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3

This is not integrable because $\int_0^1 \frac{1}{x}e^{-x} dx > \frac{1}{e}\int_0^1 \frac{1}{x} dx = \infty$

2

I believe you have shown everything you need (apart from the mistake you have made). Any elementary reflector $Q$ of the form above is symmetric and satisfies \begin{align} Q^T Q &= (I + \alpha xx^T)(I + \alpha xx^T) \\ &= I + 2\alpha xx^T + \alpha^2 (xx^T)(xx^T) \\ &= I + 2\alpha xx^T + \alpha^2 x(x^Tx)x^T \\ &= I + 2\alpha xx^T + \alpha^2 ...

1

There is no general rule, and you must carefully analyse the function. The first function is $$f(x_1,x_2) = \|(x_1,x_2)\|^2 +2x_1 -3x_2 -2$$ and it is coercive because the term $2x_1 -3x_2 -2$ grows linearly at infinity. The second example is not coercive, since $$f(0,x_2) = -x_2^2 \to -\infty$$ when $|x_2| \to +\infty$

1

If we start at $x_0=0$, then the next member of the sequence is $-\frac{f(x_0)}{f'(x_0)}=-\infty$, so the method fails. The line $y=kx$ is tangent to the curve at $x$ if $k=3x^2$ and $kx=x^3-1$, so $x=-\left(\frac{1}{2}\right)^{1/3}$. So if we take $x_0=-\left(\frac{1}{2}\right)^{1/3}$ we are also in trouble because the next approximation will be $x=0$ and ...

1

Another fun way of doing it is by means of a generating function. Define $$A(t) = \sum_{n\geq 0} x_nt^n = x_0 + x_1 t + x_2 t^2 + \dots \quad (1)$$ for real $t<1$. Then multiply your recursion formula by $t$ and sum to get $$\sum_{n\geq 0} x_{n+1} t^n = \frac{1}{2} \sum_{n\geq 0} x_n t^n + \frac{1}{2} \sum_{n\geq 0} t^n.$$ Note that the left-hand ...

1

To elaborate a little around @Winther's comments: For a first order ODE with an initial condition, if the right-hand side is Lipschitz continuous then, by the Picard-Lindelöf theorem, a unique solution exists (i.e. if you find one solution, you have found all solutions). Here you have specified two conditions on your solution. Generally speaking, if you ...

1

Euler's method: $r(0) = 2.5\, mm\\ v(0) = \frac 43 \pi 2.5^3 \,mm^3\\ A(0) = 4\pi 2.5^2 mm^2\\ v'(0) = -0.08 * A(0) = -2\pi \,mm^3 / m\\ v(0.25) = v(0) + 0.25 v'(0) = \frac{61\pi}{3}\\ r(0.25) = (\frac{3}{4\pi} 61\pi)^{\frac13} = \frac{183}{4}^{\frac13}\\ A(0.25) = (183)^{\frac23}(4)^{\frac13}\pi\\ v'(0.25) = -0.08 A(0.25)\\ v(0.5) = v(0.25) + ... 1 By definition, the absolute value of the$i$th pivot$\mu_{ii}$maximizes the absolute value of all elements in trailing submatrix. In particular, the$i$th pivot is larger than elements in the first row of the trailing submatrix. This row constitutes the upper triangular elements of the$i$th row of the matrix$U$. In short, for each row of$U$the element ... 1 I think you can do it with vectors. Define $$\mathbf{v} = \left( \begin{matrix} x & y & x' & y' \end{matrix} \right)^T$$ and set up Runge-Kutta method in the form $$\mathbf{v}' = f(t, \mathbf{v})$$ 1 I wouldn't modify the code except to suppress output on the last line of gauss3.m. All you need is a function that chops up$[a,b]$into$N$subintervals and invokes gauss3.m$N\$ times and adds up the results: % gaussq.m function y = gaussq(func,a,b,N); h = (b-a)/N; y = 0; for i = 1:N, y = y+gauss3(func,a+(i-1)*h,a+i*h); end Then you want a script ...

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