Week 2 Flashcards

Question

How to find equation of a tangent line to a function?

Answer 1

Calculate the slope f ′ (a) of the function f(x) at x=a (the derivative of f(x) at x=a). Use the point-slope form of the equation of a line: y−f(a)=f ′ (a)(x−a) Simplify to obtain the equation of the tangent line: y=f(a)+f′(a)(x−a)

Answer 2

Yes quadratic approximation : f(x)~ f(x*)+f'(x*)(x-x*) + 1/2 f''(x*) (x-x*)^2. there are more high approximations available.. third order .. fourth order.. etc.. but for most of this course we use linear approximation.. in special cases, we may need quadratic approximations..

Answer 3

If one set touched(at most at one place ) another set then it is said to be a tangent. A tangent line touches a function at atmost one point.

Answer 4

If f(x) = g(x). h(x). What is f'(x)? f'(x) = g'(x) h(x) + h'(x) g(x)

Answer 5

If f(x) - g(h(x)), what is f'(x)? f'(x) = g'( h(x)) h'(x)

Answer 6

f: R^d ---> R

Answer 7

1. A line in R^d is subset of R^d 2. A line through the point U belongs to R^d along the vector v belongs to R^d = { x belongs to R^d such that u+alpha.v for alpha belongs to R) 3. Line through u, u' belongs to R^d= { x belongs to R^d such that x=u+alpha (u' -u) for alpha belongs to R} ={ x belongs to R^d: x=(1-apha)u + alpha.u' for alpha belongs to R^d} Note: line through u along u'-u is same as line though u' along u-u'

Answer 8

A (d-1) dimensional hyperplane is subset of R^d. A hyperplane normal to the vector w belongs to R^d with value b belongs to R = {x belongs R^d such that wTx=b} ={x belongs to R^d such that summation (i=1 to d) wi. xi.=b }

Answer 9

Both points and vectors are represented as tuples in R^d, and it will only be in context we could know which is which

Answer 10

finding derivative of a single variable, keeping all other variables constant. (simply.. finding the derivative or slope of a function in a particular direction)

Answer 11

Gradients in s vector containing all the partial derivatives) Gradient = column vector (all partial derivatives)

Answer 12

Find the Gradient: Compute ∇f(a), which is a vector of partial derivatives of f with respect to each variable evaluated at a.

Answer 13

if gradient vector is constant, then the underlying function is linear.

Answer 14

Linear approximation of a multivariate function extends the concept of tangent line approximation from single-variable calculus to functions of multiple variables. It allows us to approximate a complex multivariate function f(x) near a point a using a linear function. Given a multivariate function f(x) and a point a in the domain of f: Tangent Plane: Analogous to the tangent line in single-variable calculus, the tangent plane at a is the best linear approximation of f(x) near a. Linear Approximation: The linear approximation of f(x) near a is given by: L(x)=f(a)+∇f(a)⊤(x−a) where: ∇f(a) is the gradient of f evaluated at a. (x−a) represents the vector difference between x and a. ∇f(a)⊤ denotes the transpose of the gradient vector. v belongs to Rd and x belongs to Rd f(x)~ f(v) + grad(f(v) Transpose . (x-v)

Answer 15

The graph of linear approximations is a plane that is tangent to the graph of f at the point (v, f(v))

Answer 16

The gradient of a function v is going to be perpendicular to the contour passing through that point.

Answer 17

Directional derivate of function f at point v along vector u is lim (alpha tends to zero) (f(v+alpha.u) - f(v)) / alpha

Answer 18

For any vectors u and v in an inner product space (such as Rn or Cn), the Cauchy-Schwarz inequality states that: ∣⟨u,v⟩∣ ^2 ≤⟨u,u⟩⋅⟨v,v⟩ Where: ⟨u,v⟩ denotes the inner product (or dot product) of u and v. ⟨u,u⟩ and ⟨v,v⟩ denote the inner products of u with itself and v with itself, respectively. The Cauchy-Schwarz inequality essentially states that the square of the absolute value of the inner product of two vectors is less than or equal to the product of the norms (or lengths) of the two vectors. In other words, the inner product of two vectors is maximized when the vectors are linearly dependent.

Answer 19

The directional derivative of f at p in the direction of v measures the rate of change of f at p when moving in the direction specified by v.

Answer 20

The directional derivative D1v f(p) is given by: Dv f(p)=∇f(p)⋅v where: ∇f(p) is the gradient of f at p. v is the unit vector specifying the direction.

Answer 21

gradient multiplied by norm of gradient at a point p is the direction of steepest ascent

Answer 22

how to check if a equation of the line is passing through [7,8,6] in the direction of vector [1,2,3]? Equation of line [1,2,3]+α[6,6,3] proof: equate equation of line to point , 1+6α=7 , 2+6α=8, 3+3α=6 , if the parameter α is any value other than zero then the equation satisfies the condition that it passes through the point p(7,8,6) in the direction of vector (1,2,3)

Week 2 Flashcards

(47 cards)