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Re: MATH1131 help thread $\noindent Recall that $e^{i\theta}+e^{-i\theta}=2\cos \theta$ (which follows from Euler's formula as posted above).$ (And I think you forgot to type the i's in the exponents.)

Re: MATH1131 help thread $\noindent Split up the fraction, so it becomes $\frac{2}{u^3}-\frac{1}{u^2}$, which is easy to integrate now.$

Re: MATH1131 help thread $\noindent So the integral becomes $I = \int \frac{1}{u}\times 2\left(u-1\right)\text{ d}u$. Write $\frac{u-1}{u}$ as $1-\frac{1}{u}$, so then $I = 2\int \left(1-\frac{1}{u}\right)\text{ d}u=2u - 2\ln |u| +C$. Since $u=1+\sqrt{x}$, this becomes $2+2\sqrt{x}-2\ln...

Re: MATH1131 help thread $\noindent What's the question? Usually dot product is better because it's faster to compute. The cosine of the angle between the two vectors is given by $\cos \theta = \frac{\vec{a}\cdot \vec{b}}{\left \|\vec{a}\right\|\left \|\vec{b}\right\|}$.$

Re: MATH1131 help thread $\noindent Because $\cos \pi=-1$.$

There's an easier way, just show that for large enough values of u, cos(1/u) becomes bounded below by some positive constant. From this, show that the integral has to go to infinity.

You can do it via L'Hôpital. To show the numerator goes to infinity, try bounding the integrand.

$\noindent It has to be the second one since the projectile would be thrown up, so $\frac{\mathrm{d}h}{\mathrm{d} t}$ should be positive at $t=0$. This'll happen with the second one, but not the first.$ (The physical significance of the coefficient of the t term is that it is the initial...

The question was typo'ed I think, it was meant to be h = 7 + 6t – t2 (a plus sign was written as =, which is an easy typo to make).

Re: MATH1131 help thread $\noindent Also, if we view complex numbers as being represented vectors in the plane (very common thing to do), then $z-w$ would be represented by the vector that points from $w$ to $z$. Also, $|z|$ would be the length of the vector for $z$ (which goes from the origin...

Re: MATH1131 help thread $\noindent In general, say $w$ is a fixed complex number, then $|z-w|$ is the distance between $z$ and $w$ in the complex plane. So geometrically, if $a$ is a positive constant, all the points with $|z-w|< a$ will be correspond to the set of all points in the complex...

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread I'm not the most knowledgeable.

Re: MATH1131 help thread $\noindent Since $A^8 =16I$, we have $\left(A\right)\left(A^7\right)=16I$. So dividing through by $16$, we have $\left(\frac{1}{16}A\right)\left(A^7\right)=I$. Hence the inverse of $A^7$ is $\frac{1}{16}A$, which we can write down since we're given $A$.$

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread Pascal's Triangle: 1 1 1 1 2 1 (n = 2) 1 3 3 1 (n = 3) 1 4 6 4 1 (n = 4) 1 5 10 10 5 1 (n = 5) etc. The rule was that nC2 + n+1C2 = n2. In terms of the blue numbers above, the sum of consecutive ones will be n2, where n is...

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread Yes, k = 2.

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent What it means is check up values in the third column of various rows of the Pascal triangle and see that the result holds.$ $\noindent $\Big{(}$Students are expected to know that the Pascal triangle entries are...

$\noindent So intuitively we need to show that there is some `tolerance' level $\varepsilon >0$ for which we can't get our function to go and stay within that tolerance level of the claimed limit of $1$ here. (Since if we can show this, it means our function can't get and stay arbitrarily close...

$\noindent To show that $\lim _{x\to \infty}f(x) \neq L$, we would need to show that there exists an $\varepsilon >0$ such that for any $M > 0$ there will exist an $x_0>M$ such that $|f(x_0)-L|\geq \varepsilon$. (If you know a bit about formal logic, all I've done here is take the negation of...

From a quick glance, Q.22 to Q.30 are by definition.

Such as Q.11.