differential equations – How can I find a numerical solution to the following problem:

I have this evolution equation:

$frac{d C(mu)}{dlnmu}=(y(mu))^2C(mu)$

and I have to find a numerical solution for $C$. The complication arises because I do not know the exact for of the function $y(mu)$ but I only know its evolution equation with $mu$ which looks something like:

$frac{d y(mu)}{dlnmu}=(y(mu))^3+y(mu)(f_1(mu)+f_2(mu))$

and it is complicated to analytically integrate to have a solution for $y(mu)$. But I can numerically solve it because I know the evolution equations for $f_1$ and $f_2$ and I am able to derive a closed form for these functions.
I was wondering if there is a way I can tell Mathematica to numerically solve the first equation for C, while using the numerical result from the numerical solution for the function $y$? The whole problem is that I dont know how to get an analytic form for $y(mu)$ or at least haven’t been able till now.

Thanks a lot for any help or insight!

trigonometry – The sum of the solutions of $sinleft(2xright)=frac{sqrt{3}}{2}$ over the interval$ [–π, d]$ is$–π.$ find the value of $d$

The sum of the solutions of $sinleft(2xright)=frac{sqrt{3}}{2}$ over the interval$ (–π, d)$ is$–π.$ find the value of $d$

I think the first thing I must know is the number of solutions. I know that $x=:frac{pi }{6},frac{pi }{3}$, but how can I deduce how many more? How can I utilise $-pi$ is the domain? What is the general approach to these kind of questions?

mathematics – I just wanted to prove Fermat’s last theorem true for the first 1000 numbers using a python script, but there is an error I’m not able to find

Fermat’s last theorem states that there are no natural numbers (1, 2, 3,…) x, y, and z such that x^n + y^n = z^n, in which n is a natural number greater than 2.

Here is the code, you can ignore the first 2 functions.

def diff(n, mid) :
    if (n > (mid * mid * mid)) :
        return (n - (mid * mid * mid))
    else :
        return ((mid * mid * mid) - n)

def cubicRoot(n):
    start = 0
    end = n

    e = 0.0000001
    while (True):

        mid = (start + end) / 2
        error = diff(n, mid)

        if (error <= e):
            return mid

        if ((mid * mid * mid) > n):
            end = mid

            start = mid

def proof(n):
  for a in range(1,n):
      for b in range(a, n):
          c = cubicRoot(a * a * a + b * b * b)
          if c.is_integer() and c <= n:

Although for some reason my program stops with the output ‘1,728’ in the console and doesn’t print anything further. Any help on improving the code and fixing this solution will be appreciated.

Thank you.

sql server – DBCC CheckDB : Operating System Error 27(The drive cannot find the sector requested)

So my regular DBCC check had this lovely message waiting for me in the error logs this morning. I’ve been trying to find out more about this issue and what can be done about it. Manually running DBCC CHECKDB will result in DBCC aborting with Error 6 (connection closed by remote host) and get me this error in the SQL log file. Running DBCC CHECKDB(dbname) WITH PHYSICAL ONLY or DBCC CHECKDB(dbname, NOINDEX) will complete successfully and report no errors. Chkdsk reports no errors on the drive.

How can I determine if this a corruption error in the database file or an actual “on the disk” bad sector? Can this be fixed in place or do I need a new disk and to restore from backups? While I have good backups, if I restore a recent backup, will that also “restore” the missing data/sector?

FWIW the DB is 2.2 TB so I’m rather reluctant to jump into a fix until I really know what’s going on.

backup – How do I find my OneNote notebooks so I can back them up before doing a complete system wipe?

Okay so there’s many issues with this. Ive worked mostly offline recently for OneNote. so nothings synced to the web. I use the desktop app for the program. under X account. when i try to log in to sync it all on the web it says there is no account for x account. which there should be. OneDrive has some of my notebooks. but they are not up to date. and it wont sync. ive tried going to C:Users<user>AppDataLocalMicrosoftOneNote<ver> . only thing in there is a backup folder thats empty. a cache folder with like 4 files 3 that are .bin (4kb) and a header file. and a tmp folder (empty).

I hope that’s enough information to help me.
need to find my up to date notebooks so I can back them up, wipe my system completely and reinstall. doing a system restore and keeping my files is not an option.

also wont be using that account anymore so getting it to work isnt a concern unless it gets me to where they can be backed up.

regex – Better way to find if value matches list of regular expressions

Sample code, see if a value matches any regex from a list:

    private val regex: List<Regex> = ...

    fun matches(value: String): Boolean {
        regex.forEach { re ->
            if (re.matches(value))
                return true
        return false

It works, but I’m wondering if there is a more elegant way to do this. Note that I don’t want to evaluate all of them, but stop whenever there is a match.

elementary number theory – Find the hundredth place digit in 2008^2007^2006^…^2^1.

Find the hundredth place digit of 2008^2007^2006^……2^1.
Answer: Given M=2008^2007^……^2^1, let k1=2007^2006^…^2^1, k2=2006^2005^……^2^1 , k3= 2005^2004^……^2^1 .
Then M=2008^k1, k1=2007^k2, k2=2006^k3 . Now mod(M,1000)=mod(2008^k1,1000)=mod(8^k1,1000)…(1). Since 1000=8125,we find mod(8^k1,8) and mod(8^k1,125). Mod(8^k1,8)=0=>8^k1=8m1…(2), m1 being positive integer.
Since phi(125)=100,( phi= Euler’s phi function )mod(8^100,125)=1…(3).
Now mod(k1,100)=mod(2007^k2,100)=mod(7^k2,100)…(4).
Since phi(100)=40,mod(7^40,100)=1…(5). Now mod(k2,40)= mod(6^k3,40)=16,since k3>2. =>k2=40s+16=>mod(7^k2,100)=mod(7^16,100)=1…(6). From (4) mod(k1,100)=1=>k1=100t+1=>mod(8^k1,125)=mod(8,125) using (3).
So 8^k1=125m2+8=8m1,using (2). So8(m1-1)=125m2=>8 divides m2.
=> m2=8m3. So 8^k1=125
From (1) it follows that hundredth place digit of M is 0.

Find the solution using graphical method

i am confused and not getting a way to solve this, please helpenter image description here

I’m using Python, to find all the pythagorian triplets within the number 1000, but this takes around 20 minutes to execute on my computer

If there is anyway in which I can make this code run faster, please let me know.

This is the code

for a in range(1,1001):
    for b in range(1, 1001):
        for c in range(1, 1001):
            if pow(a, 2) + pow(b, 2) == pow(c, 2):
                print(str(a) + "," + str(b) + "," + str(c))

If there is anyway in which I can make this code run faster, please let me know.

Thank you.

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