Recent content by reza

then t1=t2? right?

my problem is velocity i don't know how air rsistance affects on velocity?

if we lunch a projectile with the act of air resistance which t (time) is longer t1 =>from t=0 to arriving to the highest point t2=> from highest point to the lanch point

f(x)=ln(x+1)-x+x^2/2 f'(x)=1/(x+1)-1+x=x^2/(x+1)>0 f'(x)>0 then it is increasing So, for x > 0, we have: f(x) > f(0) <=> f(x) > 0 <=> ln (x+1)-x+x^2/2>0 <=> ln (x+1)>x-x^2/2 is this proof Ok?

excuse me i put mistakaly sinx^2~x^2 lim (cot^2 x-(1/x^2))=lim [(1/tan^2 x) - (1/x^2)]=.. and set 1/tan x^2 = (1/sin x^2)-1 is it right?

thanks then sin cos cot and tan when x tends to infinity the limit does not exist __________________________ lim (cot^2 x-(1/x^2))=lim [(cos^2 x -1)/x^2]=lim [(-2sinxcosx)/2x] = x->0 x->0 x->0 =lim (-sin2x/2x) =lim (-2cosx/2)=-1/2...

f(x) =2x + ln x f^-1(x)=? i do f(x)=2x +lnx=ln (e^2x) + ln x = ln [(e^2x)x] =>e^f(x)=e^2x(x)=> can eny body solve this equation

thanks then the answre would infinit yes? and what wold happen if x->0

thank you very much but how we can prove the other side

I think a function is integrable if it be dic-continus on countable point

yes you right x - (x^2)/2 and in the book (calculus , rechard a.silverman) has written for x>0 Do you need the adress of problem?

sorry for my bad method writing lim [(cot x^2) - (1/(x^2))] x-->1/0 1/0 means infinite i don't know how to show its sign sorry again

show that (for all posetive x) x-1/2 x^2<ln(1+x)<x i understand it in graf please prove it for my in methemathic

Lim cot x^2 - 1/x^2 x--->&

can you give me a mthemathica proof