Recent content by Yuqing

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    Cauchy Sequence

    Well, you initially begin with \left|x_2 - x_1\right|=\left|f(x_1) - f(x_0)\right|\le \lambda\left|x_1 - x_0\right| if you take one more step down the sequence, you can see that \left|x_3 - x_2\right|=\left|f(x_2) - f(x_1)\right|\le \lambda\left|x_2 - x_1\right|\le \lambda^2\left|x_1 -...
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    What pre-calculus needed for this calculus class?

    In general, mathematics used in introductory biological sciences typically involve exponential and trigonometric functions for modelling ecological models and population models. A solid foundation of general functions is of course preferred for calculus in general, but specifically a focus for...
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    Prove the following Power Series is monotonic

    Ah, true. I missed that point. The simplest way is to check the end points and verify that f(-1) >= 1.5 and f(1) <= -0.5. It should not be too difficult to do that. After which the fact that f(x)=1.5 and f(x)=-0.5 exists follows from the intermediate value theorem.
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    Prove the following Power Series is monotonic

    Like I've mentioned, the solutions exist because the series is convergent at those two values. The limit it converges to is defined as the value of the function. The monoticity of the function guarantees that the solutions are unique.
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    Convergence of a Series

    Consider doing a limit comparison test. What well known limit involving sine might help here?
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    Prove the following Power Series is monotonic

    It is single simply because it's monotonous. Suppose there were two solutions, from Rolle's theorem there must be a point where the derivative is 0 which is a contradiction.
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    Prove the following Power Series is monotonic

    The derivative does look right to me. To check the end points, substitute x=1 and x=-1 explicitly and look at the convergence of the resulting function. That's the traditional way end points are checked. Remember this when working with power series: the tests for convergence such as the ratio or...
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    How do I find the center of mass of a cone?

    Was your assumption that the CoM of the cone is vertically at the same location of the CoM of the triangle? If so then that's not right. When you rotate the triangle, more weight is prescribed to the larger end, so intuitively the CoM should move towards the larger end.
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    How do I find the center of mass of a cone?

    But the CoM of the cone is at H/4, assuming the cone is pointed up. There seems to be some miscommunication here. Edit: 1-sqrt(2) isn't even positive.
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    How do I find the center of mass of a cone?

    The median of an isosceles triangle is H/3 while the CoM of a cone is at H/4. I'm not quite sure how your method works out.
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    Vector. Is there an inverse of dot and cross product?

    Each of these ultimately boil down to solving a large numerical equation (or system of equations). The answer will generally not be unique and its rather pointless to do so. Say we are given\vec{x}\cdot(3, -2, 1) = 6then we are essentially solving3x_{1} - 2x_{2} + x_{3} = 6for arbitrary numbers...
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    Trouble understanding Symmetries

    This is not so much a question on symmetry. If you see your friend moving without any sort of external landmark to keep track of movement, then all you can say with certainty is that your friend has non-zero velocity relative to you. The question of whether you are moving or whether your friend...
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    Multivariable Limit

    How exactly would you suggest I approach this? The biggest problem I have is that the numerator is of higher overall order than the denominator. I cannot find a path which does not take me to 0.
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    Multivariable Limit

    Homework Statement Find the limit of \lim_{(x,y)\rightarrow (0,0)}\frac{x^2y^2}{x^3+y^3} Homework Equations I'd like to solve this in a rather elementary manner, so preferably only using the squeeze theorem or through proving the limit doesn't exist via multiple path approach. The...
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    Physics Tension Force Homework Problem

    Try equating the horizontal components of the tension. You know that the object is in static equilibrium and so there is no horizontal force on it. Working backwards from the fact that the horizontal components must be equal will give you an idea for the relative proportions of the tensions.