29-8-20 Class & Assignment వివరణ తెలుగులో

నమస్కారము / नमस्कार / ನಮಸ್ಕಾರ/ வணக்கம்/ নমস্কাৰ/ નમસ્તે/ ସୁପ୍ରଭାତ/ നമസ്കാരം


MFCS   Tautology   / contradiction/ well formed formulas/ substitution instance  వివరణ తెలుగులో 



Plz try to submit 
Assignments by 5-9-20:

From text book Exercise 1-2.4   1(b), 2(c), 4(a)
                                        1-2.6    1(a), 2(b), 3 
                                         1-2.8    1(b),(c),(e), 2(a)    

Upload ur answers in ur blog/ Github/ drive and send links only

ఎప్పుడైనా అర్థం కానిచో ..వీలైనంతవరకూ వెంటనే అడుగగలరు. పరీక్షలకు కొద్ది ముందు సందేహాలు అడిగితే.. చెప్పటం కుదరదు.   ఎప్పటికప్పుడు నోట్స్ వ్రాసుకోవటం/ పాట్ఃయపుస్తకంలో ఇచ్చిన ఉదాహరణలు చూసి నేర్చుకొని తర్వాత స్వంతంగా చేయటం ముఖ్యం... అన్ని సబ్జెక్ట్స్ కీ ఇది వర్తిస్తుంది.
 
THose who are willing to do voluntary work.. plz inform. 

Plz try to participate in the competitions that were announced (Project making/ SInging/ writing)
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100_working code for conjugate gradient iterative algo for finding solution of Ax=b

%%100_working code for conjugate gradient iterative algo for finding %%solution of Ax=b, where A is symmetric Positive definite %%written by vikram bhayyA on 18-6-26 ~13:19-13:59 %%general clc clear all A=input( 'enter a Symmetric Positive Definite matrix:' ); %A=[4,1,1;1,3,-1;1,-1,3]; %has to be symmetric Positve definite [m,n]=size(A); n b=input( 'enter the output vector of size n,1:' ); %b=[6,3 3]'; %x0=zeros(n,1); x0=[1 -1 0]'; %x0=[0 0 0]' ; % initial guess %initialize r0=b-A*x0; %residual if norm(r0)<0.1 display( 'solution of Ax=b is:' ) x0 ...
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100% working code for Steepest Descent iterative algo for finding solution of Ax=b

%%100_working code for Steepest Descent iterative algo for finding %%solution of Ax=b, where A is symmetric Positive definite %%written by vikram bhayyA on 18-6-26 ~14:37-14:42 %%general. clc clear all A=input( 'enter a Symmetric Positive Definite matrix:' ); %A=[4,1,1;1,3,-1;1,-1,3]; %has to be symmetric Positve definite [m,n]=size(A); n b=input( 'enter the output vector of size n,1:' ); %b=[6,3 3]'; %x0=zeros(n,1); x0=[1 -1 0]'; %x0=[0 0 0]' ; % initial guess %initialize r0=b-A*x0; %residual if norm(r0)<0.01 display( 'solution of Ax=b is' ) x0 else %p0=r0; %search direction k=1; while (k) alpha0=(r0'*r0)/(r0'*A*r0); %step size x1=x0+alpha0*r0; %update x1 r1=b-(A*x1)...
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