Unsolved analysis and number theory from other sites....

[chisigma]

Scope of this thread is to supply [when possible...] an answer to unsolved question in other sites in the field of analysis [real or complex...] and number theory, avoiding to make dispersion in different threads... The first unsolved question is 'easy enough' and was posted on www.mathhelpforum.com by the user zokomoko...

I've tried multiplying and dividing by sqrt()+1 and then used (a-b)(a+b)=a²-b² but got nowhere...

$\displaystyle \lim_{n \rightarrow \infty} \sum_{k=1}^{n} (\sqrt{1 + \frac{k}{n^{2}}}-1)\ (1)$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Scope of this thread is to supply [when possible...] an answer to unsolved question in other sites in the field of analysis [real or complex...] and number theory, avoiding to make dispersion in different threads... The first unsolved question is 'easy enough' and was posted on www.mathhelpforum.com by the user zokomoko...

I've tried multiplying and dividing by sqrt()+1 and then used (a-b)(a+b)=a²-b² but got nowhere...

$\displaystyle \lim_{n \rightarrow \infty} \sum_{k=1}^{n} (\sqrt{1 + \frac{k}{n^{2}}}-1)\ (1)$

The initial approach of zokomoko wasn't so bad... $\displaystyle \sum_{k=1}^{n} (\sqrt{1 + \frac{k}{n^{2}}}-1) = \frac{1}{n^{2}}\ \sum_{k=1}^{n} \frac{k}{\sqrt{1 + \frac{k}{n^{2}}} + 1} (1)$... and is... $\displaystyle 0 \le \frac{1}{n^{2}}\ \sum_{k=1}^{n} \frac{k}{\sqrt{1 + \frac{k}{n^{2}}} + 1} \le \frac{1}{n\ (\sqrt{1 + \frac{1}{n^{2}}} + 1)}\ (2)$

... so that the requested limit is 0...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 07 06 2013 on www.artofproblemsolving.com by the user tom-nowy and not yet solved...

For $t > 0$ please show that...

$\displaystyle \sum_{n \le t} n^{i\ t} = \mathcal {O} (\sqrt{t}\ \ln t)\ (1)$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 10 10 2013 on www.mathhelpforum.com by the user yoonsung14 and not yet solved...

Hello! I need to derive Binet's Formula! I know how to do it for one of the definitions of the Fibonacci sequence: F(1)=1, F(2)=1, F(n)= F(n-1)+ F(n-2), for all n > or = 3. However, for my assignment, I have to use an alternate definition: F(0)=1, F(1)=1, F(n)= F(n-1)+ F(n-2) for all n > or = 2.

I tried using the same method for the 1st derivation by forming a linear combination and solving for the constants, but I ended up with a formula that doesn't work for n...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 10 10 2013 on www.mathhelpforum.com by the user yoonsung14 and not yet solved...

Hello! I need to derive Binet's Formula! I know how to do it for one of the definitions of the Fibonacci sequence: F(1)=1, F(2)=1, F(n)= F(n-1)+ F(n-2), for all n > or = 3. However, for my assignment, I have to use an alternate definition: F(0)=1, F(1)=1, F(n)= F(n-1)+ F(n-2) for all n > or = 2.

I tried using the same method for the 1st derivation by forming a linear combination and solving for the constants, but I ended up with a formula that doesn't work for n...

The statement proposed by yoonsung14 leads to the difference equation...

$\displaystyle f_{n+2} - f_{n+1} - f_{n} = 0,\ f_{0}=0,\ f_{1}=1\ (1)$

The (1) is a linear constant coefficients homogenous difference equation and its solving procedure is described in...

http://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/difference-equation-tutorial-draft-part-ii-860.html

The characterisc equation is $\displaystyle r^{2} - r - 1 =0$ the solution of which is $\displaystyle r = \frac{1 \pm \sqrt{5}}{2}$ so that the general solution of (1) is...

$\displaystyle f_{n} = c_{1}\ (\frac{1 + \sqrt{5}}{2})^ {n} + c_{2}\ (\frac{1 - \sqrt{5}}{2})^ {n}\ (2)$

Taking into account the initial conditions and setting $\displaystyle \varphi = \frac{1 + \sqrt{5}}{2}$ the (2) becomes...

