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**gourish**- Replies: 0

if an electric field E and magnetic field B exists in a frame of reference F1.... for another frame of reference F2 this magnetic and electric fields and B* and E* respectively where F2 is moving with a constant speed V with respect to F1.... what is the relationship between E, E* and B, B*

well... is there a function which cannot be expressed in the form of polynomials...

well i need the Taylor series only for basic applications....

thanks bob... appreciate it i will work on the problem following your advice....

osculating polynomials? can you elaborate please

**gourish**- Replies: 9

i need to find out the formula to find the sum of sines whose angles are in A.P. in as many ways as i can... can anybody help me?

P.S. i should be apply such techniques to find the sum of cosines or tangents whose angles are in A.P.

**gourish**- Replies: 25

hi i am interested in learning Taylor series but i still haven't come across integration to a good depth can i still learn it and can anybody help me learn it because i didn't get any good sites to learn it from....

P.S. i don't even know when is the Taylor series used... i am just curious to learn about it

you guys still in school?

i thought you are in college @agnishom and niharika

thanks again shivam your explanation is much better then that available online....

P.S. if i want to find the the velocity of the particle from the frame of earth then? and can you tell me what is your last diagram for? it looks like a parabolic motion of some body i don't understand it because the particle is supposed to move "on" the surface of sphere

**gourish**- Replies: 5

a smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. find the speed of the particle with respect to the sphere as a function of the angle θ it slides..

i hope i will get a quick reply i am breaking my head over this problem....

on applying arc sine on both sides becomes pie sinθ = pie/2 -(pie cosθ ) now shift the terms on the same side and divide throught by pie

you get sinθ + cosθ =1/2 and then square both sides to get 1+2sinθ cosθ =1/4 and then solving 2sinθ cosθ =-3/4

and we already know that 2sinθ cosθ = sin2θ

Agnishom,

try using the inverse trignometric functions on the equality maybe it will help....

**gourish**- Replies: 2

hi,

i find it hard to deal with complex numbers maybe because they are just "complex" i do not want to attempt any great depth into the topic but i need to know the complex numbers right from their basics to using them in argand plane and other geometries like conic sections can you guys suggest me the right book to read?

P.S. i remember things only when i derive the result on my own or read the complete explained derivation so i would like "that" kind of book..

**gourish**- Replies: 0

potential inside a hollow conductor is the same at all points and is equal to that at its surface. This is true for all hollow conductors whether spherical, cylindrical or of any arbitrary shape. can anybody give me the proof of this statement?

hey which chapter is it form... i.e. from HC VERMA?