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34

The apparent paradox in this question arises from the conflict between the following two ideas: Our intuition about physical lamps. We expect them to have certain properties, such as being on or off, but not both or neither. A complicated function that oscillates between two values at an increasing pace, whose oscillations become arbitrarily rapid. It ...


5

This is known as Thomson's lamp. The Wikipedia article is probably less enlightening than the answers given here, and I don't think that there's much more to be said about it, but the search term will lead you to more articles if you want them. Added: Your reasoning is correct. By definition, the lamp must be in one of two states: on or off. This state is ...


4

Some infinite series converge, and some do not. In this case, because of the halving of the interval between each switching event, the total time for an infinite number of such events does converge, and converges to a small number (1 minute if you start with a 30 second gap). However the state of the lamp is represented by a different infinite series which ...


4

A subsequence is just some of the terms of the original sequence, kept in order. If your original sequence is $1,2,3,4,5,6\dots $, one subsequence is $1,3,5,7,9 \dots $ Another is $1, 2343, 23565848, 8685845855858, \dots $ Finding a subsequence is easy. Finding one that does what you want depends on what you want.


4

You are almost there $$ a_{n+1} = a_1 + nd\\ a_n = a_1 + (n-1)d $$ thus we get $$ a_{n+1}a_n = \left(a_1 + nd\right)\left(a_1 + (n-1)d\right) $$ re-arrange your given relation you find $$ a_{n+1}a_n =9n^2-21n + 10 $$ can you equate coefficients?


3

The lamp will be off. Because you switch infinite many times, and every switch takes some energy to be done, you will have exhausted all energy in the universe. Hence it is off.


2

Consider the sequence $a_n=1-2^{-n}$. The situation that you describe could be mathematically described this way: For each $x\in[0,1)$ there exists some unique $n\in\Bbb N$ such that $x\in[a_n,a_{n+1})$. Define $f(x)=1$ if this $n$ is even and $f(x)=0$ otherwise. As you can see, the value $f(1)$ remains undefined. So, assuming that the job of toggling ...


2

For your first question: If $z=\pm n$ then $1-\tfrac{z^2}{n^2}=0$, so the entire product equals $0$. To see that the product converges we remove this term from the product, and the argument below serves just as well to show that the product converges. Your second question does require some effort. I don't see how to show that ...


2

We have $$a_2=\frac{9-21+10}{a_1}\Rightarrow a_1a_2=-2\tag1$$ and $$a_3=\frac{36-42+10}{a_2}\Rightarrow a_2a_3=4\tag2$$ Since we have $a_1+a_3=2a_2$, with $(1)(2)$, we have $$a_1+\frac{4}{a_2}=2a_2\Rightarrow a_1a_2+4=2a_2^2\Rightarrow a_2=\pm1.$$ Can you take it from here?


2

No, we may not conclude that $a_n$ converges. Consider $$a_n=\sum_{j=1}^ns_jj^{-1}$$ where $s_j=\pm 1$. We can choose the $s_j$ so that they are positive until $a_n$ becomes greater than one, then negative until $a_n$ becomes less than zero, then positive until $a_n$ becomes greater than one, and so on. Then $a_n$ does not approach infinity but it never ...


2

If you have a recursion of the form $a_{n+1} =f(a_n) $, if $L = \lim_{n \to \infty} a_n $, then we must have $L = f(L)$. In your case, $f(x) = \sqrt{2+x}$. Therefore, for any limit $L$, we must have $L = \sqrt{2+L}$. Squaring, $L^2 = L+2$, which is a standard quadratic equation. Completing the square, from $L^2-L = 2$ we get $L^2-L+1/4 = 2+1/4 =9/4$, so ...


1

$$L=\sqrt{2+L}$$, squaring both side we have, $$L^2-L-2=0$$ $$(L-2)(L+1)=0$$ which gives $L=2,-1$ since $a_n\gt 0$, for all $n\in \Bbb N$, Hence $L=2$


1

With help from above comment we have: \begin{align} 2AB &= 2(1!×2!×...×1002!)(1004!×1005!×...×2006!) \\ &= 2(2^{501}×1×2×...×501)(2^{500}×502×503×...×1003)((1!)^2(2!)^2 ...(1003!)×(1005!)^2 ...(2005!)^2 \\ &= 2^{1002}(1!)^2 ...(1003!)(1003!)(1005!)^2 (1006)^2 ...(2005)^2 \\ &= 2^{1002} ... \end{align} square.


