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For (iv), differentiate the binomial theorem equation with respect to x. Then, sub. in x = 22 = 4.

For (ii), integrate the binomial theorem equation twice (use x = 0 to get the values of the constants of integration each time you integrate). Finally, plug in x = 1.

Not sure. I think it's been lagging like this for the past few days.

I think he wrote it in LaTeX, but it's just been lagging on this site recently.

Re: Several Variable Calculus $\noindent Note that this is an ``ice-cream cone'' shape where the sphere intersects the cone above the $x$-$y$ plane in the horizontal plane $z = a$ (assuming $a > 0$) and makes a circle with $x^{2} + y^{2} = a^{2}$ in this plane in the intersection. So if we...

Re: Discrete Maths Sem 2 2016 $\noindent The first highlighted part is just writing the LHS of the RTP (required to prove) statement out in full. The last term in the LHS there is $2(k+1) = 2k+2$. This means the product goes all the way up to $2k+2$, so they've just written in the last few...

Re: Discrete Maths Sem 2 2016 Because that says an even number is equal to an odd number, which we know is impossible.

If you know a place where the polynomial is positive and another place where it is negative (and the root you already found does not lie between these two places), then the other root lies between these two numbers. You can then use the bisection method (check your textbook if you haven't learnt...

Re: maths questions help .

$\noindent a) To solve this, we can use a double angle formula. We know $\cos 4x = 2\cos^{2}2x - 1$. Substituting this into the original equation will give us a quadratic equation in $\cos 2x$, and you should be able to go from there.$

$\noindent b) The solution is $3x = x + 2n\pi$ for some integer $n$ or $3x = -x + 2n\pi$ for some integer $n$. That is, $x = n \pi$ for some integer $n$ or $x = n\frac{\pi}{2}$ for some integer $n$. As $n$ varies through the integers, we observe that the solutions picked up are precisely...

Re: maths questions help Are there any examples in the textbook using bisection method? See this page if not (there's an example in it): https://en.wikipedia.org/wiki/Bisection_method .

Re: maths questions help If you know a place where the polynomial is positive and another place where it is negative (and the root you already found does not lie between these two places), then the other root lies between these two numbers. You can then use the bisection method (check your...

Re: MATH1131 help thread $\noindent Here, think of $x$ as some given number in $\left(\pi, \frac{3\pi}{2}\right)$ and $\theta$ being something depending on $x$. We can't just sub. arbitrary values for $\theta$ and $x$. Rather, $x$ is given and we are finding what $\theta$ has to be. We know...

Just find the roots of the RHS function of the DE y' = f(y). $\noindent In more detail, equilibrium solutions to an autonomous differential equation $y' = f(y)$ are just constant solutions of the form $y = y_{0}$ where $y_{0}$ satisfies $f(y_{0}) = 0$ (it is easy to see that these are...

Re: MATH1131 help thread $\noindent Actually, $\theta$ \emph{is} $\cos^{-1}(\cos x)$, which is what we want to find. This gives $\cos \theta = \cos x$, so (see e.g. Year 12 3 Unit Pender textbook), we have $\theta = x + 2k\pi$ for some integer $k$ or $\theta = - x + 2\ell\pi$ for some integer...

Re: MATH1131 help thread $\noindent That's not the right answer I'm afraid. Basically, in general, for any $u\in [-1,1]$, $\cos^{-1}u$ is the angle in the range $[0,\pi]$ whose cosine is $u$ (such an angle exists and is unique for any $u \in [-1,1]$). So $\cos^{-1}(\cos x)$ is that angle in...

Re: MATH1131 help thread $\noindent Pretty much (and because this gives the correct value). As an exercise, you may wish to try the following: simplify $\cos^{-1}(\cos x)$ for $\pi < x < \frac{3\pi}{2}$.$

Re: MATH1131 help thread $\noindent What is the question? Sketch that curve? Simplify the expression? Anyway, just note that $\sin^{-1}(\sin x) = \pi - x$ for all $\frac{\pi}{2}\leq x \leq \frac{3\pi}{2}$. To see this, you should try and make sure you understand the sine function's periodic...

Re: maths questions help $\noindent Have you learnt how to solve quadratic inequations like this? If not, you should probably learn that first, since you need to know how to do it for this question. Basically you find the roots of the quadratic $m^2 + 42m - 23$, sketch the parabola, and deduce...