User:EGM6321.f12.team7.Zhou/R5

Problem 3: Prove property of eigenvalue matrix
Report problem 5.3 from section 20-6

Given: The expression of eigenvalue matrix, the diagonal matrix and exponentiation of matrix
The exponentiation of matrix is given by,

For a matrix$$A$$,

The eigenvalue matrix is made of matrix of eigenvector as shown below. Suppose we have a matrix $$\mathbf{A}$$. The eigenvector is given by,

$$\phi$$ is eigenvector, If there are n linearly independent eigenvectors, we will have,

So the the equation can be written as,

Where,

The matrix can be diagonalizable, which is given by,

Find: The expression of exponentiation of matrix can also be diagonalizable
$$\exp \mathbf A =\mathbf \phi \, \text{Diag}[e^{\lambda_1}\cdots e^{\lambda_n}]\, \mathbf \phi ^{-1}$$

Solution

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On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. Use equation ($$) into equation($$), we have,
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Since $$\mathbf \phi \, \mathbf \phi^{-1} =1$$, we have,

According to associative property, which is given by,

where $$\mathbf A, \mathbf B ,\mathbf C \in \mathbf E^{n\times n}$$. So the equation turns into,

In the problem 2, we have already prove that,

So the equation turns into,

Problem 9: Use Maclaurin series to expand equations
Report problem 5.9 from section 64

Find: Maclaurin series expansion and definition of hypergeometric function
The Maclaurin series is an Taylor series expansion of function at x about 0. Which is given by,

Where, $$f^n(x) = \frac{d^n f(x)}{dx^n} $$

Two functions are given by,

$$(1-x)^{-a}$$

$$\frac{1}{x}\arctan (1+x)$$

The notation of hypergeometric function is given by,

Where $$-1< x < 1$$,

$$\begin{cases} a(0):=1\\a(k)=a(a+1)\cdots(a+k-1)\end{cases}$$

Solution

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On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

First function
SInce $$1-x=1$$. the expansion of this function is given by,

According to the notation,when $$ b = c=d$$, the equation satisfied, so the expression is given by,

Second function
The directly expansion of this function is wrong, since the $$\frac{1}{x}$$ is undefined at the zero point and direct derivation is complicated. But we can expand the $$\arctan (x+1)$$ first, then divide it by x.

Using the wolframe Alpha, we can get the derivatives of $$\arctan(1+x)$$, which are given by,

Where,

$$f^{(i)}= \frac{d^i f}{(dx)^i}$$

The derivatives seems like no path. But when we plug $$ x=0$$ into the equations, we will have,

So obviously at x=0, the constant doesn't fit the expressions given in the lecture note.

Personally I think when we use x=-1, we could have,

So we can get the expansion of $$ \arctan(x+1)$$, which is given by,

equation($$) is divided by x, then we get,

So the condition of the problem should be changed, then the answer is satisfied.