How to solve this nonlinear difference equation $$a_{n+1} = 2a_n + \frac{1}{a_n},\quad a_1 = 1.$$
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For $n\in\mathbb N$, $$a_{n+1}=2a_n+\frac{1}{a_n},\quad a_1=1. $$ Can any one give an explicit formula for all $a_n$? If such an explicit general formula doesn't exist, please explain it. I've tried to figure out the $n$-iterated function $f^{(n)}$ where $f(x)=2x+1/x$ or even $f(\tan(t))$. But in either cases, I failed.Since the recurrence isn't linear nor homogeneous,the generating function method doesn't apply here.
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Let $a\ge1$ and define the sequence $(x_n)$ recursively by: $$x_1 = \sqrt{a}$$ $$x_{n+1}= \sqrt{a+x_n}$$  Here's what I did:  Plugging in some values makes it seem as if the sequence is increasing. I first solve the equation $t=\sqrt{a+t}$, equivalently $t^2 -t -a =0$, to get an idea what's going on. The roots are $$t_{1,2}= \frac{1 \pm \sqrt{1+4a} }{2}.$$ Therefore, between the roots, that is for $$t\in \left( \frac{1 - \sqrt{1+4a} }{2}, \frac{1 + \sqrt{1+4a} }{2}\right)$$  we have $t^2-t-a \le 0 \iff t \le \sqrt{a+t}$. Here's my attempt at showing inductively that $$\forall n\in \mathbb{N}, \ x_n \le \frac{1 + \sqrt{1+4a} }{2}$$  For $n=1$, the statement is true (simple arithmetic). Suppose it's true for $n$. Consider  $$x_{n+1}= \sqrt{a+x_n}\le \sqrt{a+\frac{1 + \sqrt{1+4a} }{2}}=\frac{1 + \sqrt{1+4a} }{2},$$  because $\frac{1 + \sqrt{1+4a} }{2}$ satisfies $x=\sqrt{x+a}$.  So this establishes the sequence $(x_n)$ is bounded. Now, by induction again, I try to show it's monotonically increasing. The base case $x_1\le x_2$ is obvious. Assume (strong induction) $x_{n-1}\le x_n$ for all $n&gt;1$. But since we have $$x_1\le x_n \le \frac{1 + \sqrt{1+4a} }{2}, $$   then $x_n$ lies in the appropriate interval, so $x_n \le \sqrt{a+x_n}=x_{n+1}.$  Is this proof correct? Can the condition on $a$ be relaxed to $a&gt;0$? Maybe even small negative $a$? And could someone provide me with another proof of convergence of $x_n$? I absolutely hate this one!
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Let $a_n$ be the sequence in $\mathbb{R}$ given by:  $$ a_1 = \sqrt{2}, \; \; a_{n+1} =\sqrt{2+ \sqrt{a_n} }. $$  Show that this sequence converges and find the limit.  I have already proven that this sequence is increasing and bounded by $2$, so the limit exists. Now, when I'm trying to find the limit, say  $L$, I get that $L$ satisfies the equation: $$L^2 -2 - \sqrt{L}=0, $$ but I don't actually know how to solve this equation, so I would like to know if there is an  easier way to find the limit, and in case I must solve the last equation, how can I solve it?  Thanks in advance!
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Let $x_n$ be the unique non-negative real solution of a equation $x^n+x^{n-1}+x-1=0$. Prove the sequence {$x_n$} is increasing and converges to 1.  I found out that $ 0 &lt; x_n &lt; 1$ but just that..
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The only solution that I have been able to come up with is $d=1$, but because $d&gt;2$, I can't seem to figure out how to find a solution.
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Determine the general form of $u_0, u_1 ~\text{and} ~ u_2$ if a system of difference equations of the form  $$x_{n+1} = Ax_n  + Bu_n,$$  where:  $$A = \begin{pmatrix}   3 &amp; 2 &amp; 2  \\   -1 &amp; 0 &amp; -1 \\    0 &amp; 0 &amp; 1   \end{pmatrix}$$  and:  $$B = \begin{pmatrix}   0 &amp; 0  \\   0 &amp; 1 \\   1 &amp; 0   \end{pmatrix}$$  is to be controlled for $x_0 = 0 ~ to ~ x_3 = [2, 1, 2]^T$ .  Show this target could have been achieved at  $x_2$   Solution So far I have caculated the controlability matrix to be $$ C =\begin{pmatrix} 0&amp;0&amp;2&amp;2&amp;6&amp;6\\ 0&amp;1&amp;-1&amp;0&amp;-3&amp;-2\\ 1&amp;0&amp;1&amp;0&amp;1&amp;0 \end{pmatrix}. $$ Thus the system is controlable Now putting Cv=x3 i have the 3 equations $$ 2c+2d+6e+6f=2\\ b-c-3e-2f=1\\ c+e+f=2\\ $$ which i have then put into augmented matrix row echleon form which i have found to be $$ a-d-2e-3f=1\\ b+d+f=2\\ c+d+3e+3e=1\\ $$  How do I now solve with so many unknowns? also can you please check my working so far is correct. many thanks
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As suggested on Mathoverflow () I am transfering this question to math-stackexchange:  I am looking for integer solutions of the following algebraic equations:  $ 2a^{2} + 2b^{2} - c^{2} - d^{2} = 0 $ Preferentially the solutions should obey $a+b+c+d=0$.   By inspection I found the following solutions: $(a,b,c,d)=(1,0,1,1)$ , $(a,b,c,d)=(0,1,1,1)$, $(a,b,c,d)=(1,1,-2,0)$  and $(a,b,c,d)=(1,1,0,-2)$.   Additional solutions can be generated by swapping the sign of $a,b,c,d$ or  by scaling the solutions that I have given by an integer.   As a theoretical physicist I am rarely working with these diophantine equations and hence I am wondering what other solutions of the equation exist that I have not taken into account.   I am looking forward to your responses.