$\displaystyle f_{n} = \frac{\varphi^{n} - \varphi^{- n}}{\sqrt{5}}\ (3)$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 11 24 2013 on Math Help Forum - Free Math Help Forums by the user Haven and not yet solved...

... find the general solution of...$\displaystyle y^{\ ''} + x^{2}\ y = 0,\ 0 < x < \infty$

I have no idea what to do and any CAS I plug this in returns Bessel Functions. Is there a more elementary solution?...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 11 24 2013 on Math Help Forum - Free Math Help Forums by the user Haven and not yet solved...

... find the general solution of...$\displaystyle y^{\ ''} + x^{2}\ y = 0,\ 0 < x < \infty$

I have no idea what to do and any CAS I plug this in returns Bessel Functions. Is there a more elementary solution?...

The ODE...

$\displaystyle y^{\ ''} + x^{2}\ y = 0\ (1)$

... is linear with non constant coefficients and the general solution is written on the form...

$\displaystyle y(x) = c_{1}\ u(x) + c_{2}\ v(x)\ (2)$

... where u(*) and v(*) are two independent solution of (1), $c_{1}$ and $c_{2}$ two arbitrary constants. Observing (1) we realize that a solution can be either an even or an odd function , so that it is convenient to choose u(*) as even function and v(*) as odd function. Now we search u(*) and v(*) in the form...

$\displaystyle u(x) = \sum_{n=0}^{\infty} a_{n}\ x^{n}$

$\displaystyle v(x) = \sum_{n=0}^{\infty} b_{n}\ x^{n}\ (3)$

... and we do that writing sequencially the derivatives of the y for n>1...

$\displaystyle y^{(2)} = - x^{2}\ y$

$\displaystyle y^{(3)} = - x^{2}\ y^{(1)} -2\ x\ y$

$\displaystyle y^{(4)} = - x^{2}\ y^{(2)} - 4\ x\ y^{(1)} - 2\ y$

$\displaystyle y^{(5)} = - x^{2}\ y^{(3)} - 6\ x\ y^{(2)} - 6\ y^{(1)}$

$\displaystyle y^{(6)} = - x^{2}\ y^{(4)} - 8\ x\ y^{(3)} - 12\ y^{(2)}$

$\displaystyle y^{(7)} = - x^{2}\ y^{(5)} - 10\ x\ y^{(3)} - 20\ y^{(3)}$

$\displaystyle y^{(8)} = - x^{2}\ y^{(6)} - 12\ x\ y^{(4)} - 30\ y^{(4)}$

$\displaystyle y^{(9)} = - x^{2}\ y^{(7)} - 14\ x\ y^{(5)} - 42\ y^{(5)}\ (4)$

...

Observing (4) we realize that in x=0 is...

$\displaystyle y^{(n)} (0) = h_{n}\ y^{(n - 4)} (0)\ (5)$

... where $\displaystyle h_{n} = - (n-3)\ (n-2),\ n \ge 4$. The (4) combined with the Taylor theorem allows us finally to find u(x) as solution of (1) with the conditions $y(0)=1,y^{\ '} (0) = 0$ ...

$\displaystyle u(x) = 1 + \sum_{n=1}^{\infty} (-1)^{n} \frac{(4\ n - 3)\ (4\ n - 2)}{(4 n)!}\ x^{4 n} = 1 - \frac{1}{12}\ x^{4} + \frac{1}{1344}\ x^{8} - ...\ (6)$

... and v(x) as solution of (1) with the conditions $y(0)=0,y^{\ '} (0) = 1$ ...

$\displaystyle v(x) = x + \sum_{n=1}^{\infty} (-1)^{n} \frac{(4\ n - 2)\ (4\ n -1)}{(4\ n + 1)!}\ x^{4 n + 1} = x - \frac{1} {20}\ x^{5} + \frac {1}{8640}\ x^{9} - ...\ (7)$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 10 16 2013 on http://www.mymathforum.com by the user aaron-math and not yet solved...

Evaluate the residue of...

