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$\noindent If this was the intention of the solution method, maybe the Q. was typed up incorrectly? The Q. asks to find $\left(\alpha+\beta - \gamma \right)\left(\beta + \gamma -\alpha \right)\left(\gamma +\alpha - \beta \right)$, where $\alpha,\beta,\gamma$ are the roots of $x^3 -2x^2 +...

$\noindent Note that clearly $f(x)>0$ whenever $x\geq 0$. Note $f(x) = \frac{1-x^{2n+1}}{1-x}$ by the GP sum formula for any $x\neq 1$. For negative $x$, we have $x^{2n+1}<0$, so $1-x^{2n+1}>0$, and $1-x>0$, so $f(x) > 0$. So $f(x)>0$ for any real $x$.$

$\noindent By the way, once you've proven the formula for $m_{X}\left(u\right)$ (or even if you don't manage to prove it), the one for $Y$ is easy (since the one for $X$ is given), it's just $m_{Y} \left(u\right)= m_{X} \left(2pu\right)$. In general $m_{aX}\left(u\right)$ will be just $m_{X}...

That one is also fairly tedious I think. I might have a look at it later.

$\noindent Linear interpolation essentially works as follows. Say we know the values $y_{1} = f\left(x_1\right)$ and $y_{2}=f\left(x_2\right)$ of a function $f$ at two particular points $x_1$ and $x_2$, where $x_{1}<x_{2}$, and we need to estimate the value of $y^{\star}$ of the function $f$ at...

I think you forgot to use product rules or quotient rules etc. where needed. Like for the (∂/∂r)[∂ƒ/∂θ* (sin(θ)/r)] term in that second line, you should use product rule because the second function here (sin(θ)/r) is dependent on r (the sin(θ) part is a constant wrt r, but the r isn't of course...

$\noindent Here would be a way to find $z_{0.475}$ from a standard normal table. This is slightly tricky (i.e. not as trivial as just reading off a table) because such tables normally only list $\Phi \left(z\right)$ values for $z>0$, so since clearly $z_{0.475}<0$, we'll need to use symmetry...

$\noindent Let the voltage r.v. (random variable) be called $X$. Let its mean be called $\mu$ (unknown). We are given $\sigma _{X}\equiv \sigma = 2$ (the standard deviation). Let the $n$ readings be $X_{1},X_{2},\ldots, X_{n}$, then $\overline{X}_{n} = \frac{1}{n}\sum _{i=1}^{n}X_{i}$. We want...

$\noindent Yes, if $f(t) \leq g(t)$ for all $t\in \left[a,b]$, then $\int _a ^b f(t)\text{ d}t \leq \int _a ^b g(t)\text{ d}t$.$

Remember h is the interval width. Have you tried using the triangle inequality as leehuan suggested? If so, what have you managed to get up to using that?

Remember that m1 is the midpoint of the interval. If we integrate a linear function over an interval and the function takes value 0 at the midpoint, the area above and below the curve on either side is equal, so the integral is 0. (Alternatively, just integrate it normally and remember m1 is...

It depends on which where you want x to start (at t = 0). Note that cos(t) starts at its maximum (and begins moving to the minimum extreme) whilst -cos(t) starts at its minimum (and begins moving towards the maximum extreme). So if you wanted a Simple Harmonic Oscillator to start at its...

$\noindent For the MVT for integrals one, recall that this theorem states that if $F$ is continuous on $[a,b]$ and $G$ is an integrable function that doesn't change sign on this interval, then there is a $c \in (a,b)$ with $\int _a ^ b F(x) G(x)\text{ d}x = F(c) \int _a ^b G(x)\text{ d}x$. So...

For the Gamma function one: $\noindent Let $I = \int _{0}^{\infty} t^{s-1} e^{-t}\text{ d}t$, where $s>0$ is fixed. Sub. $t = u^{-1}$, and $\mathrm{d}t = -u^{-2}\text{ d}u$. Updating the limits, this gives$ $$\begin{align*}I &= \int _0 ^\infty u^{1-s} e^{-\frac{1}{u}}\times \frac{1}{u^2}\text{...

Yeah it's essentially a comparison test. For all t large enough, the integrand in the second integral is greater than the first (since t^2 + t > 2t for all t large enough, say all t > 1). Since the former integral taken from 1 to oo diverges to +oo (from previous part of Q.), by the comparison...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Essentially, use trig. identities. $\noindent For example, for (a), get everything in terms of just $\cot ^2 \theta$ or just $\csc^2 \theta$, using the trig. identity $\cot^2 \theta + 1 = \csc^2 \theta$. Say you did...

The HSC has been known to have physically incorrect answers as their stated correct answers in the past for HSC Physics.

$Let the error on the $i$th reading be $\varepsilon _{i}$ (this is a random variable with a uniform distribution on the given interval). Note that the overall absolute error in the sum is $\left |\sum _{i=1} ^{n} \varepsilon _{i} \right|$. Recall from uniform distribution formulas that for each...

This isn't CLT either. It's simply induction using previous results (similar to the Bonferoni's Inequality induction method I mentioned in another thread if you remember it).

This doesn't really have anything to do with CLT. $\noindent This result is getting us to show that the sum of two independent Poisson r.v.'s is Poisson, with parameter equal to the sum of the two original ones' parameters (and it follows from this by induction that we can extend the result...