Mathematics desk
< March 7	<< Feb \| March \| Apr >>	March 9 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

March 8

Pascal Matrix

Is there a way to invert the Pascal matrix U_infinity? Black Carrot (talk) 02:16, 8 March 2009 (UTC)[reply]

The exponential definition should give a simple formula for the inverse. Upper triangular matrices behave reasonably well, even in infinite dimensions (even over non-discrete linear orders). If you work a few finite examples, the pattern for the full inverse should become clear too. JackSchmidt (talk) 03:10, 8 March 2009 (UTC)[reply]

And for the symmetrical one - since S_n = L_nU_n, this means that

S_{n}^{-1}=U_{n}^{-1}L_{n}^{-1}

. עוד מישהו Od Mishehu 10:20, 8 March 2009 (UTC)[reply]

Formula for nxn Singular Matrices

Is there a formula to calculate any arbitrary nxn sized singular matrices, and if yes, what?~~The Successor of Physics~~ 03:56, 8 March 2009 (UTC)[reply]

I think you can create one by creating an arbitrary (n-1)×n matrix and then add any two rows to give you the final row. This should yield a matrix whose determinant is zero. More generally, you can calculate

Row_{n}=\sum _{i=1}^{n-1}a_{i}\cdot Row_{i}

for arbitrary

a_{i}\,\!

values. -- Tcncv (talk) 05:59, 8 March 2009 (UTC)[reply]

Thanks!~~The Successor of Physics~~ 13:10, 8 March 2009 (UTC)[reply]

Add a row of zeros. Add rows of zeros. Add all zeros. —Preceding unsigned comment added by 69.72.68.7 (talk) 03:11, 10 March 2009 (UTC)[reply]

True, but if we take that approach, we might as well make an entire matrix of zeros. However, I suspect the OP was looking for matrices where the singular nature was not obvious – possibly as a test question or homework problem. -- Tcncv (talk) 03:35, 10 March 2009 (UTC)[reply]

It seems to me that the OP was looking for a way of generating all singular matrices. In that case, the method above (take an arbitrary (n − 1)×n matrix, add an extra row which is linear combination of the others) almost works, except that you have to allow the extra row to be put in an arbitrary position in the matrix, not just the last row. — Emil J. 11:09, 10 March 2009 (UTC)[reply]

You're right. Defining

Row_{j}=\sum _{i=1}^{n,i\neq j}a_{i}\cdot Row_{i}

could generate additional cases not reproducible in my original formula if

a_{n}=0\,\!

. Question: If we applied the technique to columns instead of rows, would we generate the same set? My gut tells me yes. -- Tcncv (talk) 03:13, 11 March 2009 (UTC)[reply]

Yes, singularity can be defined either in terms of rows or in terms of columns, it makes no difference. — Emil J. 11:23, 12 March 2009 (UTC)[reply]

Hard to compute mathematical constant

Is there a prominent example of a mathematical constant which is so hard to compute, that (relatively) few decimal places are known? I'm aware that you easily can construct numbers to be hard to compute, but I'm asking for mathematical constant which is reasonanly important in its field and is hard to compute "by accident" and not by design. --Pjacobi (talk) 13:49, 8 March 2009 (UTC)[reply]

Off the top of my head, I'd suggest looking at Feigenbaum constants. Apparently, only a thousand digits have been worked out of δ. Pallida Mors 14:32, 8 March 2009 (UTC)[reply]

Chaitin constant is hard for sure, even if not by accident... --pma (talk) 16:29, 8 March 2009 (UTC)[reply]

A hard to compute function is the zeta function. If you "compute" that, expect to get a fields medal.

What's so hard about computing the (presumably Riemann) zeta function? People do it all the time. Algebraist 12:07, 9 March 2009 (UTC)[reply]

The inverse to the Riemann zeta function is difficult to compute (which is, presumably, what the person without a signature meant), but I don't think that's really relevant to this discussion. --Tango (talk) 13:39, 9 March 2009 (UTC)[reply]

If you can compute n digits of a monotonic function in O(f(n)), shouldn't you be able to compute n digits of the inverse in O(n(f(n))? (Assuming you have finite bounds on the value of the inverse.) Eric. 131.215.158.184 (talk) 18:15, 9 March 2009 (UTC)[reply]

Of course, you can define constants that are currently impossible to compute at all, such as C=0 if Goldbach's conjecture is true, otherwise C=the least even number that is not the sum of two primes; or the lowest counterexample to the Collatz conjecture; or the position of the first sequence of a googol consecutive zeros in the decimal expansion of pi. AndrewWTaylor (talk) 12:36, 9 March 2009 (UTC)[reply]

Chaitin's constant beats them by being uncomputable period, not just currently uncomputable. It's less contrived, too. Algebraist 12:45, 9 March 2009 (UTC)[reply]

Percentage-wise I'd imagine it's fairly hard to compute the digits of Graham's number. You'd probably run out of atoms to write the digits down with before you got any significant percentage of the overall number. Readro (talk) 13:18, 9 March 2009 (UTC)[reply]

Brun's constant is only known to a dozen or so decimal places. Here, "known" actually means more like "made an educated guess based on a massive computation and unproved conjectures", IIUIC there is not even an explicit upper bound on the constant which would be rigorously proved. There's also a discussion in [1]. — Emil J. 13:59, 9 March 2009 (UTC)[reply]

Good example. Eric. 131.215.158.184 (talk) 18:15, 9 March 2009 (UTC)[reply]

How about the area of the Mandelbrot set? Fredrik Johansson 15:15, 9 March 2009 (UTC)[reply]

Compute the largest number of consecutive primes via a polynomial of an order less than the number of primes. —Preceding unsigned comment added by 69.72.68.7 (talk) 03:18, 10 March 2009 (UTC)[reply]

conditions for a map to be a homomorphism

if a map between two groups preserves group structure on some non-trivial normal subgroup of the domain, is this enough to say that it is a homomorphism. If the answer is no are there any further conditions under which this is the case.

No, it's not sufficient. Consider the map

f:\mathbb {Z} _{2}\times \mathbb {Z} _{2}\rightarrow \mathbb {Z} _{4}

that maps (0,0) to 0 and (1,0) to 2. That's a homomorphism on a normal subgroup of the domain. Then map (0,1) to 1 and (1,1) to 3. You how have f(2(0,1))=f(0,0)=0, but 2f(0,1)=2*1=2, so not a homomorphism everywhere. I don't know if there is a similar result that works, with just a few extra conditions, I doubt it, though. --Tango (talk) 15:23, 8 March 2009 (UTC)[reply]

what about if the map is surjective and the normal subgroup maps once over injectively onto the image

The counter-example I gave is bijective. --Tango (talk) 16:37, 8 March 2009 (UTC)[reply]

Note that if any automorphism of a group, G, fixes a subgroup H of G, H is said to be characteristic in G and we write H char G. If every inner automorphism of G fixes H, H is normal in G. In particular, every characteristic subgroup of a group is normal. Examples of characteristic subgroups are: the center of a group, the whole group, and the trivial group. A question you might like to investigate is whether any automorphism of a normal subgroup of a group G, extends to an automorphism over the whole group. If not, what additional conditions can you impose? --PS T 11:44, 9 March 2009 (UTC)[reply]