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13

In general, it holds that $$\sqrt{n(n-1)+\sqrt{n(n-1)+\sqrt{n(n-1)}}}=n-\frac{1}{8n^2}+O\left(\frac1{n^3}\right)$$ and that $$\sqrt{n(n-1)-\sqrt{n(n-1)-\sqrt{n(n-1)}}}=(n-1)+\frac{1}{8n^2}+O\left(\frac1{n^3}\right)\,$$ for all $n\geq 1$. Hence, their difference is $$1-\frac1{4n^2}+O\left(\frac1{n^3}\right)\,.$$ In particular, for $n=5$, the difference ...

7

The rational function $$\frac{18x}{x^2+4x+1}-2$$ is an extremely good approximation on $[\tfrac12,2]$ and substituting $x=2$ (or $x=\tfrac12$) gives $\tfrac{10}{13}$. I stumbled upon this by noting that $$\frac1{1-\cos(\log(x))}= \frac2{(x-1)^2}+\frac2{x-1}+\frac13+O((x-1)^4)$$ and therefore that $$\frac2{(x-1)^2}+\frac2{x-1}+\frac13$$ is a good ...

7

Even without a calculator, you can do the numerics fairly easily. 20 is about halfway between 16 and 25, so $\sqrt{20} \approx 4.5$. So $20\pm\sqrt{20}$ is about 15.5 and 24.5, respectively. These in turn have square roots of about (a little less than) 4 and 5. This leaves $\sqrt{25}−\sqrt{16}\approx 1$. The "little less than" might contribute significant ...

5

If we consider an infinite chain. Suppose $x = \sqrt{20 +\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{\cdots}}}}}$ $x = \sqrt{20 +x}\\ x^2 = 20 + x\\ x^2 - x - 20 = 0\\ (x-5)(x+4) = 0$ $x$ must be greater than $0, x = 5$ and $y = \sqrt{20 -\sqrt{20-\sqrt{20-\sqrt{20-\sqrt{\cdots}}}}}$ $y = \sqrt{20 - y}\\ y^2 + y - 20=0\\ y = 4$ $x-y = 1$ As we add more terms ...

5

We will make some extra assumption on $f(x)$ and shows that under such assumption, the limit diverges unless $f(0)+f(1) = 0$. Let $f : [0,1] \to \mathbb{R}$ be any $C^2$ function on $[0,1]$, i.e. twice differentiable and the $2^{nd}$ derivative $f''(x)$ is continuous. For any $0 \le a < b \le 1$, let $h = b - a$ and consider following integral $$\... 3 Use the classical approximation:$$\sqrt{a^2 + b} \approx a + \frac{b}{2a}$$With a = \sqrt{20} and b = \sqrt{20 + \sqrt{20}} we have$$\sqrt{20 + \sqrt{20 + \sqrt{20}}} \approx \sqrt{20} + \frac{\sqrt{20 + \sqrt{20}}}{2\sqrt{20}} = \sqrt{20} + \frac{\sqrt{400 + 20\sqrt{20}}}{40} Now use the same classical approximation again, this time working with ... 2 One possible way is to look at ratios: about a factor of 1.4 from 3500 to 5200, about 2.3 from 5200 to 120000. If those two factors had both been the same, you could use them to guess the next term. Alas, they're pretty different. One thing that might help is if we knew something about the source of the numbers. If they're the number of thousands of shares ... 2 By binomial formulas and cancellation, you get that \begin{align} ...&=\frac{\sqrt{20+\sqrt{20}}+\sqrt{20-\sqrt{20}}}{\sqrt{20+\sqrt{20+\sqrt{20}}}+\sqrt{20-\sqrt{20-\sqrt{20}}}} \\&=\frac{ \sqrt{5+\sqrt{\frac54}}+\sqrt{5-\sqrt{\frac54}} }{ \sqrt{5+\sqrt{\frac54+\sqrt{\frac5{64}}}}+\sqrt{5-\sqrt{\frac54-\sqrt{\frac5{64}}}} } \end{align} As ... 2 Repeatedly using \sqrt{1+x} \approx 1+x/2, \begin{array}\\ d(a) &=\sqrt{a+\sqrt{a+\sqrt{a}}}-\sqrt{a-\sqrt{a-\sqrt{a}}}\\ &=(\sqrt{a+\sqrt{a+\sqrt{a}}}-\sqrt{a-\sqrt{a-\sqrt{a}}})\dfrac{\sqrt{a+\sqrt{a+\sqrt{a}}}+\sqrt{a-\sqrt{a-\sqrt{a}}}}{\sqrt{a+\sqrt{a+\sqrt{a}}}+\sqrt{a-\sqrt{a-\sqrt{a}}}}\\ &=\dfrac{(a+\sqrt{a+\sqrt{a}})-(a-\sqrt{a-\... 1 Setup hint: Start with finding a suitable matrix A^{2\times2} such that AA^T=\Sigma. Then write X=AU+\mu where U=(U_1,U_2)^T is a random vector such that U_1,U_2 are iid and have standard normal distribution. Then X has the distribution that you mention and:\Pr\left(X\in C\right)=\Pr\left(\left(X-\mu\right)^{T}\left(X-\mu\right)\leq R^{2}\...