1

There are many ways to approximate (even very accurately) functions close to a point. The simplest is Taylor expansion; in the case of the tangent, assuming $b<<a$, the expansion is $$\tan(a+b)=\tan (a)+ \left(\tan ^2(a)+1\right)b+ \left(\tan ^3(a)+\tan (a)\right)b^2+$$ $$ \left(\tan ^4(a)+\frac{4 \tan ^2(a)}{3}+\frac{1}{3}\right)b^3+ \left(\tan ...


1

$$ a_{n+1}=\frac{9n^2-21n+10}{a_n}\\a_{n+1}a_{n}=9n^2-21n+10\\(a_1+nd)(a_1+(n-1)d)=9n^2-21n+10\\a_1^2+(2n-1)a_1d+(n^2-n)d^2=9n^2-21n+10\\n^2d^2+n(-d^2+2a_1d)+(a_1^2-a_1d)=9n^2-21n+10$$ and now $$\left\{\begin{matrix} d^2=9 & \\ -d^2+2a_1d=-21& \\ a_1^2-a_1d=10& \end{matrix}\right. $$ you can find $d=\pm 3$ first ,and put d to find $a_1$


1

The odds converge, since $a_{2n+15}$ converges. Let their limit be $x$ and let the limit of $a_{2n+3}^3 - a_{2n+4}^3$ be $y$. Then, by algebra of limits, $$a_{2n+4} = (a_{2n+3}^3 - (a_{2n+3}^3 - a_{2n+4}^3))^{1/3} \rightarrow (x^3 - y)^{1/3}.$$ Thus the evens also converge. Let $b_n = a_{n^3}$, and $z$ be its limit. Consider the subsequences $b_{2n}$ and ...


1

A sequence of real numbers $(s_n)$ is: monotone if $s_n \leq s_{n+1}$ for all $n \in \mathbb{N}$ or $s_n \geq s_{n+1}$ for all $n \in \mathbb{N}$. strictly increasing if $s_n < s_{n+1}$ for all $n \in \mathbb{N}$. strictly decreasing if $s_n > s_{n+1}$ for all $n \in \mathbb{N}$. As $s_n < s_{n+1}$ for all $n \in \mathbb{N}$ implies $s_n \leq ...


1

You have that $$ y_j = \sum_{k=1}^m a_k x_j^{b_k} $$ So if we define the matrix $M$ such that $M_{j,k} = x_j^{b_k} $, $a=(a_1,\cdots,a_m)^T$ and $y=(y_1,\cdots,y_n)^T$ we have the linear system $$y = Ma$$ And from here it's "simple" to solve.


1

Note $$1<1+x+x^2+\cdots+x^{n-1}<n,x\in(0,1)$$ so $$x^n-1<\dfrac{nx^n}{1+x+\cdots+x^{n-1}}-1<nx^n-1$$ and we know $\sum_{n=1}x^n,\sum_{n=1}^{+\infty}nx^{n}$ are converge then you know?


1

To answer that question, you must first answer the question Is Infinity Even or Odd? Assume the lamp is initially on. After one minute, if you flip the switch an even number of times, the lamp will be on. However, if after that minute, you had flipped the switch an odd number of times, the lamp will be off. So if you flip the switch an infinite number of ...


1

okay I tried to solve this one, $$ [a+(n-1)d] r^{n-1} $$ is term used to show nth term of Arithmetico-geometric sequence if I take $1/2$ common from the whole series I get $$ 0 * 1 + 1 * \frac{1}{2} + 2 * \frac{1}{4} + 3 *\frac{1}{8} + ....$$ which gives me$$ a = 0, d=1, r=1/2$$ now the sum is $$\lim_{n \to \infty}S_{n} = ...


1

Suppose we seek to evaluate $$\sum_{k=0}^n {n\choose k} k^p.$$ Introduce $$k^p = \frac{p!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} \exp(kz) \; dz.$$ This yields for the sum $$\frac{p!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} \sum_{k=0}^n {n\choose k} \exp(kz) \; dz \\ = \frac{p!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} (1+\exp(z))^n \; ...



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