Messages posted by vinu

Hey ..this is serioulsy unfair !this is an expert Q n i believed i'd surely get an answer !!But , this is ridiculous , i've waited for 20 days and still no anwer to this ?

Hello..

Can anyone tell me any practical application(s) of "groups/rings/fields ", the bulding blocks of ABSTRACT algebra ?

I know they are merely unifying threads that connect different areas in Mathematics,

but as a seperate entity ..... where can they be of any significance ?

Is it that they are only developed to deeply dig into the definition of them or do they help in solving a particular pratical situation ?

as in .... for example a matrix ...its totally a mathematical tool which can used in many applications...viz; solving a set of linear equations , defining optimistic solutions ,etc.

For a layman, matrix or a set is understandable ..like set of integers,states in a country or matrix of coefficients of numbers etc .. heir definiton has been precise to be used in numerous ways .....

whereas groups/rings /fields ..they themselves are inter-connected , how do we explain these to a layman for him to understand ?

I'm sorry , i'm totally confused at this ...help me!

Hello to goiit ,

I've been a member of goiit since 2006 ....it has been such a wonderful experience posting things n answering queries ...

I seldom come online these days , but whenever i come, i try to check out hows the website tht aided me for my IIT preparation 3 yrs ago ...

Right now , i'm busywith soem college work....i'm planning to organise a mathematics event in my college's tech fest ...

have been browsing a lot of websites for ideas or problem statements ...

just then i thought ...y shudnt i post it here too ...

i hope someone can help me with some suggestions for the maths event ....something tht can really interest people to participate in it ..

So,get ready ...rack ur brains too ...

i hope this website comes to my aid now also ...

Thank you !

nice 1 shubhu.....i miss ur {jokes} punch lines ! ;)

congrats bro....

Hope u cross 10,000 too with success!!

dito is same as -do- ;
i.e,u use less than in between as used in the prev inequality...

YEESSS !!!!!!!!!!!
Btw..this is the 300th post of this topic...
shubham(topic creator) u've made a triple century mannn!!
congratsssssss.......!

Hi pav,

u know.._[x]

_[0] (1+x)^1/x = e ;_[x]

_[a] (f(x))^g(x)= e^{lim(as x->a) [f(x) - 1]*g(x)} where g(x) tends to infinity and f(x)->1 for x->a ;

given..f(x) = cosx ; g(x) = cot^2 x ;

so,limit = e^{[ - 2sin^2(x/2) ] * [ cos^2(x) / 4sin^2(x/2)cos^2(x/2) ]} ;

as x->0 ; lim = 1/

e.....ans

Ho ho ho...I'M THE NETA :)

Hie shubhu,

3) [xf(a) - af(x)] = [xf(a) -a{f(x) - f(a)} - af(a)] ;

u know , _[x]

_[a] f(x) - f(a) / (x-a) = f ' (a)

so, limit = { _[x]

_[a] [(x - a)f(a)] / (x-a) } - af ' (a) + af ' (a) = f(a) .....ans.

1] given a(1+r) = 1 ;

sum (S) = a/(1-r) ;

now , u've a = 2[S-a] ;

so, r = 1/3 ;

hence,frm the first eq,......a = 3/4..ans.

2]

for the 5th Q,

u can simply use L'Hospital's rule....jus one step solu...

4.) Use L'Hospital's rule.....

limit as x->0 = 0......ans.

6.)

u know ,

_[x]

_[a] [f(x)]^g(x) = e^{lim [f(x) - 1] *g(x) as x->a};

if g(x)->infinity , f(x)->1 as x->a ;

so,given = e^{lim {(a^x+b^x+c^x - 3) / 3} * (2/x)};

= e^{{(2/3) * lim {(a^x-1)+(b^x-1)+(c^x -1) / x}};

now , lim as x->0 (a^x-1)/x = log a ;

So, given lim as x->0 = e^{(2/3) * log(abc)}= (abc)^2/3............ans.

5.)

u know..._[x]

_[0] [log(x+1)] / x =1 ;

given = (1/e)* _[x]

_[e] [log(x/e)] / [(x/e)-1] ;

=(1/e)*_{[(x/e) - 1][0]} [log(x/e)] / [(x/e)-1] ;

= 1/e.................ans.

Hi shubhu,

2)

num = sinx[5-14cosx+3(3-4sin²x)];

= sinx[14(2sin²x/2)-12sin²x];

now limit = 7{ (sin²x/2) / (x/2)²} - 12{sin²x/x²} ;

as x->0 , lim->-5.........ans.

Hi shubhu,

4)its very easy....

jus differentiate 'y' two times and divide both sides by 'y'...u get the result...

2)this one too...yr = a^r * cos[ax+(rpi/2)] ;

u'll observe tht all the rows in the det r same takin a³,a⁶ common frm 2nd,3rd rows resp.....

hence,det = 0 .

4) Put z = lzlcis

;

so LHS = lcis

-1l = 2*l sin

/2 l * l i*cis

/2 l..........{also, l i*cis

/2 l = 1 } ;

= 2*lsin

/2l ;

<= l

l = l arg(z) l.............ans.

3) given : x+i(y+2)+

*

[(x-1)²+y²] = 0........

also,given tht

is a real no...

so, y = -2 ; x+

*

[(x-1)²+y²] = 0 ;

i.e; x²(

²-1) - 2x

²+ 5 = 0 ;

x is real...so, discriminant >=0.........

⁴-5

²+ 5 >= 0 ...u can find range of

frm this.......ans.

Hi shubham,

2)if p were tht real root...then p²+

p+

= 0 ;

so, Im(

) / Im(

) = -p ; p²+p*Re(

) = -Re(

) ;

substitute value of p frm 1st relation into the 2nd relation....ans.

1) if u meant lz+1/zl = p(say)....

then check out the ans....

u know lz+1/zl <= lzl+1/lzl ....{let lzl be x}

so,x²-px+1 >= 0

therefore min value of x is ....u find it out

Hey taruntanuj,

u needn't sweat urself so much on Q1....

its simple...check this out...

1) t_r = tan^-1{1/[r²+(3/4)]} = tan^-1 {(r+1/2) - (r-1/2) / [1+(r²-1/4)]} ;

= tan^-1 (r+1/2) - tan^-1 (r-1/2) ;

S_n = sum to n terms = _[r=1]

^[n] t_r = tan^-1(n+1/2) - tan^-1(1/2) = tan^-1{n/[1+(2n+1)/4]} ;

as n->infinity ; (div num & denom by n) ;

S = tan^-12 . ...............ans.

Hi iit2008,

i made a small mistake in putting values 4 p....

neways,i edited the solu....check it out..

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