Let S={0,1,2,3}. (a) Find the sample vectors corresponding to the functions 1 , cosπx, cos2 πx,

Let S={0,1,2,3}. (a) Find the sample vectors corresponding...

Key Concepts

Sample Vectors

A sample vector is constructed by evaluating a function at a specified set of points in its domain. This transforms the function into an ordered collection of numbers, allowing one to work with it using linear algebra techniques. It encapsulates the behavior of the function over the predetermined sample set, turning the infinite-dimensional function space into a finite-dimensional vector space when the domain is finite.

Function Representation in Finite-Dimensional Vector Spaces

When a function is defined on a finite set, it can be represented as a vector in a finite-dimensional space. Each component of the vector corresponds to the function's value at a specific point in the set. This perspective facilitates the analysis, combination, and transformation of functions using the tools and concepts of linear algebra.

Basis Functions

Basis functions are a set of functions that, when combined linearly, can represent a wide variety of functions within a given function space. In the context of sampling and finite-dimensional spaces, selecting an appropriate set of basis functions allows one to represent any function in the space uniquely with the correct coefficients. They serve as the fundamental building blocks of the function space in consideration.

Uniqueness of Function Determination via Samples

The idea of whether a function is uniquely determined by its sample values relates to the concept of interpolation and the function space's dimensionality. A finite set of sample values may belong to many different functions, particularly when the underlying space of possible functions is infinite-dimensional. This concept is critical in understanding the limitations of sampling, where additional structure or constraints are needed to guarantee uniqueness.

Transcript

00:01 In this problem of vector valid function we have given the equation of space curve which is r t is equals to 2 cos t multiplied with i plus 2 sine t so plus 2 sine t multiplied with j plus t divide with 2 multiplied with k.

00:18 And now we have to find the vectors t and n so vectors t and n that is tangential unit tangent vector and unit principal unit normal vector and the bina which is b is defined as t cross product with n for the vector valued function at t not is equals to pi divided with two so first we would find t so for this we have to differentiate it so t say r dash t is equal to differentiation of this term differentiation of course is minus sine so minus 2 sine t multiplied with i plus differentiation of sign is cause so two cost t multiplied with j definition of t divided with 2 is equal to simply half so half multiplied with key now t t so t is given by r dash t divided with the magnitude of r d now when we take this term so magnitude is equal to this is 2 square which is 4 plus 4 so this would be 8 and plus 1 under root of 17 divide with 17 and here we left that same terms of minus 2 sine t multiplied with i plus 2 cost t multiplied with j plus 1 divide with 2 and this is multiplied with k now we can find n t so n t would be equals to say this is t dash t so t d d divided with the magnitude of t so this is equal to the differentiation of k term is equal to 0 and differentiation of this term after solving we get minus cost t multiplied with i and minus sine t multiplied with this would be j say here j actually so this is j now we can find here r of pi divided with two so r of pi divided with two so putting t is equal to pi divided with two so putting t is equal to pi divided with 2 we get r of pi divide with 2 so cos pi divide with 2 is equal to 0 sine pi divide with 2 is equal to 1 so this would be 2 j and putting t is equal to pi divide with 2 so this is pi divided with 4 so pi divided with 4 multiplied with k similarly we can find t of pi divide with 2 putting t is equal to pi divide with 2 all 2 terms here so this would be 2 under root of 17 divided with 17 and here sine pi divide with 2 is 1 cost pi divide 2 is equal to 0 so this would be minus after solving this term so we get this would be minus 2 so this is minus 2 and here this would be 1 divided with 2 so 1 divide with 2 and we have to find n of pi divide with 2 so n of pi divided with 2 so here this term is equals to 0 and here we have minus of 1 so minus of 1 multiplied with j is simply j now we can find b so b of pi divided with 2 is equals to t of pi divide with 2 cross product with this is n of pi divide with 2 so this would be here cross product is this is i j and k now we have to write the coefficient of i j and k for t pi divide with 2 so writing it so this would be minus 2 multiply with terms of minus 4 under root of 17 divide with 17.

04:14 J term is 0 and here this would be simply under root of 17 divide with 17...

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