# Tag Info

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The way to do this without generating functions is to start with the ansatz that $$a_n = x^n$$ satisfies the recursion but possibly not the starting points at $n=0$ and $1$. If we have that solution then any $a_nkx^n$ satisfies the recursion as well. And if we have two such solutions $x^n$ and $y^n$ then any linear combination $a_n=kx^n+my^n$ will ...

8

$$0,1,\frac 1 2, \frac 3 4, \frac 5 8, \frac{11}{16}, ...$$ Now, the trick is that if you take the common differences: $$1, -\frac 1 2, \frac 1 4, -\frac 1 8, \frac 1 {16}$$ It's a geometric series of first term $1$ and ratio $-\frac 1 2$. Thus, this sequence is just a sequence of partial sums of this geometric sequence, so we just use the partial sum ...

5

This is called a linear homogeneous recurrence relation. If we look at the recursive case, we find that the coefficient of $a_{n-1}$ is $10$ and the coefficient of $a_{n-2}$ is $-21$. This means the "characteristic polynomial," which is basically the polynomial which tells us what the bases of the explicit formula will be, is like this: $$x^2-10x+21$$ Notice ...

5

Interpreted as linear recurrence with constant coefficients $$x_n = \frac{1}{2} x_{n-1} + \frac{1}{2} x_{n-2}$$ with characteristic polynomial $$p(t) = t^2 - \frac{1}{2} t - \frac{1}{2}$$ where we deduce the roots from $$0 = \left( t - \frac{1}{4} \right)^2 - \frac{1}{2} - \frac{1}{16} \iff \\ t = \frac{1 \pm 3}{4}$$ so we have the solutions $$x_n = ... 3 This is a homogeneous linear recurrence relation with constant coefficients. From$$ a_n = 10 a_{n-1} -21 a_{n-2} $$you can infer the order d=2 and the characteristic polynomial:$$ p(t) = t^2 - 10 t + 21 $$Calculating the roots:$$ 0 = p(t) = (t - 5)^2 - 25 + 21 \iff t = 5 \pm 2 $$this gives the general solution$$ a_n = k_1 3^n + k_2 7^n $$The two ... 3 Where did you find that equation? It's quite different from what I got, which I shall explain now. First, a common power series is$$ \frac{1}{1-x} = \sum_{i\geq0} x^{i}. $$Using the substitution x=-t^{n},$$ \frac{1}{1+t^{n}} = \sum_{i\geq0} (-t^{n})^{i} = \sum_{i\geq0} (-1)^{i}t^{in}. $$Then,$$ \int_{0}^{x} \frac{1}{1+t^{n}} dt = \int_{0}^{x} \...

3

If a sequence is convergent then we know that $\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} a_n = L$. take the limit of both sides and you will get: $$\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \frac{2a_n}{7+a_n} = \frac{2 \cdot \lim_{n \to \infty} a_{n}}{7 + \lim_{n \to \infty} a_n} \implies L = \frac{2L}{7+L}$$ This should be an easy ...

3

The "standard" example is the sequence $\{g_n\}$ defined by $$g_1=1_{[0,1]},\; g_2=1_{[0,\frac{1}{2}]},\;g_3=1_{[\frac{1}{2},1]},\; g_4=1_{[0,\frac{1}{4}]},\;g_5=1_{[\frac{1}{4},\frac{1}{2}]},\dots$$ where $1_A$ is the indicator function of the set $A$, i.e. $1_A(x)=1$ if $x\in A$ and $1_A(x)=0$ otherwise. Note that $$\lim_{n\to\infty}\int_0^1g_n(x)\;dx=... 2 Since you say you can evaluate the remaining infinite sum in the d=2 case, I gather that reducing the multiple sum to a single infinite sum would already be sufficient progress. Then you want to count the number \pi_d(n) of ways in which a positive integer n can be written as an ordered product of positive integers. Find the prime factorisation, n=\... 2 (Just for the cauchy part) For some integers m,n where m>n>0, |x_n-x_{n+1}|=|x_n-\frac12(x_n-x_{n-1})|=\frac12|x_{n-1}-x_n|=...=\frac1{2^{n-1}}|x_1-x_2|=\frac1{2^{n-1}}\\ |x_n-x_m|=|(x_n-x_{n+1})+(x_{n+1}-x_{n+2})+...+(x_{m-1}-x_m)|\\ \leq \frac1{2^{n-1}}+\frac1{2^n}+...\\ =\frac1{2^n}\to 0\text{ as }n\to\infty 2 If we start from$$ \cos(x) = \prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2 \pi^2}\right) \tag{1}$$we have:$$ \log\cos(x) = -\sum_{n\geq 0}\sum_{m\geq 1}\frac{4^m x^{2m}}{m(2n+1)^{2m} \pi^{2m}}=-\sum_{m\geq 1}\frac{(4^m-1)\zeta(2m)\,x^{2m}}{m\pi^{2m}}\tag{2} $$hence:$$ \sum_{l\geq 1}\log\cos\left(\frac{x}{3^l}\right)=-\sum_{m\geq 1}\frac{(4^m-1)\zeta(2m)}{...

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