$\displaystyle f(z) = \frac{e^{- z}}{(z -1)^{4}}$

... at its isolated pole singularity...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 10 16 2013 on http://www.mymathforum.com by the user aaron-math and not yet solved...

Evaluate the residue of...

$\displaystyle f(z) = \frac{e^{- z}}{(z -1)^{4}}$

... at its isolated pole singularity...

The function has a pole of order four in z=1 and the most simple way to compute the residue into expand f(z) in Laurent series around z=1 and to find the coefficient of $z^{- 1}$. With little effort one find the Taylor expansion of $e^{- z}$ around z=1...

$\displaystyle e^{- z} = e^{-1}\ \sum_{n=0}^{\infty} (-1)^{n}\ \frac{(z-1)^{n}}{n!}\ (1)$

... so that is...

$\displaystyle f(z) = \frac{e^{- z}}{(z-1)^{4}} = e^{-1}\ \sum_{n=0}^{\infty} (-1)^{n} \frac{(z-1)^{n-4}}{n!}\ (2)$

... and the residue is $\displaystyle R \{f(z)\}_{z=1} = - \frac{e^{- 1}}{3!}$...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 11 23 2013 on www.mymathforum.com by the user Ryan_1993 and not yet solved...I tried to solve the integral of $\displaystyle x^{x}$ without limits (indefinite integration) and I came up with an answer which I believe is correct. I even put in the limits from 0 to 1 and came up with Sophomore's dream which I didn't even know about at the time. Would solving this integral be useful in mathematics?...

Kind regards$\chi$ $\sigma$

 

[chisigma]

Posted the 11 23 2013 on www.mymathforum.com by the user Ryan_1993 and not yet solved...I tried to solve the integral of $\displaystyle x^{x}$ without limits (indefinite integration) and I came up with an answer which I believe is correct. I even put in the limits from 0 to 1 and came up with Sophomore's dream which I didn't even know about at the time. Would solving this integral be useful in mathematics?...

We can use the series expansion...

$\displaystyle x^{x} = e^{x\ \ln x} = \sum_{n=0}^{\infty} \frac{(x\ \ln x)^{n}}{n!}\ (1)$

... and with the results of...

http://mathhelpboards.com/math-notes-49/integrals-natural-logarithm-5286.html

... we arrive to write...

$\displaystyle \int_{0}^{x} \xi^{\xi}\ d \xi = \sum_{n=0}^{\infty} x^{n+1} \sum_{k=0}^{n} \frac{(-1)^{k}}{(n+1)^{k+1}}\ \frac{\ln ^{n-k} x}{(n-k)!}\ (2)$

For x=1 is...

$\displaystyle \int_{0}^{x} \xi^{\xi}\ d \xi = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(n+1)^{n+1}} \sim .783431\ (3)$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 12 05b 2013 on www.artofproblemsolving.com by the user Kingofmath101 and not yet solved...... for $x \ne 2$ and $x \ne -2$ find the general solution to ...

$\displaystyle (x^{2} - 4)\ y^{\ '} + x^{2}\ y = x^{4} - 16$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 12 05b 2013 on www.artofproblemsolving.com by the user Kingofmath101 and not yet solved...... for $x \ne 2$ and $x \ne -2$ find the general solution to ...

$\displaystyle (x^{2} - 4)\ y^{\ '} + x^{2}\ y = x^{4} - 16$

Kind regards

$\chi$ $\sigma$

Deviding both term by $x^{2}-4$ You obtain...

$\displaystyle y^{\ '} + P(x)\ y = Q(x),\ P(x) = 1 + \frac{4}{x^{2} + 4},\ Q(x) = x^{2} + 4\ (1)$

... and the solution is given from the formula...

$\displaystyle y\ e^{\int P\ dx} = \int Q\ e^{\int P\ dx}\ dx + c\ (2)$

With a little of patience [or if You have no patience using 'Monster Wolfram' (Smile)...] You compute...

