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Differentiator function Lagrange theorem Mathematics Assignment Sample

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Question 1:

(a)f(x) = sin^3x(1-cosx)

The task is to get linear approximation at x= 0

Linear approximation formula is = L(x) ≈ f (x0) +f′(x0) (x−x0).

At the given point function value is 0 (Cai et al. 2019).First derivative format is,

f′(x) = (−3 (cos?(x)−1)cos?(x)+sin2?(x))sin2?(x)

At x=0, the value of first derivative is 0

So the approximation exists and the value of approximation is 0

(b)Second stage differentiation is = 6 cos^2x sin x - 6 cos^3x sin x +10 sin^3 x cos x -3 sin^3 xSecond stage differentiation is = 6 cos^2x sin x - 6 cos^3x sin x +10 sin^3 x cos x -3 sin^3 x

The value of this at x=0 is 0

So the second derivative exists for this function and the value of the second derivative is 0.

(c)Inverse function is = sin^3(1/x) ((1-cos (1/x))) this does not exist at x=0 as there 1/0 format can be found in this section..

Question 2:f(a+v)= f(a)+ Df(a)v +(½)v.Bv +o(?v?)^2

Here the main task is to make the two continuous differentiation of this function. Here two matrices are involved.

L(x) = Bx = b11 b12 · · · b1k b21 b22 · · · b2k . . . . . . . . . . . . bm1 bm2 · · · bmk x1 x2 . . . xk

x = x1 x2 . . . xk ∈ IRk .

The two matrices are B1 and B2.

Question 3:

(a)Let function f: R2→ R be given by,

F (x1, x2) = x1x2/(x12+x22) not equal to (0,0) and 0 at the value of (0,0)

The function is not continuous at (0,0) (Cui and Hong, 2020). If x2 is considered as mx1 then f (x1, mx1) → m/(1+m2) for x → 0

On the other hand, if x2 is x12, then f (x1,x12) → 0 for x → 0

So, as the value of function is changing with the value of x2, this function is not continuous at (0,0)

(b)Partial derivative is done based on x1 and x2.

= ((x12 +). x2 - x1x2(2x1))/()

At (0,0) , = 0 and = 0

= ((x12 +). x1- x1x2(2x2))/()

At (0,0) , = 0 and = 0

So partial derivatives of this function exist at (0,0) (Grigoryeva and Ortega, 2019). For x2= mx1 and for x2= x1 the value of both the derivatives are 0

As this function value is same for 0+ and 0- , so these partial derivatives are continuous at (0,0)

(c)For checking differentiability, limit h, k

(f(0+h,0+k)- f(0,0) - hfx1 (0,0) - k fx2 (0,0))/ √(h^2+k^2)

= f (h,k)/ √(h^2+k^2)

= hk / (h^2+k^2)√(h^2+k^2)

For k=h, value of this above function does not exist as this is a 1/0 format

On the other hand, for k =mh. Above function does not exist because it is a 1/0 format

So this function f (x1,x1) is not differentiable at (0,0)

Question 4:

(a)g(x)= 6x1^2 + 5x1x2/√2+ 7x2^2- 5x1x3/√2+2x2x3+7x3^2

Here three variables are present. These are x1, x2 and x3 (Kim et al. 2020). Normal to the surface is = (12x1, 14x2, 14x3)

Let the position of point E is (a, b, c)

The plane equation is , 12(x1-a)+ 14(x2-b)+14(x3-c) = 0

(b)Maximum point is (a,b,c) if the value of f (0,0,0) is greater than equals f (x1,x2,x3).

(c)Minimum point is (a,b,c) if the value of f (0,0,0) is lower than equals f (x1,x2,x3).

g’(x1) = 12x1+5 x 2/√2 +5x3

g’(x2)= 5 x 1/√2+ 14x2+ 2x3

g’(x3) = 5x1/√2+ 2x2+ 14x3

Question 5:

d((a1,a2),l)= min x ∈ l√((a1-x1)^2+(a2-x2)^2)


Question 6:


g(x) = {0,1,2,3}

f (x) = {-2, -3, -4, -5, -6}


f (x1,x2) = x1-2x2-3

After putting the values of f(x), four equations are found. These are:

x1= 2x2, x1-2z2=-2, x1-2x2=1 and -x1+2x2=1.

All the lines are plotted on the graph. The diagram is shown below.

Derivative with respect to x1 = 1

Derivative with respect to x2 = -2

So the critical point is (1,-2) for this function.

In this point the value of the function is = 2


This critical point is situated on the point (1,-2). This is situated in the fourth quadrant.

Reference list


Cai, A., Chavaudret, C., You, J. and Zhou, Q., 2019. Sharp Hölder continuity of the Lyapunov exponent of finitely differentiable quasi-periodic cocycles.Mathematische Zeitschrift,291(3), pp.931-958.

Cui, J. and Hong, J., 2020. Absolute continuity and numerical approximation of stochastic Cahn–Hilliard equation with unbounded noise diffusion.Journal of Differential Equations,269(11), pp.10143-10180.

Grigoryeva, L. and Ortega, J.P., 2019. Differentiable reservoir computing.J. Mach. Learn. Res.,20(179), pp.1-62.

Kim, W., Kanezaki, A. and Tanaka, M., 2020. Unsupervised learning of image segmentation based on differentiable feature clustering.IEEE Transactions on Image Processing,29, pp.8055-8068.

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