Dear SciFace, It looks like there is an infinite loop in MuPAD 3.1. Your product returns instantly correct answers to the following sums sum(n, n= 1..n); sum(1/n, n= 1..n); sum(1/2^n, n= 1..n); 1/2*n*(n + 1) EULER + psi(n + 1) 1 - (1/2)^n However, if the user makes an attempt to calculate sum(1/n!, n= 1..n); MuPAD 3.1 keeps running after 40,000 seconds. Secondly, clicking at the Stop button produces no effect, and the only way to continue the calculations is to save the notebook and restart it. The same bug manifestation is present in MuPAD 3.0. We enjoy MuPAD over years and have not a vestige of doubt that your analysts will localize the origin of the problem soon. Best wishes, Cyber Tester http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ ..................................................................

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12/19/2004 3:09:20 AM

Hi All, I've been taking a look at DB Designer 4, and looking through the documentation (http://www.fabforce.net/dbdesigner4/doc/index.html) I am a little unclear on some of their nomenclature: '1:1' - Ok, one to one. Got it. '1:1 generalization' - Don't know this. Obviously different somehow from one to one, but how? '1:n' - One to many, I assume. '1:n non identifying' - Nonidentifying? What does this mean? 'n:m' - Many to many? Again, not sure. Can anyone help clarify? Thanks! -Josh Joshua Beall wrote: > I...

Hi, Anybody knows how to create this matrix but without using any loops? a=[1 1 1 1 1 ;2 2 2 2 2 ;3 3 3 3 3 ;.......;n n n n n ] Thank you. Hana. Hana wrote: > Hi, > Anybody knows how to create this matrix but without using any loops? > a=[1 1 1 1 1 ;2 2 2 2 2 ;3 3 3 3 3 ;.......;n n n n n ] HELP REPMAT - Randy Hana wrote: > > > Hi, > Anybody knows how to create this matrix but without using any > loops? > a=[1 1 1 1 1 ;2 2 2 2 2 ;3 3 3 3 3 ;.......;n n n n n ] > > Thank you. > Hana. Hope this isn't homework. >> repmat([1:n]'...

Ideally I would like to see someone explain why I get the following results they might also be of interest to others. For all a, x, and y Mathematica gives the following where c is the constant given by the convergent series c=NSum[(-1)^n*(n^(1/n)-1),{n,Infinity}] = 0.18785964246206... = the MRB constant. Regularization is used so sums that formally diverge return a result that can be interpreted as evaluation of the analytic extension of the series: NSum[(-1)^n*(n^(1/n)-a),{n,Infinity}] gives c-1/2*(1-a). NSum[(-1)^n*(x*n^(1/n)+y*n),{n,Infinity}] gives (c-1/2)*x-1/4*y. NSum[(-1)^n*(x*n^(1/n)-a),{n,Infinity}] gives (c - 1/2)*x + 1/2*a. ...

#1 #n+1 #2n+1 #2 #n+2 #2n+2 #3 #n+3 #2n+3 #n #2n #3n #1 #2 #3 #n #n+1 #n+2 #n+3 #2n #2n+1 100 Gbps aggregate stream of 64/66b words virtual lane markers virtual lane 1 virtual lane 2 virtual lane 3 virtual lane n Simple 66-bit word level round robin distribution Transmit PM Transmit PM Receive PM Receive PM Transmit PC Receive PC 1 0.0 1.0 0.1 0.2 0.3 1.2 1.2 1.1 0.0 1.0 0.1 1.1 2.0 3.0 4.0 6.0 7.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0 17.0 18.0 19.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0 16.0 17.0 1...

Hi! Often, I have some situations when I have to convert a vector of 10 elements to a vector of 30 elements or vice-versa. (the ratio is 1-3) I want to use the std::transform do to these tasks, but I end up using std::for_each Is a special/custom iterator the only way to do that with std::transform? Example: vector<int> source, destiny; .... now source has 30 elements destiny.resize(10); transform(source.begin(),source.end(),destiny.begin(),ThreeToOneConverter()); ** how to write the ThreeToOneConverter functor in order to process and skip three elements at a time instead of one...

I have a Conjecture: For every n > 1 there is always at least one prime p such that (n-1)n/2 < p < n(n+1)/2 n = 2; b = 1; While[NextPrime[b] < (b = n (n + 1)/2), n++] n $Aborted 14394105 In this diagram,there is at least one small point(prime) between every two medium points(triangle number ). In[58]:= start = 1; n = 400; pl = Table[{Prime[i], n}, {i, start, n}]; tl = Table[{i (i + 1)/2, n}, {i, start, n}]; ListLinePlot[Prime[Range[start, n]], Epilog -> {PointSize[Medium], Point[tl], PointSize[Small], Point[pl]}] The still unsolved Legendre's conjecture asks whether for every n > 1, there is a prime p, such that n^2 < p < (n + 1)^2. Comparing, (n+1)^2-n^2=2n+1, while n(n+1)/2-(n-1)n/2=n Clear[n]; FindInstance[ NextPrime[n (n - 1)/2] > n (n + 1)/2 && 0 < n < 2^2^20, {n}, Integers] FindInstance::nsmet: The methods available to FindInstance are insufficient to find the requested instances or prove they do not exist. On Feb 1, 7:13 pm, a boy <a.dozy....@gmail.com> wrote: > I have a Conjecture: For every n > 1 there is always at least one prime p > such that (n-1)n/2 < p < n(n+1)/2 > > n = 2; b = 1; > While[NextPrime[b] < (b = n (n + 1)/2), n++] > n > > $Aborted > > 14394105 > > In this diagram,there is at least one small point(prime) between every tw= o > medium points(triangle number ). > > In[58]:= st...