$\displaystyle \int P\ dx = x + \ln \frac{2 - x}{2 + x} \implies e^{\int P\ dx} = e^{x}\ \frac{2-x}{2 + x}\ (3)$

$\displaystyle \int Q\ e^{\int P\ dx}\ dx = \frac{32}{e^{2}}\ \text{Ei}\ (2 + x) - e^{x}\ (x^{2} - 6 x + 18)\ (4)$

... so that the general solution of (1) is... $\displaystyle y = 32\ \frac{2 + x}{2 - x}\ e^{- (2 + x)}\ \text{Ei}\ (2 + x) + c\ \frac{(2 + x)\ (x^{2} - 6 x + 18)}{2 - x}\ (5)$

I reported this first order ODE because its solution is in some way interesting. Considering that if You set $f(x)= x\ \text{Ei}\ (2)$ is $f(0)=0$, all the solution of (1) have a singularity in $x=2$ and a zero in $x=-2$... Kind regards $\chi$ $\sigma$

 

[chisigma]

Posted the 12 11 2013 on www.artofproblemsolving.com by the user vadim and not yet solved...

... let $a_{0}, a_{1}$ two positive real numbers. Prove that the sequence solution of $\displaystyle a_{n+2} = \frac{2}{a_{n+1} + a_{n}}$ converges...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 12 11 2013 on www.artofproblemsolving.com by the user vadim and not yet solved...

... let $a_{0}, a_{1}$ two positive real numbers. Prove that the sequence solution of $\displaystyle a_{n+2} = \frac{2}{a_{n+1} + a_{n}}$ converges...

Let's write the difference equation as...

$\displaystyle \Delta_{n} = a_{n+1} - a_{n} = \frac{2 - a^{2}_{n} - a_{n}\ a_{n-1}}{a_{n}+ a_{n-1}} = f(a_{n},\ a_{n-1})\ (1)$

The (1) has two constant solutions $ a_{n}= 1$ and $a_{n}= -1$ and taking into account the condition of the problem we consider only the first. Is $\displaystyle f(x,y)= \frac{2 - x^{2} - x\ y}{x+y}$ and $\displaystyle f_{x} (x,y) = - \frac{2 + (x + y)^{2}}{(x + y)^{2}}$, so that is $\displaystyle f(1,1)=0$ and $f_{x}(1,1) = - \frac{3}{2}$ and the point [1,1] is an attractive fixed point. Finally for x > 0 and y> 0 is $\displaystyle f_{x} (x,y) < -1$ so that any solution of (1) converges to 1 oscillating... View attachment 1769

Merry Christmas from Serbia

$\chi$ $\sigma$

 

[chisigma]

Posted the 01 14 2013 on www.mathhelpforum.com by the user sbhatnagar and not yet solved...

... show that $\displaystyle \sum_{n = 1}^{\infty} \frac{\zeta (2\ n)}{n\ 2^{\ 2\ n}} = \ln \frac{\pi}{2}$... https://www.physicsforums.com/attachments/1769

Merry Christmas from Serbia

$\chi$ $\sigma$

 

[chisigma]

Posted the 01 14 2013 on www.mathhelpforum.com by the user sbhatnagar and not yet solved...

... show that $\displaystyle \sum_{n = 1}^{\infty} \frac{\zeta (2\ n)}{n\ 2^{\ 2\ n}} = \ln \frac{\pi}{2}$...

Start from the [not well know...] series expansion... $\displaystyle \pi\ x\ \cot (\pi\ x) = 1 - 2\ \sum_{n = 1}^{\infty} \zeta (2\ n)\ x^{\ 2\ n}\ (1)$

... first we derive...

$\displaystyle \sum_{n=1}^{\infty} \zeta (2\ n)\ x^{2\ n - 1} = \frac{1 - \pi\ x\ \cot (\pi\ x)}{2\ x}\ (2)$... and then... $\displaystyle \sum_{n=1}^{\infty} \frac{\zeta (2\ n)}{n} (\frac{1}{2})^{\ 2\ n} = \int_{0}^{\frac{1}{2}} \frac{1 - \pi\ \xi \cot (\pi\ \xi)}{\xi}\ d \xi = \ln \frac{\pi}{2}\ (3)$
https://www.physicsforums.com/attachments/1769

Merry Christmas from Serbia

$\chi$ $\sigma$

 

[chisigma]

Posted the 12 13 2013 on www.artofproblemsolving.com by the user Cerberos and not yet solved...

