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## #1 Re: Help Me ! » how to find the sum of sines whose angles are in A.P. » 2014-05-28 20:34:15

i am still stuck with it i don't see any relevance other then the sin {a+(n-1)d/2} that comes in the sum of sines i am not getting to the proof....

## #2 Re: Help Me ! » Taylor series » 2014-05-28 20:29:47

well i am stuck with my post on sum of sines with angles in A.P. i still didn't get to prove it can you help me on that?

## #3 Re: Help Me ! » Taylor series » 2014-05-28 20:20:35

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

## #4 Re: Help Me ! » Taylor series » 2014-05-28 20:11:32

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

## #5 Re: Help Me ! » how to find the sum of sines whose angles are in A.P. » 2014-05-28 02:13:18

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

## #6 Re: Help Me ! » Taylor series » 2014-05-28 02:11:21

osculating polynomials? can you elaborate please

## #7 Re: Help Me ! » how to find the sum of sines whose angles are in A.P. » 2014-05-27 20:28:54

we have this series sin a + sin a+d + sin a+2d + sin a+3d+.......+sin  a+(n-1)d = [sin (nd/2) sin{a+(n-1)d/2) ] over sin (d/2)... this is given in my book... but i don't trust them i want to find out the answer to this series on my own... and the angles need not necessarily add up to 180 degree.... bob can you give any advice on this? a quick one if you are in a hurry

## #9 Help Me ! » how to find the sum of sines whose angles are in A.P. » 2014-05-27 20:15:14

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.

## #10 Re: Help Me ! » Taylor series » 2014-05-27 19:59:45

can i apply it to find the sum of sines where the angles are in Arithmetic progression? i mean i don't even know how to find the sum of such series by any method.... so the Taylor series can be helpful (if it can be used..)

## #11 Help Me ! » Taylor series » 2014-05-27 18:52:51

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

## #12 Re: Science HQ » volume content » 2014-05-27 18:46:34

you guys still in school?
i thought you are in college @agnishom and niharika

## #13 Re: Science HQ » work and energy problem » 2014-05-27 18:43:18

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

## #14 Re: Science HQ » work and energy problem » 2014-05-24 03:18:18

i found the solutions online but they are hardly clear it' all getting straight to answer i need to know how to get to the solution... people on this forum usually explain why a certain step is taken to find the answer so i wanted an answer on the forum but i can wait for the answer... thanks Shivam...

## #15 Science HQ » work and energy problem » 2014-05-24 02:44:10

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....

## #16 Re: Help Me ! » (Less than) 21 Trig Questions from Amrita » 2014-05-17 04:56:49

the equality sin(pie sinθ ) = cos(pie cosθ )
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

## #17 Re: Help Me ! » (Less than) 21 Trig Questions from Amrita » 2014-05-17 04:52:11

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

## #18 Re: Help Me ! » (Less than) 21 Trig Questions from Amrita » 2014-05-17 04:08:57

shouldn't sin2θ= -3/4 cause that's quite easy to solve by using inverse trignometric functions (i assumed the angles to be in the principal branches) then what's the problem in finding it?

## #19 Re: Help Me ! » a good book can solve my problems » 2014-05-10 18:32:51

well i used the mac graw hill's book but it's just having bunch of equations and no proof's... plus it's not helping me visualize a geometrical interpretation of problems on complex numbers.... any suggestions on how to deal with this?

## #20 Help Me ! » a good book can solve my problems » 2014-05-10 05:13:48

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..

## #21 Science HQ » electrostatics... » 2014-05-10 05:09:53

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?

## #22 Re: Science HQ » vector » 2014-05-04 03:05:20

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

## #23 Re: Science HQ » mirrors... » 2014-04-22 02:21:52

i didn't understand your question.... the "What transformation takes the original shape onto the final image?" part is bit confusing what do i need to find here...?

## #24 Re: Science HQ » mirrors #2 » 2014-04-22 02:20:24

but it can take a lot of time to find the exact number of images being formed isn't there a better way then this that i can use? other then the formula that i mentioned because it doesn't work when the mirrors are the 3 sides of an equilateral triangle....

## #25 Re: Science HQ » root mean square value » 2014-04-21 06:43:42

well  i=3 sin Δ t + 4 cos Δ t differentiate this with respect to time (after all it's a wave which is variable in space with time) doing this i get diff(i) w.r.t.time  = 3cosΔt-4sinΔt

for maxima diff(i) w.r.t.time=0 so 3cosΔt-4sinΔt=0
which gives tanΔt=3/4 now we can put it back into the original equation to find the answer to my question which is what i did