IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

arnold

_{[ n]}

_{[ 1]} 1+(-1)^n /1-n^2

harshiiit

cos pi= -1
therefore the question reduces to lim n tends to 1 1+(cospi)^n/1-n^2
apply demovire's law then L hospital rule da ans will come out to be 0

please rate if correct

arnold

i didn't remember the answer but ur method seems correct so i am giving u

a salute.......

nitin62225

ans is i*(

/2)

just apply l-hospital's rule

u'll get -(1/2)*ln(-1) .i.e= ln(i)=i*(

/2)

and correction in the earlier soln posted.........

after applying d-moveire's theorem u'll be getting 1+cis(n

) in the numerator and on seperating u get real part tending to 0 but the imaginary part as i

/2

plz rate......

harshiiit

yes nitin u are correct i did a very silly and unimaginable mistake
shit man

harshiiit

nitin
d/dx apowerx = apowerx log x only when a>0 and how did u write lni = i pi/2

nitin62225

ln x is defined when x>0 until u r not concerned with imaginary values.

root(x) is defined when x>=0 but we can calculate it for x<0 also but the values we get is imaginary.

we can differentiate a^x for a<0 also

and ln (a+ib)=1/2 ln(a^2+b^2)+ itan(inverse) (b/a)

so ln i=i(pi/2)

harshiiit

can u explain
ln (a+ib)=1/2 ln(a^2+b^2)+ itan(inverse) (b/a)
also can u tell me the expansion of lnx

nitin62225

a+ib = Re^(i

)

take log on bot sides u'll get the result.

ln x =ln (1+(x-1))=(x-1)-(x-1)^2/2............................

harshiiit

nitin
ialso thought the same but expansion for lnx does not satifies taylor's series

ariyam66

(1+(-1)^n + n - n ) / ( 1 - n^2 ) = 1/(1+n) + ( n - 1 ) / ( 1 - n^2 ) = 0

for Lt n ---- 1

So, my answer comes to be zero.

swaroopcool21

answer is zero

i agree with harshjeet

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