... find all functions $\displaystyle f (x), 0 \le x \le 1$ for which is ...

$\displaystyle f^{\ '} (x) + f(x) + \int_{0}^{1} f(t)\ dt = 0\ (1)$

https://www.physicsforums.com/attachments/1769

Merry Christmas from Serbia

$\chi$ $\sigma$

 

[MarkFL]

I hope you don't mind me chiming in on this one...:D

I would differentiate (1) with respect to $x$ to obtain:

\(\displaystyle f''(x)+f'(x)=0\)

whose solution is:

\(\displaystyle f(x)=c_1+c_2e^{-x}\)

Substituting this into (1), we find:

\(\displaystyle -c_2e^{-x}+c_1+c_2e^{-x}+\left[c_1t-c_2e^{-t} \right]_0^1=0\)

\(\displaystyle c_1+\left(c_1-\frac{c_2}{e}+c_2 \right)=0\)

\(\displaystyle 2c_1=\frac{c_2}{e}-c_2=\frac{c_2(1-e)}{e}\)

\(\displaystyle c_1=\frac{c_2(1-e)}{2e}\)

Hence:

\(\displaystyle f(x)=\frac{c_2(1-e)}{2e}+c_2e^{-x}=\frac{c_2}{2e}\left(1-e+2e^{1-x} \right)\)

Define \(\displaystyle C\equiv \frac{c_2}{2e}\) and we have:

\(\displaystyle f(x)=C\left(1-e+2e^{1-x} \right)\)

 

[chisigma]

Posted the 12 13 2013 on www.artofproblemsolving.com by the user Cerberos and not yet solved...

... find all functions $\displaystyle f (x), 0 \le x \le 1$ for which is ...

$\displaystyle f^{\ '} (x) + f(x) + \int_{0}^{1} f(t)\ dt = 0\ (1)$

A simple solution that doesn't increase complexity is to consider like a constant the term...

$\displaystyle a= - \int_{0}^{1} f(t)\ d t\ (1)$

... so that the ODE becomes...

$\displaystyle f^{\ '} + f = a\ (2)$

... the solution of which is...

$\displaystyle f(x) = a + c\ e^{- x}\ (3)$

Now we have to impose that...

$\displaystyle \int_{0}^{1} (a + c\ e^{- x})\ dx = a\ (4)$

... and solving the (4) in c we obtain...

$\displaystyle f(x) = a\ (1 - \frac{2\ e^{- x}}{1 - e^{-1}})\ (5)$ https://www.physicsforums.com/attachments/1769

Merry Christmas from Serbia

$\chi$ $\sigma$

 

[chisigma]

Posted the 11 09 2013 on www.mathhelpforum.com by the user Suvadip and not yet solved...

... how many solutions does the differential equation...$\displaystyle \frac{d y}{d x} = 60\ y^{\frac{2}{5}}\ ;\ x > 0;\ y(0)=0$... have?...

https://www.physicsforums.com/attachments/1769

Merry Christmas from Serbia

$\chi$ $\sigma$

 

[chisigma]

Posted the 11 09 2013 on www.mathhelpforum.com by the user Suvadip and not yet solved...

... how many solutions does the differential equation...$\displaystyle \frac{d y}{d x} = 60\ y^{\frac{2}{5}}\ ;\ x > 0;\ y(0)=0$... have?...

Let's consider the first order ODE... $\displaystyle y^{\ '} = 60\ y^{\frac{2}{5}}\ ;\ y(0)=0\ (1)$

The 'standard' solution, obtained separating the variables, is ...

$\displaystyle y (x)=\begin{cases} 216\ 6^{\frac{1}{3}}\ x^{\frac{5}{3}} &\text{if}\ x \ge 0\\ 0 &\text{if}\ x<0 \end{cases}$ (2)

With a more carefully analysis however we discover that we are in the same situation described in...

http://mathhelpboards.com/differential-equations-17/interesting-ordinary-differential-equation-3684.html#post16353

... where $y^{\ '} (x)$ is function of the y alone, so that if y(x) is solution [for example the solution given in (2)...], then y(x - a) being a any positive real number is also solution. The conclusion is : the number of solutions of (1) is infinite...
https://www.physicsforums.com/attachments/1769

Merry Christmas from Serbia

$\chi$ $\sigma$

 

[chisigma]

Posted the 12 19 2013 on www.matematicamente.it by the user Shika93 [original in Italian...] and not yet solved...

