Trignometry - Limit

21

tapanmast Vora ·2009-03-13 07:37:32

Pl. reply Priyam/akand

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33

Abhishek Priyam ·2009-03-13 07:58:51

[3]

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21

tapanmast Vora ·2009-03-13 07:56:59

OH!!!

he he.....

i thot the first pink Akand got was for Q1....

NIce GOOGLY huh!!! [9]

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33

Abhishek Priyam ·2009-03-13 07:54:57

and ya Q1 mein zero isliye aaya na kyunki Sum of n Roots of unity = 0 (for n>1

Q.1

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21

tapanmast Vora ·2009-03-13 07:52:46

are upar me .. sin () cos() hai usko infinity se divide karnege to zero hi naa aega...

[7] wat? kis post ki baat kar raheho priyam

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33

Abhishek Priyam ·2009-03-13 07:50:05

are upar me .. sin () cos() hai usko infinity se divide karnege to zero hi naa aega...

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21

tapanmast Vora ·2009-03-13 07:46:09

Yah true.....

I gave the ans for n+1 terms.... with a lil calc mist. LOL.....

and ya Q1 mein zero isliye aaya na kyunki Sum of n Roots of unity = 0 (for n>1)

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33

Abhishek Priyam ·2009-03-13 07:43:13

zero... for first... in book it was none of these :P

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33

Abhishek Priyam ·2009-03-13 07:42:25

thoda alag hai...

cos(A+((n-1)B/2))sin(nB/2)
sin(B/2)

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21

tapanmast Vora ·2009-03-13 07:42:19

ek aur dbt.

Q1 mein dont v hav ne specific answer?

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33

Abhishek Priyam ·2009-03-13 06:45:03

cosA+cos(A+B)+cos(A+2B)+....+cos(A+(n-1)B)=?

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21

tapanmast Vora ·2009-03-13 07:19:37

DUDEsss : is da ans to 2nd part = \cos ((2A + nB)/2)*\sin ((n+2)B/2)/sin(B/2)

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1

Philip Calvert ·2009-03-13 06:58:19

yes bhaiya i understand........ (sorry who ek or thread .....
wo patanahi main kuch or soch raha tha.)

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1

Akand ·2009-03-13 06:57:51

yup...i learnt it from nishant bhaiyya long back.......hehe

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62

Lokesh Verma ·2009-03-13 06:57:18

Philip this is a standard trick... try to see that you understand it :)

It just kills some of the dirty tough looking summations. :)

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1

Akand ·2009-03-13 06:55:29

y philip is it wrong???

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1

Akand ·2009-03-13 06:53:36

NEXT WALA........

Z=real(Î£e^i(A+rB))...r=0 to n-1......

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1

Akand ·2009-03-13 06:51:45

so....Sn=real(eⁱ/1-eⁱ)
so the limit becomes...
I=real(eⁱ/1-eⁱ)/n
=(cos1-cos²1-sin²1)/n(1+cos²1+2cos1+sin²1)
= (cos1-1)/n(2+2cos1)
..........then i dont kno........ wel thanks to PRIYAM...
I forgot 1/âˆž=0...hehehehehe

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33

Abhishek Priyam ·2009-03-13 06:48:47

Cool :)

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1

Akand ·2009-03-13 06:46:38

e^iÎ¸=cosÎ¸+isinÎ¸
so......
eⁱ+e²ⁱ...eⁿⁱ=cos1+cos2+...cosn+i(sin1+sin2...sinn)
real(Î£e^ir)=Sn

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