Recent content by chisigma

Of course the first order DE... $\displaystyle y^{\ '} = \frac{y}{x},\ y(x_{0}) = y_{0}\ (1)$... can be solved separain the variables and its soltion is $\displaystyle y=c\ x$...if Youy want to use the Frobenious method however, You must hypotize that $y(x)$ is analitic in $x_{0}$...Kind...

First we write the PDE in the more 'conventional' form... $\displaystyle x\ u_{x} + y\ u_{y} = \alpha\ u;\ u(x,1)= g(x)\ (1)$ Applying the standard Method of characteristic curves You arrive to... $\displaystyle \frac{d x}{x} = \frac{dy}{y} = \frac{d u}{\alpha\ u}\ (2)$ ... which is a...

The difference between integrating in $\mathbb {R}$ and integrate in $\mathbb {C}$ is that in the second case it must be generally specify not only the limits of integration a and b, but also the path that connects a and b... Let's consider the following illustrative example taking $f(z) =...

If the integral $\displaystyle \int_{a}^{b} f(z)\ d z$ does not depend on the path connecting a and b in the complex plane, then f(z) is analytic in the whole complex plane and effectively is $\displaystyle \int_{a}^{b} f(z)\ d z = - \int_{b}^{a} f(z)\ d z$... ... otherwise, i.e. if f(z) has...

All right!... You can find a solving the cubic equation... $\displaystyle a^{3} - 1 = (a - 1)\ (a^{2} + a + 1) \equiv 0\ \text{mod}\ 24\ (1)$ Now $\displaystyle a - 1 \equiv 0\ \text{mod}\ 24$ has the only solution $\displaystyle a\ \equiv 1\ \text{mod}\ 24$ and that means a=1, 25, 49, 73...

Because 24 is even and 23 is odd, a must be odd and is a<5... then 1 is solution and 3 isn't solution... Kind regards $\chi$ $\sigma$

By sight You notice that x = 1 and a solution of the equation... $\displaystyle P(x) = x^{3} - 5\ x^{2} + 8\ x - 4 = 0\ (1)$ ... so that (x-1) divides P(x)... then divide P(x) by (x-1) and obtain a second order polynomial that possibly can be further factorized... Kind regards $\chi$...

What is required is the range of uniform convergence and not simply the range of convergence, that You have found... now the function $\displaystyle \zeta(x)= \sum_{n=1}^{\infty} \frac{1}{n^{x}}$ has a singularity in x=1 and that means that the range of uniform convergence is $s \le x < \infty$...

The man was born in the year 1991 so that in the year 2011 his age was 1 + 9 + 9 + 1 = 20 years ... Kind regards $\chi$ $\sigma$

On this occasion I am afraid that the first 'logical step' is not true ... ... more precisely if You have a series $\displaystyle \sum_{n=1}^{\infty} a_{n}$ and You demonstrate that it converges, i.e. is $\displaystyle \lim_{k \rightarrow \infty} \sum_{n=1}^{k} a_{n}$ exists and it is equal to...

If the series 1 - 1 + 1 - 1 +... diverges, then writing S= 1 - 1 + 1 - 1 +... is a nonsense because no real number S exists that satisfies such a requirement... Kind regards $\chi$ $\sigma$

Welcome on MHB Peterw222!... ... the function... $\displaystyle f(x,r) = \sqrt{x} - r\ x\ (1)$ ... is linear in r and non linear in x... what You have to do is, given r, find the value x* that maximizes (1) and construct f*(r) = f(x*,r)... Kind regards $\chi$ $\sigma$

Because $e^{- y}$ never vanishes you can divide by it and the conditions fom maximum or minimum become... $\displaystyle 2\ x = x^{2}-2 = 0\ (1)$ ... and there is no value of x satisfiyng (1)... Kind regards $\chi$ $\sigma$

Your matrix is [4 3 2; 1 7 8; 3 9 3]... a) divide the first row by 4 and subtract it fron the second row obtaining [4 3 2; 0 25/4 15/2; 3 9 3]... b) multiply the first row by 3/4 and subtract it from the third row obtaining [4 3 2; 0 25/4 15/2; 0 27/4 3/2]... c) multiply the second row by...

The homogeneous equation is... $\displaystyle a_{n+2} + 4\ a_{n} = 0\ (1)$ ... and its solution is... $\displaystyle a_{n} = 2^{n}\ \{ c_{0}\ i^{n} + c_{1}\ (- i)^{n}\}\ (2)$ The particular solution we have found is... $\displaystyle w_{n} = 8\ 2^{n}\ (\chi_{0} + \chi_{1}\ n)\ (3)$ ...