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Suppose that $a_{1}&lt;a_{2}&lt;…&lt;a_{n}$. Prove that the equation  $ \frac{1}{x-a_{1}}+\frac{1}{x-a_{2}} +…+\frac{1}{x-a_{n}}=c$ has exactly $n-1$ roots if $c=0$ and $n$ roots if $c\neq 0$.    I am absolutely not sure but I think I need to use Rolle's theorem on this. Multiply both sides by $x-a_{1}$ so we get:  $1+\frac{x-a_{1}}{x-a_{2}}+…+\frac{x-a_{1}}{x-a_{n}}-c(x-a_{1})=0$. Now it is defined for $a_{1}$ but not for $a_{2},…,a_{n}.$. If we consider the interval $(a_{1}-\varepsilon,a_{n}+\varepsilon), \varepsilon&gt;0$ we get..nowhere. I want to create an interval with endpoints that are solutions to the equation. Yet I don't know how to prove that there are exactly $n$ roots(Fundamental theorem of algebra can't be used).
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I'm given two sequences:  $$a_{n+1}=\frac{1+a_n+a_nb_n}{b_n},b_{n+1}=\frac{1+b_n+a_nb_n}{a_n}$$  as well as an initial condition $a_1=1$, $b_1=2$, and am told to find: $\displaystyle \lim_{n\to\infty}{a_n}$.  Given that I'm not even sure how to approach this problem, I tried anyway. I substituted $b_{n-1}$ for $b_n$ to begin the search for a pattern. This eventually reduced to:  $$a_{n+1}=\frac{a_{n-1}(a_n+1)+a_n(1+b_{n-1}+a_{n-1}b_{n-1})}{1+b_{n-1}+a_{n-1}b_{n-1}}$$  Seeing no pattern, I did the same once more:  $$a_{n+1}=\frac{a_{n-2}a_{n-1}(a_n+1)+a_n\left(a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})\right)}{a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})}$$  While this equation is atrocious, it actually reveals somewhat of a pattern. I can sort of see one emerging - though I'm unsure how I would actually express that. My goal here is generally to find a closed form for the $a_n$ equation, then take the limit of it.  How should I approach this problem? I'm totally lost as is. Any pointers would be very much appreciated!  Edit:  While there is a way to prove that $\displaystyle\lim_{n\to\infty}{a_n}=5$ using $\displaystyle f(x)=\frac{1}{x-1}$, I'm still looking for a way to find the absolute form of the limit, $\displaystyle\frac{1+2a+ab}{b-a}$.
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Help me understand problems of this type a bit more intuitively.   The solution $C(3+11−1,11)$ seems simple enough, but I got stuck thinking about how many integers you are choosing from within $x_1$, $x_2$, and $x_3$ and I was trying to get an $n$ - as in $C(n+r-1,r)$ - this way. Also, I though the $r$ was 3 since you have $x_1$, $x_2$, $x_3$ choices. Basically, I went wrong at every step. Help!
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(Fitzpatrick Advanced Calculus 2e, Sec. 2.4 #12)  For $c \gt 0$, consider the quadratic equation $x^2 - x - c = 0, x &gt; 0$.  Define the sequence $\{x_n\}$ recursively by fixing $|x_1| \lt c$ and then, if $n$ is an index for which $x_n$ has been defined, defining  $$x_{n+1} = \sqrt{c+x_n}$$  Prove that the sequence $\{x_n\}$ converges monotonically to the solution of the above equation.  Note: The answers below might assume $x_1 \gt 0$, but they still work, as we have $x_3 \gt 0$.    This is being repurposed in an effort to cut down on duplicates, see here: .  and here: .
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Inspired by        Can we solve the equation $$1=\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\cdots +\frac{1}{a_n}$$ where $a_1,a_2,a_3,\cdots ,a_n$ are distinct prime powers (primes are allowed) ?   I tried a modified greedy-algorithm with different starting vectors and brute force with $6$ entris with limit $200$, but did not find a representation.
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$x_{n+2} = x_{n+1} + 20x_n + n^2 + 5^n \text{ with } x_0 = 0 \text{ and } x_1 = 0$  How would you solve this recursive relation? I have the homogenous solution, but am having issues with the in-homogenous one.
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$$(a_1 + a_2 + · · · + a_n)( 1/a_1 +1/a_2+ · · · +1/a_n) ≥ n^2$$  My TA mentioned he solved this using Cauchy Schwartz inequality, but I don't see how. All I've managed to do is make a summation of $1$, $n$ times, which ends up being $n ≥ n^2$.
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Solve the equation with Matlab  Given: $a_{n+1} = 4a_{n} + n +1$  Start condition: $a_{i}=0$  Questions:  a) Find x,y for: $b_{n} = a_{n}+xn+y$  such that: $b_{n+1}=4b_{n}$  b) Find an explicit formula for $b_{n}$  c) Write recursive function with input n   I'll be very grateful for your help!
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Find the general solution of $xy''-(1+x)y'+y=x^2$ knowing that the homogeneous equation has the following solution: $e^{ax}$, where $a$ is a parameter you have to find.  I have found that $a=1$ or $a=1/x$.
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How can I prove that this iterative sequence converges to $2$? Can I use the convergence definition?  $$a_{n+1} = \frac{a_n}{2} + \frac{2}{a_n},\qquad  a_0 = 4  $$  Thanks for the help.