Hi, I've gotten into the habit of mentally converting big expressions to parse tress (started doing this after reading some c.l.c posts on parse trees). Can someone verify if the following assumptions are correct? 1) For ++n, one can us (n=n+1) as it's equivalent: | = | / \ | n + | / \ | n 1 2) For assignment operators of the form n op= b, the equivalent tree is: | = | / \ | n op | / \ | n b 3) For n++, one can use n, but with a side condition that (n=n+1) will occur somewhere between now and the ear...

Hello, I was trying to show that this limit was 'e' But when I try large numbers I get errors def lim(p): return math.pow(1 + 1.0 / p , p) >>> lim(500000000) 2.718281748862504 >>> lim(900000000) 2.7182820518605446 !!!! What am i doing wrong ? Regards On Mon, Nov 9, 2015 at 11:21 PM, Salvatore DI DIO <artyprog@gmail.com> wrote: > I was trying to show that this limit was 'e' > But when I try large numbers I get errors > > def lim(p): > return math.pow(1 + 1.0 / p , p) > >>>> lim(50000000...

Notation: R(a,k) denotes the rising factorial a(a+1)...(a+k-1 ...the certificate being 2 2 2 (5 m - 12 m p - 2 m + 8 p + 3 n m - n - 4 n p + 1 + 4 p) n -------------------------------------------------------------, QED . ... Musatov ...

HELLO!! :) need help people!! imagine: M=[1,2,3 ; 4,5,6 ; 7,8,9] now i say: x=2 (needs to be to any x; in this case x between 1 and 3!) the new matrix should be: NEW_M=[ 1,0,2,3 ; 0,0,0,0 ; 4,0,5,6 ; 7,0,8,9 ] I tried a lot, but i have problems putting the values: 2,3,4 and 7(for this especific case). It appears zero! My code is: for hh=1:aux_invB for j=1:aux_invB hh2=hh; j2=j; if ...

Please give me a direction - am I totally lost in my own world and have spent 6 months in vain trying to figure this out, or is there a fruit of truth hidden somewhere? Excuse me for rapid end of some sentences and chaotic appearance of equations- i am just tired. I have spent 1 month practically thinking only about 1 thing. I need some rest and support. On other hand, this chaotically ordered text is good as it will cause confusion and will stretch imagination:) Approximately 50% of the following text is obviously right > 37,5%-is subjective <12,5%- optional Approximately 50% of peop...

I sincerely hope You will read the text bellow, get intrigued and discuss with some of Your colleagues. I have a feeling Your INTUITION will not allow You just to dismiss my thoughts in a trash bin. What if I give up? Please for give the messy form - form represents content, messy form allows to represent a developing idea. Rigid form can only represent dead , already solved idea- why would You want to see that? Best regards, Ivars Fabriciuss Dear Friends, 0 has infinity potential dimensions all of whom has potential to materialize -it encompasses the beginning of everything;when used...

While I was playing around with probabilities, I noticed that I wasn't able to get Maple 12 to figure out that (1-p)^n = 1-n*p has a solution at n = 1. Of course, it's trivial to verify by substitution that this is indeed a solution for any p, but I wasn't able to get solve() to cough it up: > assume(p > 0, p < 1, n > 0): > interface(showassumed = 0): > solve((1-p)^n = 1-n*p, n); 0 OK, n = 0 certainly is a solution, but I'd have liked the other one too. Let me see if solving for all variables would do a...

C:\Apache\Apache2\htdocs\online\gallery\session.php on line 63 ... (N F. Ti,Tjli=1 I K ~a(NP)XN iT-Ck I(Y? (A-3) p=l i=1 k=1 j=1 Cij J ... n Ck (-1) i (A-13) k=1 Ci I Equation (A-13) = Equation (A-4) Q.E.D. ... ...

Dear all, I make a loop to pull out the data and save to files, but I got the error messages. ??? In an assignment A(I) = B, the number of elements in B and I must be the same. What's wrong on my code? a = (1:5)' b = (5:10)' for n = 1:5 c(n) = cat(1,a(n,:),b(n,:)) end I want to save [1 6], [2 7], [3 8], [4 9] and [5 10] at each looping to different file named 1.txt, 2.txt, 3.txt, 4.txt and 5.txt, respectively. For example: c(1) = [1 6] --------> then save to 1.txt c(2) = [2 7] --------> then save to 2.txt c(3) = [3 8] --------> then save to 3.t...

