Recent content by absolute76

Ok that's true i have 40 bits now. and because of : rand('state',100); randn('state',200); a=rand(1,x)>0.5 ...this line will generate 0,1 with equal probility so my 40 bits just now are being generated as 0 and 1 and through that i want to partition it into 5...

Ok, let say i have 40 bits.. My question now is..how do i do 5 partition for the 40 bits (each partition consists of 8 bits)so that everytime i simulate it,it will randomly pick 8 bits from the 40 bits?? but i have to use this code in it: x=40 rand('state',100); randn('state',200)...

I need help in partioning bits of random number in matlab! Hey guys, I really need your help here...What i wanted to do actually is to partition the bits into 4 smaller group.. If anyone of you can give me a basic coding of it or maybe u can check if there anything wrong wit my coding...

owh ok..i think i get it... anyway thanks a lot for your help! i really appreciate it..

ok, so if i want to find the general solution, should i just let it be in terms of convolution plus the homogenous equation right? that should be my final general solution.

Owk now is this correct? u~(k)=e^[(k²)-(2 ln k)] ...u~(k)=e^(k²)/e^(2 ln k) then i transform it using the table?? right?

Ok my mistake again, so now the equation will be u~(k)[-2k²+2]=-k du~/dk so u~(k)= [-k/(-2k²+2) du~/dk Is it correct if i diffrentiate towards k on the right side?

i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right] owk i think i got it, is it correct...i(iu~(k)-k²u~(k)) that will give me -u~(k)+ik²u~(k)---->u~(k)[-1+ik²]?? owk how do i separate u~(k) so that it don't cancel each other(left and right side)??

:confused: error? error in which part?? yes i got the answer of 2k³-4k²-k/(-2k²+1)² using...(u'v-uv')/v² am i right??

ok thank you!, so now i rearrange it, u~(k)=(-k/-2k²+1)(du~/dk) is this correct if i diffrentiate towards k on the right side? that will give me u~(k)=2k³-4k²-k/(-2k²+1)²?

u~[-2k² + 1] = i/2∏ [d/dk u~(k)] I transfer, xu’ to the right side It this correct?? doesnt [d/dk u~(k)] =iku~(k) is the same?

If u don’t mind, can I ask one more question.. 2 d²u/dx² - x du/dx + u =0 For this question, I already Fourier transform both sides which give me, u~(k)[-2k² - xik + 1]=0… but it is impossible if just u~(k)=0 do you have any idea of solving it?

please correct me if I am wrong, for f(x)=x^5 let say i take -∞<y<0 (first) then i substitute inside---->∫(x-y)^5 e^-a|y| dy y=0---> (x)^5 [(e^0)=1] which give us when y=0 x^5 will this solution conclude that f(x)=x^5 is convergence??

oh sorry! my mistake! i get it already!..thank you.. anyway I am aware that for the second part f(x)=x^5, 1/2a ∫(x-y)^5 e^-a|y| dy -∞<y<∞ right?? I am aware that i have to use convergence test..isn't it the same? i have to integrate it and substitue the range?

oh issit that way?? i thought a=(λ^2)^1/2 and will give us a=λ because we cancel the 2x(1/2)??