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$\noindent 2) Suppose $\lambda$ is a real eigenvalue of $T$ corresponding to invertible eigenvector (which is a matrix here) $X$. Then $AX = \lambda X$. Taking determinants of both sides, we have $\left(\det A\right)\left(\det X\right) = \lambda^{3} \det X$. Since $X$ is invertible, its...

3) Remember, geometrically, eigenvectors are those special directions (really vectors) that only get scaled by T, without having their direction changed (a typical vector won't have this property as it would normally get rotated a bit too). And the factor an eigenvector gets scaled by (which is...

Re: MATH1231/1241/1251 SOS Thread $\noindent They just typoed the integral that they wrote in the answer, but their final answer is correct (so just a pure typo for the first fraction in their integral, which should have instead been $\frac{-x}{x^{2}+2}$).$ $\noindent So they were meant to...

$\noindent At one stage in your working you had a $\sqrt{\sin^{2} x}$ that you simplified to $\sin x$, but it actually is $\left|\sin x\right|$, because $\sqrt{a^2} = |a|$. In our interval of integration, $\sin x$ will be negative sometimes so it is crucial to include the absolute value signs...

$\noindent We need to integrate something like $\sqrt{1+\cos x}$. To do this, we can recall that $1+\cos x = 2\cos^{2}\frac{x}{2}$. Taking square roots of both sides yields $\sqrt{1+\cos x} = \sqrt{2}\left| \cos \frac{x}{2}\right|$. So the function we'll end up integrating is (a multiple of)...

Re: MATH1231/1241/1251 SOS Thread Replace all sin^2 (theta) there with (1 – cos^2 (theta)) and expand and simplify and it should come down to your one. (But it'll be pretty tedious by hand so if you want to confirm it, better to use a computer to do it).

No, you did same thing as Mr_Kap haha. Need to use absolute values.

See here for an example etc. for how to find arc length in polar coordinates: http://tutorial.math.lamar.edu/Classes/CalcII/PolarArcLength.aspx . $\noindent The length is given by $L = \int \, \mathrm{d}{s} = \int_{0}^{2\pi} \sqrt{r^{2} + \left(...

Re: MATH1231/1241/1251 SOS Thread Well you didn't give the full question haha. Here we just needed Ab, not A itself. When you asked, you asked about finding the actual matrix iirc.

Re: MATH1231/1241/1251 SOS Thread $\noindent We can show the equivalence between two apparently different answers by using the fact that $1-\sin^2 \theta = \cos^2 \theta$. Using your method, we have$ $$\begin{align*}\int \sin^3 \theta \cos^5 \theta \, \mathrm{d}\theta &= \int \sin \theta...

$\noindent If $x \ln x = 0$, we obviously have convergence (as the series is just summing $0$ then), so assume now $x \ln x \neq 0$. Let $a_{k} = \left( \frac{x \ln x}{k}\right)^{k}$, then$ $$\begin{align*}\left|\frac{a_{k+1}}{a_{k}}\right| &= |x \ln x| \frac{k^k}{(k+1)^{k+1}} \\ &= |x \ln x|...

If you meant it didn't matter exactly what polynomial in ln(k) was in the numerator, then yes, that's right, it'd work with any polynomial in ln(k) (or any finite sum of powers of ln(k)), as I explained in the later post. It's not fudging because it's something we can prove.

Which part is fudging? Note that for all k sufficiently large, (ln k)^3 < k^{1/6}, so the summand is smaller than something asymptotic to 1/(k^{3/2 – 1/6}), and the sum of 1/(k^{3/2 - 1/6}) converges by the p-test.

It'll converge (absolutely in fact) for any x > 0, because for any fixed x, the denominator k will eventually become bigger than |x*ln(x)| and the series can then be compared to a GP with common ratio smaller than 1. All the terms in the given series will eventually become less than GP terms...

Have you tried looking at the ratio test limit?

Re: MATH1231/1241/1251 SOS Thread (The "Ab stuff" just refers to multiplying the matrix A by the vector b.)

Re: MATH1231/1241/1251 SOS Thread $\noindent Write $\vec{b}$ as a linear combination of $\vec{v}_1$ and $\vec{v}_2$ and then use the given eigenvalue-eigenvector relations to work out the answer. I.e. find (using row reduction or inspection or whatever method you like) the values of...

$\noindent In fact, any polynomial in $\ln k$ will eventually become less than any fixed positive power of $k$ (in fact, become `little-o' of any such fixed power of $k$). So for any sum of the form $\sum \frac{P\left(\ln k\right)}{f(k)}$, ($P(\ln k)$ a polynomial in $\ln k$) where the...

$\noindent a) Convergent. The summand is asymptotically equivalent to $\frac{1}{k^{2}}$ as $k \to \infty$, and $\sum \frac{1}{k^{2}}$ converges. Hence the given series converges.$ $\noindent b) Convergent. Let $a_{k} = \frac{\ln k!}{k^3}$. Note $a_{k}\ge 0$ for all $k\geq 1$. Also, $k! \leq...

By the way, in a conditionally convergent series, the sum of positive terms in the series will diverge and the sum of negative terms will also diverge.