... I have to compute...

$\displaystyle \int_{\Gamma} \frac{d z}{1 - \cos z}\ (1)$

... where $\Gamma$ is the unit circle...https://www.physicsforums.com/attachments/1769

Merry Christmas from Serbia

$\chi$ $\sigma$

 

[chisigma]

Posted the 12 19 2013 on www.matematicamente.it by the user Shika93 [original in Italian...] and not yet solved...

... I have to compute...

$\displaystyle \int_{\Gamma} \frac{d z}{1 - \cos z}\ (1)$

... where $\Gamma$ is the unit circle...

Inside the unit circle f(z) has the only singularity in z=0 so that is...

$\displaystyle \int_{\Gamma} f(z)\ dz = 2\ \pi\ i\ r\ (1)$

... where r is the coefficient of degree -1 of the Laurent expansion of f(z) around z=0 and it can easily found as follows...

$\displaystyle \frac{1}{1 - \cos z} = \frac{2}{z^{2} - \frac{z^{4}}{12} + ...} = \frac{2}{z^{2}} + \frac{1}{6} + ...\ (2) $

... so that is...

$\displaystyle \int_{\Gamma} \frac{d z}{1 - \cos z} = 0\ (3)$https://www.physicsforums.com/attachments/1769

Merry Christmas from Serbia

$\chi$ $\sigma$

 

[chisigma]

Posted the 12 22 2013 on www.mathhelpforum.com by the user benbas and not yet solved...

... You have the integral...

$\displaystyle I = \int_{0}^{2 \pi} \sqrt{1 + \sin^{2} x}\ d x $

https://www.physicsforums.com/attachments/1769

Merry Christmas from Serbia

$\chi$ $\sigma$

 

[chisigma]

Posted the 12 22 2013 on www.mathhelpforum.com by the user benbas and not yet solved...

... You have the integral...

$\displaystyle I = \int_{0}^{2 \pi} \sqrt{1 + \sin^{2} x}\ d x $

It is not difficult to see that the integral represents the length of curve $\displaystyle y = \cos x$ along its period and for symmetry is...

$\displaystyle I = 4\ \int_{0}^{\frac{\pi}{2}} \sqrt{1 + \sin^{2} x}\ dx\ (1)$

The only answers the post received has been something the 'classical': '... the integral is Elliptic of the second type, so that it is not solvable in terms of elementary functions...'. Since MHB is a forum 'problem solving' I don't think that such answer is appropriate and some effort has to be made in order to solve the integral. Fortunately for this type of elliptic integrals a series expansion is known...

$\displaystyle E(k) = \int_{0}^{\frac{\pi}{2}} \sqrt{1 - k^{2}\ \sin^{2} x}\ dx = \frac{\pi}{2}\ \sum_{n = 0}^{\infty} \{ [\frac{(2 n -1)!}{(2 n)!}]^2\ \frac{k^{2 n}}{1 - 2\ n}\}\ (2)$

... so that is...

$\displaystyle \int_{0}^{\frac{\pi}{2}} \sqrt{1 + \sin^{2} x}\ dx = E(i) = \frac{\pi}{2}\ \sum_{n = 0}^{\infty} \{ [\frac{(2 n -1)!}{(2 n)!}]^2\ \frac{(-1)^{n}}{1 - 2\ n}\}\ (3)$

The (3) is a good result because a simple computer program allows to compute the integral with good accuracy. But if computer program is not allowable then, adding fortune to fortune, the (3) admits a closed form...

