Discuss Infinite Dimensional And Finite Dimensional Vector Spaces.

Finite Dimensional: A vector space is finite-dimensional if it has a basis with only finitely many vectors. (One reason for sticking to finite-dimensional spaces is so that the representation of a vector with respect to a basis is a finitely-tall vector, and so can be easily written.) We shall take the term "vector space" to mean "finite-dimensional vector space". Other spaces are interesting and important, but they lie outside of our scope.

Infinite Dimensional: The process of extending the algebraic and geometrical methods of linear algebra from matrices to differential or integral operators consists of going from a finite dimensional vector space, typically Rⁿ , to an infinite dimensional vector space, typically a function space. However, a vector space of functions has certain idiosyncrasies precisely because its dimension is infinite. These peculiarities are so important that we must develop the framework in which they arise.

Discuss Implicit Differentiation.

In calculus, a method called implicit differentiation makes use of the chain rule to differentiate implicitly defined functions.  y can be given as a function of x implicitly rather than explicitly. When we have an equation R(x, y) = 0, we may be able to solve it for y and then differentiate. However, sometimes it is simpler to differentiate R(x, y) with respect to x and y and then solve for dy/dx.
Example: Assume that y is a function of x . Find y' = dy/dx for x³ + y³ = 4 .
SOLUTION 1 : Begin with x3 + y3 = 4 . Differentiate both sides of the equation, getting

D ( x3 + y3 ) = D ( 4 ) ,

D ( x3 ) + D ( y3 ) = D ( 4 ) ,

(using the chain rule on D ( y3 ) .)

3x2 + 3y2 y' = 0 ,

so that (Now solve for y' .)

3y2 y' = - 3x2 , and

y' = -3x²/3y² = -x²/y² Ans.

 

Prove that ~[r ∨ (q ∧ (~r →~p))] ≡ ~r ∧ (p∨ ~q) by using a
series of logical equivalences.

Take RHS ~r ∧ (p ∨~q)

≡ ~r ∧ (~q ∨ p) Commutative law applied

≡ (~r ∧¬q) ∨ (~r ∧ p) Distributive law applied

≡ (~r ∧~q) ∨ ((~r ∧ ~r) ∧ p) Idempotent law applied≡ (~r ∧¬q) ∨ (~r ∧ (~r ∧ p)) Associative law applied

≡ ~r ∧ (~q ∨ (~r ∧ p)) Distributive law applied

≡ ~r ∧ (~q ∨ ~ (r ∨ ~p)) De Morgan’s law & double negation applied

≡ ~r ∧ ~ (q ∧ (r ∨~p)) De Morgan’s law applied

≡ ~r ∧ ~ (q ∧ (~~r ∨ ~p)) Double negation law applied

≡ ~r ∧ ~ (q ∧ (~r → ~p)) Conditional rewritten as disjunction applied≡ ¬[r ∨ (q ∧ (¬r → ¬p))] De Morgan’s law applied

Proved that ~ [r ∨ (q ∧ (~r →~p))] ≡ ~r ∧ (p∨ ~q)

Discuss about the LU Factorizations.

In linear algebra (Lu-factorization)  LU decomposition factors a matrix as the product of a lower triangular matrix and an upper triangular matrix. The product sometimes includes a permutation matrix as well. Let A be a square matrix. An LU factorization refers to the factorization of A, with proper row and/or column orderings or permutations, into two factors, a lower triangular matrix L and an upper triangular matrix U,

A=LU