Math Is Fun Forum

"A polynomial p(x) of degree two or less which takes values y0 y1 and y2 at three distinct values x0 x1 x2 respectively is given by {the la grange's interpolation polynomial}" use this data to answer
1.) A polynomial of degree 2 which takes values y0 , y0 , y1 at points x0 x0+t x1 t is not equal to 0 is given by?
2.)  A polynomial of degree 2 and "a" not equal to 0 or 1 then p(x) is....
3.) in last question if a tends to 0 then p(x) is given by

i don't know what this " la grange's interpolation polynomial is" can somebody tell me about it and explain a way to answer these question. i did not have this in my class so i find it hard to understand

"A tower has the following shape: a right circular truncated right circular cone (one with radii 2R(the lower base) R(the upper base), and height R bears a right circular cylinder whose radius is R and height is 2R. Finally a semisphere of radius R is mounted on the cylinder. suppose the cross sectional area S of the tower is given by f(x) where x is the distance of the cross section from the lower base of the cone"
define the function.
i know that the function is (pie)*R^2 for the interval R≤ x≤ 3R
but what about the other intervals and yeah the domain of the function is is from 0 to 4R. i am facing problems with relating the area of the cross section with the radii and x of both the cone and semisphere help me please!

The work done on a particle of mass m by force=k*[x/(x^2+y^2)^3/2 i(cap)+y/(x^2+y^2)^3/2 j(cap)]
k is a constant and the particle is taken from (a,0) to (0,a) along a circular path of radius a about the origin in x-y plane is

well i let x^2+y^2=r^2 and then found the force as k[x/r^3 i(cap)+y/r^2 j(cap)]
the problem is i found in my course sheet that take x*dx+y*dy=r*dr now that is the MATH part and i am confused
P.S. bob here is that problem which i was discussing with u on my other post and any math+physicist please help me out

hi i am back with another question on equations Let p(x) = x8 − x6 + x5 + x4 + x2 − x + 1=0 then p(x) has roots:
A)atmost 4 positive real roots
B)atmost 2 negative real roots
C)maximum of 6 real roots
D)minimum of 2 complex roots
answer is all of the above as explained below but i have no idea how it is justified can anybody please explain me what is going on below
or just tell me how can the answer be found

"Let p(x) = x8 − x6 + x5 + x4 + x2 − x + 1 = 0
Here, the coefficients in p(x) have 4 changes in sign so p(x) = 0 have atmost 4 positive real roots.
Now, p(−x) = x8 − x6 − x5 + x4 + x2 − x + 1 = 0
Since the coefficients in p (−x) has 2 changes in sign, so p(−x) = 0 have atmost 2 negative real roots.
Thus, p(x) = 0 can have a maximum of (2 + 4) = 6 real roots.
The coefficients in p(x) = 0 are real, therefore, imaginary roots occurs in conjugate pair.
Since p(x) = 0 can have a maximum of 6 real roots, so the minimum number of complex roots of p(x) = 0 is 2."

i just accepted it as the truth but now i am curious... how did we get that rule i just need a simple proof...