$\displaystyle E(i) = \frac{2\ \Gamma^{4} (\frac{3}{4}) + \pi^{2}}{2\ \sqrt{2\ \pi}\ \Gamma^{2} (\frac{3}{4})}\ (4)$ https://www.physicsforums.com/attachments/1769

Merry Christmas from Serbia

$\chi$ $\sigma$

 

[alyafey22]

If you allow me I want to show how to solve the integral

Assume the following

\(\displaystyle E(k) = \int^{\frac{\pi}{2}}_0 \sqrt{1-k^2 \sin^2(\theta)}\, d\theta \)

Use the substitution \(\displaystyle x=\sin (\theta) \)

\begin{align}E(k) = \int^{1}_0 \frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}} \, dx &= \int^{1}_0 \frac{1-k^2x^2}{\sqrt{1-x^2}\sqrt{1-k^2 x^2}} \, dx\\ &= \int^{1}_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2 x^2}} \, dx-k^2\int^{1}_0 \frac{x^2}{\sqrt{1-x^2}\sqrt{1-k^2 x^2}} \, dx \end{align}

Hence we have

\(\displaystyle E(i) = \int^{1}_0 \frac{dx}{\sqrt{1-x^4}} \, dx+\int^{1}_0 \frac{x^2}{\sqrt{1-x^4}} \, dx \)

The rest is a beta integral easy to solve using

\(\displaystyle \int^1_0 x^{a-1} (1-x)^{b-1} \, dx=B(a,b)\)

 

[chisigma]

Posted the 11 29 2013 on http://www.mymathforum.com by the user rody and not yet solved...

... I'm looking for a bibliography on the resolution of differential equation with constant delay of the type...

$\displaystyle y^{\ '} (t) + a\ y(t) + b\ y(t - t_{0}) = c$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 11 29 2013 on http://www.mymathforum.com by the user rody and not yet solved...

... I'm looking for a bibliography on the resolution of differential equation with constant delay of the type...

$\displaystyle y^{\ '} (t) + a\ y(t) + b\ y(t - t_{0}) = c$

May be it is wise to proceed 'step by step' starting with the particular case $a = 0$ so that the DE becomes...

$\displaystyle y^{\ '} (t) + b\ y(t - t_{0}) = c\ (1)$

Writing (1) in terms of Laplace Transform we obtain...

$\displaystyle s\ Y(s) - y(0) + b\ e^{- s\ t_{0}} = \frac{c}{s}\ (2)$

... and from (2)...

$\displaystyle Y(s) = \frac{ y(0)\ s + c}{s^{2}\ (1 + \frac{b}{s}\ e^{- s\ t_{0}})}= \frac{y(0)\ s + c}{s^{2}}\ \sum_{n=0}^{\infty} (-1)^{n}\ (\frac{b}{s})^{n}\ e^{- s\ n\ t_{0}}\ (3)$

... so that is...

$\displaystyle y(t) = y(0)\ \sum_{n= 0}^{\infty} \{(-1)^{n}\ \frac{b^{n}\ (t- n\ t_{0})^{n}}{n!}\ \mathcal {U}\ (t - n\ t_{0})\} + c\ \sum_{n= 0}^{\infty} \{(-1)^{n}\ \frac{b^{n}}{(n+1)!}\ (t- n\ t_{0})^{n+1}\ \mathcal {U}\ (t - n\ t_{0})\}\ (4)$

... where $\displaystyle \mathcal{U}\ (*)$ is the so called 'Heaviside Step Function'. A DE that is almost unfeasible with traditional approach becomes almost trivial with the Laplace Transform!...Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 12 31 2013 on www.matematicamente.it by the user N56VZ [original in Italian...] and not yet solved...

... I searched on Internet how to solve this ODE but I didn't find anything...

$\displaystyle u^{\ '} = u^{2}\ \sin u,\ u(0)= - \frac{3}{2}\ \pi$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 12 31 2013 on www.matematicamente.it by the user N56VZ [original in Italian...] and not yet solved...

... I searched on Internet how to solve this ODE but I didn't find anything...

$\displaystyle u^{\ '} = u^{2}\ \sin u,\ u(0)= - \frac{3}{2}\ \pi$

In the site this problem has been proposed this ODE has been classified as 'Riccati equation' and the conclusion has been 'unsolvable'. With a little more precise evaluation it is easy to see that the ODE belongs to the most 'comfortable' first order ODE : the separable variables type. The solving procedure is well known...