1

Actually, it's the mid-point rule for integrals \begin{align*} \int_{a}^{b} f(x)\, dx &= (b-a)f\left( \frac{a+b}{2} \right)+ \frac{(b-a)^{2}}{24} [f'(b)-f'(a)] \\ &\quad \: - \frac{1}{6} \int_{a}^{\frac{a+b}{2}} (t-a)\left( t-\frac{a+b}{2} \right) \left( t+\frac{b-3a}{2} \right) f'''(t) \, dt \\ & \quad \: - ...

1

Usually, you reduce the partition problem to the bin packing problem as follows: Let an instance of the partition problem be given by integers $a_1, \ldots, a_n$. Define an instance of the bin packing problem by integers $a_1, \ldots, a_n$ and bin size $\frac{1}{2} \sum_{i = 1}^n a_i$. Then the partition problem has a solution if and only if the bin packing ...

1

The stopping condition used in the video is a bit unusual for bisection, but would be very common for other techniques. He is stopping once $|g^2-x|<\epsilon$ where $g$ is the current guess. He is not stopping once $|g-\sqrt{x}|<\epsilon$, even though in bisection one can proceed this way. Here $|g^2-x|$ is called the forward error and $|g-\sqrt{x}|$ ...

1

As in @mweiss 's answer we use repeatedly the approximation $$\sqrt{a^2+b}\approx a+{b\over 2a}\qquad(|b|\ll a^2)\ .$$In this way we obtain on the one hand \eqalign{ \sqrt{20}&=\sqrt{25-5}\approx 5-{1\over2},\quad 20+\sqrt{20}\approx25-{1\over2},\cr \sqrt{20+\sqrt{20}}&\approx5-{1\over20},\quad 20+\sqrt{20+\sqrt{20}}\approx 25-{1\over20},\cr \sqrt{...

1

Hint: As long as there are $z_n \in D$ with $|f_n(z_n)| \ge 1$, say, the convergence can't be uniform in $D$.

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This is not an answer: Welcome to Berkeley Logo version 5.5 ? sqrt 378 You don't say what to do with 19.4422220952236

1

If I understand correctly you want to approximate $e^1$ by the value of the function $f(x)=e^x$ which has the value $f(0)=1$. By the Taylor theorem you have that $f(1)=\sum_{n=0}^5 \frac{f^{(n)}(0)}{n!}(1-0)^n+R_5(f)$. Evaluating $R_5(f)$ is by the Lagrange form of the reminder which is the value of the next factor in the sum $\frac{f^{(6)}(c)}{6!}(1-0)^... 1 I'll try to address some of the misunderstandings I think you have based on the question and the comments. Someone else can tackle the problem given in$(3)\$ of explaining everything about Taylor polynomials and Taylor series from the ground up if they want (though I doubt anyone will). First off, it sounds like you're conflating Taylor polynomials with ...

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