Recent content by Antiderivative

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    Taylor's Theorem for Vector-Valued Functions (Real Analysis)

    I can derive the first-order version of the inequality, which is \left|\mathbf{f}(b) - \mathbf{f}(a)\right| \leq (b - a)\left|\mathbf{f}'(x)\right| As I said this is a "mean value theorem" for vector-valued functions. How would I extend this to an n^{th} order inequality so as to "follow from...
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    Solve Integral of y dx: Step-by-Step Guide

    My fault, I wasn't aware of that in the PF guidelines. I'll refrain from posting fully-worked-out solutions in the future. Sorry about that guys.
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    Solve Integral of y dx: Step-by-Step Guide

    I see you're talking about a well-known and widely-used single-variable calculus textbook here, OP. I presume it's for Calc II or a high school AP Calculus course? In any case, perhaps your teacher/professor did not cover it yet, but the best way (and the easiest way) to solve such things is...
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    How can I accurately read mathematical symbols to understand 3D space?

    Indeed, HallsofIvy is correct. The book is talking about the set of all points in 3D space and how one can associate ordered triples of the form (a,b,c) with each such point. The book's statement also implies that these points (a,b,c) are unique in describing a single location in 3D space (hence...
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    Taylor's Theorem for Vector-Valued Functions (Real Analysis)

    I see. So we're trying to make a relationship for f(t), not just the single-variable t itself. As for the length, what justifies using that as the manner in which one orders these vectors? I mean I can do the math (it's just z = \sqrt{z\overline{z}}), but why are we allowed to? Or is it...
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    How can I accurately read mathematical symbols to understand 3D space?

    You're right in that you need to be able to learn what mathematical symbols mean as a way to determine shorthand. It's very important in set theory (and real analysis, which is a total pain but appreciable nevertheless). Your definition states that "The Cartesian product of R, R, and R is equal...
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    Taylor's Theorem for Vector-Valued Functions (Real Analysis)

    Homework Statement "Formulate and prove an inequality which follows from Taylor's theorem and which remains valid for vector-valued functions." Homework Equations I know that Taylor's theorem generally states that if f is a real function on [a,b], n is a positive integer, f^{(n-1)} is...
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    Projection of space curves onto general planes

    Yep I just used that method after the orthonormal one. In the end it's kind of the same thing because you'll end up getting v – (n•v/n•n) n, which is like the Gram-Schmidt process for creating an orthonormal basis. Analogous procedures! Geometric versus algebraic arguments I suppose.
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    Projection of space curves onto general planes

    Okay just for completion I want to post that I officially figured it out. Using the change of basis, I ended up rewriting the old x,y,z in terms of the new ones, and deriving an equation for this "flattened" helix on the plane: \left( \begin{array}{c} \hat{x} \\ \hat{y} \\ \hat{z}...
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    Projection of space curves onto general planes

    Okay so I figured out that the three basis vectors in my new coordinate system would be \left( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right), \left( \frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}} \right), and \left(...
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    Projection of space curves onto general planes

    Hi tiny-tim! Hmm... okay so the normal vector to the plane is N(1,1,1), and the plane goes through A(1/√2,0,-1/√2). Using that, I guess I can try to find an orthonormal basis. Clearly A•N = 0, so A and N are normal. The other vector can be found by finding N x A presumably. I'm getting (-1/√2...
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    Projection of space curves onto general planes

    So I've encountered many "what is the projection of the space curve C onto the xy-plane?" type of problems, but I recently came across a "what is the project of the space curve C onto this specific plane P?" type of question and wasn't sure how to proceed. The internet didn't yield me answers so...
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