Question

2 easy questions
1) what does this notation mean ??
\(\Large (\dfrac{\partial u}{\partial x})_y \)

2) will ask later in comments.

Answer 1

i have
u=lx+my,v=mx-ly

and i need to prove
\(\Large (\dfrac{\partial u}{\partial x})_y \times \Large (\dfrac{\partial x}{\partial u})_v = \dfrac{l^2 }{l^2+m^2}\)

proving part will be easier once i know what that notation means!

Answer 2

also i already know
\(\Large (\dfrac{\partial u}{\partial x})\)
means differentiating 'u' partially w.r.t x

Answer 3

would it be
\(\Large \dfrac{\partial }{\partial y} \Large (\dfrac{\partial u}{\partial x})\)
?

Answer 4

And my 2nd Question was :
Solve : \(x^6-x^5+x^4-x^3+x^2-x+1 =0 \)
I think i need to use DeMovire's Theorem, just need a start.
(or any other method that does not involve use of calculator)

Answer 5

\[\large \left(\dfrac{\partial x}{\partial u}\right)_v \]

paritual u keeping v constant ( y depends on v)

Answer 6

\(u=lx+my, \\ v=mx-ly\)

from 1st equation :
\[\large u = lx + m\dfrac{(mx-v)}{l} \]

\[\large u = x(l + m^2/l) - mv/l \]

\[\large 1 = (l + m^2/l) \left(\dfrac{\partial x}{\partial u}\right)_v - 0 \]

\[\large \dfrac{l}{l^2+m^2} = \left(\dfrac{\partial x}{\partial u}\right)_v \]

Answer 7

u,v,x,y are not independent, they depend on each other..
the subsript(v) makes it explicit what variable we're keeping constant

Answer 8

nice!
got it :)
how about 2nd question ?

Answer 9

@SithsAndGiggles

Answer 10

the second question looks hard...

Answer 11

a similar example in notes was
\(x^7+64x^4+64x^3+64^2=0\)
but that was easily factorable...this isn't...

Answer 12

Notice that
\[x^6-x^5+x^4-x^3+x^2-x+1=\sum_{n=0}^6(-x)^n\]
There's a formula for finite geometric series, but it escapes me at the moment...

Answer 13

I think it's something like
\[\sum_{n=0}^6(-x)^n=\frac{1-(-x)^{n+1}}{1-(-x)}\]

Answer 14

So when setting this equal to zero, it's the same as finding the 7th roots of \(x^7=-1\), excluding \(x=-1\) itself because that would make the expression undefined.

Answer 15

brilliant xD

Answer 16

so i just prove
( x^6-x^5+x^4-x^3+x^2-x+1)(1+x) = x^7+1

Answer 17

which would be easy
and then exclude x=-1

Answer 18

Yeah pretty much :)

Answer 19

nice!
any other method comes to mind ?

Answer 20

Newton's method? or any approximation technique.

Answer 21

Though I doubt that work due to the complex answers.

Answer 22

well then, would use your method...thanks! :)

Answer 23

yw!