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probably another reason why some people prefer Leibniz's notation. it reads very nicely i just thought of a few questions that i think are interesting. hopefully others find them interesting/surprising(?) too. i'll wait a few days then answer them if no one else has (but hopefully this won't...

still sounds quite hand-wavy. perhaps it can be formalised in terms of nonstandard analysis..? ah this is more like it. though admittedly I've always found that first formula utterly indecipherable (this incarnation looks especially nasty) a few more qns to hopefully get people...

interesting thread! something that used to annoy me: If dy/dx isn't a fraction, then why is it that \frac{dy}{dx} \times \frac{dx}{dy} = 1 \text{ and } \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} (inverse function theorem and chain rule) What are precise statements and how...

this isn't true as stated… 2 + √1 = 1 + √4 so 2 = 1 and 1 = 4, right? you need a, b, c, d to be rational and one of b and d to not be a perfect square edit: take a look at the post above

Re: MX2 Integration Marathon \\\text{Let } I(m,n) = \int^{1}_{0} x^m(1-x)^n \, \mathrm{d}x \text{ where } n \text{ and } m \text{ are positive integers.}\\\begin{align*}\text{Integrating by parts with } u &= (1-x)^n \hspace{7mm} \text{ and } \hspace{6.5mm} \mathrm{d}v = x^m \...

Re: MX2 Integration Marathon \int^{\pi}_{0} \frac{\mathrm{d}x}{1 + \cos^2(x)}

Re: HSC 2013 4U Marathon Hint: Choose \alpha as the largest root(s) and consider the discriminant of \frac{x^3 + ax^2 + bx + c}{x - \alpha}

Re: HSC 2013 4U Marathon oops haha yep

Re: HSC 2013 4U Marathon \\\text{Let } a, b, c \text{ be real numbers such that the roots of the cubic equation }\\[2pt]x^3 + ax^2 + bx + c = 0 \text{ are real}\\[5pt]\text{Prove these roots are bounded above by } \frac{2\sqrt{a^2 - 3b} - a}{3}

Re: HSC 2013 4U Marathon n(n+1)n! lol

There should really be absolute value signs around that: \left | p - q \right | = \left |1 + pq \right |  (unless some other condition is placed on p and q), because the angle between two lines takes the absolute value: \tan\left ( \theta \right ) = \left | \frac{m_{1} - m_{2}}{1 +...

Centrifugal force does exist. (however centrifugal force does not exist in an inertial frame of reference) There's a decent explanation near the end of The Student's Guide to HSC Physics under 'extra content' p.s. i think this comic is relevant

Did you get –80/27?

\begin{align*}\alpha^3 + \beta^3 &= (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2)\\&=(\alpha + \beta)\left (( \alpha + \beta)^2 - 3\alpha\beta \right )\\&=\cdots\end{align*}

\begin{align*}\text{Height} &= 3 + 2\frac{2}{5} + 1\frac{23}{25} + \cdots\\&=3 + 3\cdot\left (\frac{4}{5} \right ) + 3\cdot\left (\frac{4}{5} \right )^2 + \cdots\end{align*} which is an infinite geometric series, with a = 3 and r = 4/5 Using the limiting sum formula (noting that |a| < 1)...

Mathematica refuses to integrate it so I highly doubt it can be found in terms of elementary functions

for future reference... note that z^{n} + z^{-n} = 2\cos(n\theta) then \begin{align*}z^6 + z^3 + 1 &= \left (z^2 -2z\cos\left ( \frac{2\pi}{9} \right ) + 1\right )\left (z^2 -2z\cos\left ( \frac{4\pi}{9} \right ) + 1\right )\left (z^2 -2z\cos\left ( \frac{8\pi}{9} \right ) + 1\right...

here's another way \begin{align*}\sqrt{n+1} &> \sqrt{n}\\\frac{n+1}{\sqrt{n+1}} &> \frac{n}{\sqrt{n}}&&\text{rationalising numerator}\\\frac{\sqrt{n}}{\sqrt{n+1}} &> \frac{n}{n+1}\\&= 1 - \frac{1}{n+1}\\\frac{\sqrt{n}}{\sqrt{n+1}} + \frac{1}{n+1} &> 1\\\therefore \sqrt{n} +...

It helps a lot for the 3rd step if you state what you wish to prove, in this case: 2(k+1) + 2 < 2^{k+1} We then work the RHS (since it will be easier to apply the hypothesis) \begin{align*}2^{k+1} &= 2\times 2^k\\&> 2\Big ( 2k + 2\Big ) &&\text{using the induction hypothesis}\\&= 4k + 4\\&>...

okay so you're concerned about this step: 2\times3^k < 3\times 3^k I want to emphasise the 'less than'; the 3rd line is less than the 2nd (not equal to), allowing us to 'replace' the 2. \begin{align*}2 &< 3\\2 \times 2^k &< 3 \times 2^k &&\text{multiplying both sides by } 2^k \quad...