Hi all. I recently came across a recurrence of the form: T(0) = a T(n) = c^n + T(n-1) Of course, when c is 2, T(n) = (a)*2^n. But, is there a general closed form for arbitrary c, or at least for when c is an arbitrary positive integer? Thanks in advance for any response. -Dan > T(0) = a > T(n) = c^n + T(n-1) > > Of course, when c is 2, T(n) = (a)*2^n. But, is there a general closed form > for arbitrary c, or at least for when c is an arbitrary positive integer? Sorry, I just noticed a cognitive misfire in my original post. When c is 2, the closed form is: T(n) = a*2^(n...

Hi, is there a "simple kernel based" routine to do the scalar product of two one dimensional vectors, producing the m,n Matrix: v1[m,1].v2[n,1] -> mat[m,n] Thanks E.F. EF wrote: > is there a "simple kernel based" routine to do the scalar product of two > one dimensional vectors, producing the m,n Matrix: > > v1[m,1].v2[n,1] -> mat[m,n] ?Outer -- Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/ San Jose, CA, USA && 37 18 N 121 57 W && AIM/Y!M/Skype erikmaxfrancis It...

P=((n*n)-2)*(n*n))-1P Q.E.D. Q.E.D,: n=22 . Then 22*24-1 =527=31*17 where are the 5's and 10's 42*44-1 =1763=41*43 And: 10002*10004-1= 100060007 = 1279.81~78233 Martin Musatov Solved P==NP with this post. ($1MM Clay Millennium Prize to be donated to cure childhood cancer) On Sep 23, 9:19=A0pm, "M.MichaelMusatov" <marty.musa...@gmail.com> wrote: =A0P=3D((n*n)-2)*(n*n))-1P =A0Q.E.D. Q.E.D,: n=3D22 . Then 22*24-1 =3D527=3D31*17 where are the 5's and 10's 42*44-1 =3D1763=3D41*43 And: 10002*10004-1=3D 100060007 =3D 1279.81~78233 Martin Musatov Solved P...

Does anyone know why the remainder and the division operations are incorrect with negative numbers in C++? Is it about assembly? OuFeRRaT -- [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] OuFeRRaT schrieb: > Does anyone know why the remainder and the division operations are > incorrect with negative numbers in C++? Is it about assembly? In how far is this incorrect? The modulo operation and division must be defined such that (x/n) * n + (x % n) = x (1) holds, and this should be true (-1 / N =...

On Fri, 28 Jan 2011, EF wrote: > Hi, > > is there a "simple kernel based" routine to do the scalar product of two > one dimensional vectors, producing the m,n Matrix: > > v1[m,1].v2[n,1] -> mat[m,n] > Hi, like this: m = 3; n = 2; v1 = RandomReal[{0, 1}, {m, 1}]; v2 = RandomReal[{0, 1}, {1, n}]; Dimensions[v1.v2] Oliver > > > Thanks E.F. > > ...

When reading a binary input stream with fread() one can read N bytes in two ways : count=fread(buffer,1,N,fin); /* N bytes at a time */ or count=fread(buffer,N,1,fin); /* 1 buffer at a time */ I would assume the latter form would be faster, or at least less of a load on the CPU. That's just an assumption though, is it typically true? Speed matters here, but the form which (I suspect) is faster can't handle a partial input block. If the input file isn't a multiple of N it will give an error on the final read, and fread() provides no possible way of figuring how many, ...

Hi: Why are allocatable indices in main different from those in subroutine? module defXY contains subroutine Bar(x,n) real*4, allocatable :: x(:) integer :: n n=3 allocate(x(n)) do i=1,n x(i)=i end do print *,'in sub:','n=',n do i=1,n print *,'in sub:',x(i) end do end subroutine Bar end module defXY program main use defXY real*4, allocatable :: x(:) integer :: n=3 call Bar(x,n) print *,'in main:','n=',n do i=1,n print *,'in main:',x(i) end do stop end Results: in sub n=3 in sub:1. in sub:2. in sub...

Consider a simple onkeypress script that alerts with messages like "You typed a A!" I have written a small function that considers when it should be "You typed an A!" where the variable string is preceded by "an" rather than "a", and would appreciate comments, since English is not my mother language and I don't know exactly when you say "an" rather than "a". If I feed whole words to this function, is a test for the first letter enough? function atrailn(n,u){ var a = ['a','e','f','h','i',...

Any more UNCOMPUTABLE<-/->FUNCTIONS need programming while I'm here! CLAY send the $8,000,000 Cheque to Kings Beach Queensland! ********************HALT(n)********************************** T |- HALT(Prog1) = Prog1|-H1() ^ Prog1|-H2() ^ .. ^ Prog1|-Hn() CONCURRENCY RULE ALL(m) ALL(n) NOT( P_m|-HALT() ^ P_m|-Hn()) PROGRAM SPECIFICATION HALT() IF NOT(EXIST-FUNCTION( Hn() ) THEN DIE i.e. a Halt function will only run if 1 or more Test Harnesses are within the program_to_test which is not the_program_doing_the_testing. ********************PRV(n)**********************************...

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