$\displaystyle \int \frac{d u}{u^{2}\ \sin u} = t + c\ (1)$

Unfortunately the integral in (1) 'cannot be expressed in term of elementary functions' so that an alternative solution has to be tried. We can use the Laurent expansion...

$\displaystyle \frac{1}{u^{2}\ \sin u} = \frac{1}{u^{3}} + \frac{1}{6\ u} + \frac{7}{360}\ u + \frac {31}{15120}\ u^{3} + \mathcal {O}\ (u^{5})\ (2)$

... and integrate (2) 'term by term' but now we meet a new criticity: the (2) converges for $\displaystyle 0 < |u| < \pi$, so that it's impossible to compute the initial condition $t(0) = u (- \frac{3}{2}\ \pi)$. Fortunately we can overcome all that searching as solution a function t (u) that is analytical in $\displaystyle u = - \frac{3}{2}\ \pi$. Setting $\displaystyle v= u + \frac{3}{2}\ \pi$ the ODE becomes...$\displaystyle \frac{d t}{d v} = - \frac{1} {(v - \frac{3}{2}\ \pi)^{2}\ \cos v},\ t(0)=0\ (3)$

... where...

$\displaystyle t(v) = \sum_{n=1}^{\infty} a_{n}\ v^{n}\ (4)$

The coefficients of (4) can be found as follows...

$\displaystyle t^{\ '} = - \frac{1} {(v - \frac{3}{2}\ \pi)^{2}\ \cos v} \implies t^{\ '} (0) = - (\frac{2}{3\ \pi})^{2} = a_{1}$

$\displaystyle t^{(2)} = - \frac{(v - \frac{3}{2}\ \pi)\ \tan v + 2}{(v - \frac{3}{2}\ \pi)^{3}} \implies t^{(2)} (0) = 2\ (\frac{2}{3\ \pi})^{3} \implies a_{2} = (\frac{2}{3\ \pi})^{3}$

$\displaystyle t^{(3)} = \frac{\{(\frac{3}{2}\ \pi)^{2} - 3\ \pi\ v + v^{2} + 6\}\ \cos 2 v - 3\ \{(\frac{3}{2}\ \pi)^{2} - 3\ \pi\ v + v^{2} -2\}}{ \cos^{3} v\ (\frac{3}{2}\ \pi - v)^{4}} \implies t^{(3)} (0) = \frac{6 - (\frac{3}{2}\ \pi)^{2}}{(\frac{3}{2}\ \pi)^{4}} \implies a_{3}= \frac{1 - \frac{1}{6}\ (\frac{3}{2}\ \pi)^{2}}{(\frac{3}{2}\ \pi)^{4}}$

Of course, if necessary [... with a little of patience (Wasntme)...] other terms can be computed...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 01 03 2014 on www.artofproblemsolving.com by the user Pirkuliyev Rovsen and not yet solved... ... calculate $f^{(n)} (0)$ of $f(x)= (\sin^{-1} x)^{2}$... Kind regards $\chi$ $\sigma$

 

[chisigma]

Posted the 01 10 2014 on www.artofproblemsolving.com by the user carlover1 and not yet solved...

... find the sum...

$\displaystyle S = \sum_{k=1}^{\infty} \frac{1}{k\ (n+1)^{k}}$

... where n is a natural number greater than 1...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 01 10 2014 on www.artofproblemsolving.com by the user carlover1 and not yet solved...

... find the sum...

$\displaystyle S = \sum_{k=1}^{\infty} \frac{1}{k\ (n+1)^{k}}$

... where n is a natural number greater than 1...

For |x|< 1 is...

$\displaystyle \sum_{k=1}^{\infty} \frac{x^{k}}{k} = \ln \frac{1}{1 - x}\ (1)$

... so that setting $\displaystyle x = \frac{1}{n+1}$ we have... $\displaystyle S(n) = \ln \frac{1}{1 - \frac{1}{n+1}} = \ln (1 + \frac{1}{n})\ (2)$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 01 14 2014 on www.mathhelpforum.com by the user Latsabb and not yet solved... ... we have just started working on the Taylor series, and one of our problems is to find the MacLaurin series for $\displaystyle \frac{\cos (x^{3})}{x}$...Kind regards $\chi$ $\sigma$