matrix 2-norm condition number
i wanted to compare the result of the fucntion [cond();] which retrun the matrix 2-norm condition number as a ratio between the largest and the smallest singular value using the [isequal();] function with the result "segma" matrix of the svd out of [svd();] fucntion as:
a = [1 2 5;4 8 7;8 9 5]; % example
[u,s,v] = svd(a);
c\_num = cond(a);
sing\_ratio = s(1,1)/s(end,end);
isequal(c\_num, sing\_ratio);
and the reslut was [0] although they are the same, and sadly this isn't the first tine that the equal function fails with me and don't know the reason.
sadly this isn't the first tine that the equal function fails with me and don't know the reason.
the isequal() function didn't fail. the two results are different ... just not by very much:
>> percent\_difference=abs(c\_num - sing\_ratio)./c\_num*100,

percent\_difference =

 2.4864e-14
if you want to know how cond() is computed, you can just read it:
 >> type cond.m
welcome to mrexcel!

it actually works for me (see below). this appears to be a kind of rounding error. the way excel stores numbers causes some issues when you have decimals. it is a bit rare when comparing a cell to itself, but it can happen. you can try to use sumproduct instead, see column h.

efgh1valuescountifranksumproduct248737.76222336113.11333454287.76111f2:f4f2=countif($e$2:$e$4,">"&e2)+1g2:g4g2=rank.eq(e2,$e$2:$e$4)h2:h4h2=sumproduct(--(round($e$2:$e$4,2)>round(e2,2)))+1 finding the correct answer here is important, but perhaps equally important is finding it using the method of matrix reduction, since those are the instructions.the augmented matrix is4 -2 -5 -514 -1 -2 -364 -2 4 -24you'll have to forgive the poor formatting. this answer box doesn't have a proper math symbol editor.using elementary row operations, the reduced row echelon form of the augmented matrix is1 0 0 -60 1 0 60 0 1 3from which we conclude that the system has a unique solution (-6,6,3). Definition used:  

Congruence.  

Let m be an integer greater than 1. If xand yare integers, then x is congruent to y modulo m if x — y is divisible by m. It can be represented as \(\displaystyle{x}={y}{\left({b}\ mod\ {m}\right)}\). This relation is called as congruence modulo m.  

Calculation:   

Consider the values and congruence classes modulo m,  

\(\displaystyle{\left[{13}\right]}^{{{6}}}\) and m = 15.  

It is evident that [x]=[y] if \(\displaystyle{\left[{x}={y}{\left({b}\ mod\ {m}\right)}\right]}\). Since \(\displaystyle{13}=-{2}{\left({b} \ mod \ {15}\right)}\).  

Then,  

[13]=[-2]  

\(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[-{2}\right]}^{{{6}}}\)  

\(\displaystyle={\left[{\left(-{2}\right)}^{{{6}}}\right]}\)  

=[64]  

Since, by the principal or congruence \(\displaystyle{64}={4}{\left({b}\ mod \ {15}\right)}\). Therefore, \(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[{4}\right]}\) in \(\displaystyle{Z}\_{{{15}}}\).  

Therefore, the value of the expression is [4]. not sure i understood. are you having problems with this code?

if yes, i think you’re using the wrong comparison operator (not = but ==). to find: limitsnote: first check the form of imit. used this method if the limit is satisfying any one from 7 indeterminate form.in this case, indeterminate form is formula used: = 1so, by using the above formula, we have = = therefore, = The problem is that your pair's closure must return two different types of objects: int if you ask for "con", and the next closure on the line if you ask for "crd".


Here's one possible implementation:


Use a speaker so you can come back.



```
public static Func<string, dynamic> Pair(int x, Func<string, dynamic> y)
{
    Func<string, dynamic> pair =
           (a) =>
           {
               if (a == "con") return x;
               if (a == "crd") return y;
               throw new Exception();
           };

    return pair;
}

public static void Print(Func<string, dynamic> pair)
{
    while (true)
    {
        var next = pair("crd");
        Console.WriteLine((next == null ? "-":"+") +"> " + pair("con"));
        if (next != null)
        {
            Console.WriteLine("|");
            pair = (x) => next(x);
            continue;
        }
        break;
    }
}

        

        
        
        
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```


  




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Test:



```
var pair = Pair(1,
            Pair(2,
                  Pair(3,
                        Pair(4, null))));

Print(pair);

        

        
        
        
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```

output:



```
+> 1
|
+> 2
|
+> 3
|
-> 4

        

        
        
        
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``` see: <https://feedback.photoshop.com/conversations/lightroom-classic/lightroom-classic-when-trying-to-add-files-to-collections-i-get-this-error-0-attempt-to-compare-two-nil-values/5f9091ea35f40c2520b9ecc4> Your function works, but both the IF and the ELSE return an alert message that says its invalid.  

The first part - the IF - should return in the alert box "Valid SSN"  

I am assuming your input should be like this 123-45-1234


but do you need to tell the user the input is valid? Surely the alert should only be displayed if the input is invalid? So the if should return nothing. imho 5.23.9 = 43 therefore, 5.2 : 3.9≠ 3:4 .hence, false. common factor of 4 and 5 is only 1∴ is in standard formcommon factor of 5 and 4 is only 1∴ is in standard formcommon factor of 16 and 25 is only 1∴ is in standard formhence, is equivalent to