09. Therefore, the total area which encompasses the area under the curve is: 1.09 + 0.6 + 0 = 1.7917, much lower than the actual area calculated above (2.5451). So, how do we solve for the area inside of these designs? It may seem a tedious task, but to be exact and accurate we must calculate the areas underneath the smaller curves of the design to eventually find the total area the design covers. In order to do this, we will have to put the design on to a grid and label the coordinates.
Good luck!!
Hello, Everyone!! Its been a while, but I have decided to start a series on Calculus topics, especially since I will be taking Cal BC this coming school year, I thought it would be cool to refresh these topics in my mind and give you some info on how Calculus can help you out. Really. I may not be in Calculus already, but I have been exposed to many calculus related problems during my ninth and tenth school years in Canada. Calculus is so amazing, and broad but at the same time, can be used to find volumes of solids of various different kinds. So, don't worry if you don't know the formula for finding the volume of a sphere during a test or quiz. Calculus has got you.
Lets do an example together. Let's find the volume of a sphere (the midpoint being 0) from the values of a and -a. lets first find the graphic formula for the area of the cross-sectional surface: x^2 + r^2 = a^2 <--- finding area through the circle formula - step 1 r = √(a^2 - x^2) <--- solving for r - step 2 S(x) = π(a^2 - x^2), -a ≤ x ≤ a <--- finding area of the circle, domain - step 3 V = ∫(from a to -a) [π(a^2 - x^2) dx] <--- set volume to definite integral - step 4 = π[ a^2x - 1/3x^3](from a to -a) <--- solve - step 5 = π[4/3a^3] = 4/3a^3π So, we finally derived the volume formula of a sphere, and since it is a generalized formula, a can be any value (radius). Good luck,
I have always wondered about this question: How do you derive the volumes of shapes? Why is the volume of a square based pyramid 1/3b^2(h)?? Why is the volume of a sphere 4/3(pi)r^3? Actually, in order to prove that these are the volumes of the sphere and square based pyramid, we have to use calculus. Calculus? Yup. In order to prove that 4/3(pi)r^3 is the volume of a sphere, you have to use integrals and basic knowledge of volumes. If you do not know what integrals are, that's okay! Watch out for my future blog posts on integrals! First, you have to find the cross-sectional volume of the shape. For instance, if the shape was a sphere, you find the cross-sectional volume of the 'disc' inside the sphere. This allows you to find one portion of the volume. You then integrate the prticular volume to a certain limit (of the 3d whole shape). Good luck,
Have you wondered the same thing? I was learning about SA and Volume of 3 dimensional composite shapes at school when I thought, 'Why is the Surface Area of a Sphere 4(pi)r^2?' How did mathematicians derive this formula? Why does this formula make sense? There is actually a simple mathematic logic for the derivation of this particular formula. The formula was first found by Archimedes (my mathematician buddy). Archimedes proved that the surface area of a sphere is equal to the lateral surface area of a cylinder. The lateral surface area of a cylinder does not include the area of the two circular ends. For a sphere to nicely fit inside a cylinder, the height of the cylinder should be twice the sphere's radius, 2r. If the height of the cylinder is 2r, then its lateral surface area is: circumference x height = 2(pi)r x 2r = 4(pi)r^2 Therefore, the surface area of a sphere is equivalent to the formula above, or 4(pi)r^2. Are you mind-blowned like me? I know, mathematics has its own way of surprising us with its theories. Good luck,
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