Evaluating A Septic Algebraic Expression

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Solving a Heptic Equation Without Using the Heptic Formula:

LECTURE:
PLAYLISTS 🎵 :

[morteza]

Square both sides of the given equation;
(z+1/z)=sqrt(2)
Z²+1/z²=0 so z⁴=-1 and z⁸=1
And we know
Z⁷+1/z⁷=z⁸/z+z/z⁸
Placement z⁸=1
1/z+z=sqrt(2)

[alextang]

2nd method is the general way to solve this kind of expression especaily for the high degree for example z^99+1/z^99. Very nice!!! 😋😋😋😋😋😋

[pauselab]

There is a general way to find all solutions for expressions like this. Multiply the expression to remove the division and turn it into a polynomial. Then apply quadratic equation and finally use circle of unity to find all solutions

[cblpu]

This can also be solved using newtons identities. We know z and 1/z are roots of the quadratic equation x^2-√2x+1=0 the recurrence will then be A_n-√2A_(n-1) + A_(n-2) =0 where A_n = z^n + 1/z^n
We need A_7 and already have the value of A_1 = √2 and A_0=2. This gives us A_2=0, A_3=-√2, A_4=-2, A_5=-2√2+2 and A_7 =√2

[adandap]

A faster variant on the first method is to note that squaring the given equation gives z^2 + 1/z^2 =0, or z^4 = -1. Then z^7 = - z^3 and 1/z^7 = -1/z^3, so we only need the cube of (z+1/z)

[sohampine]

I think after this equation I need an antiseptic.. (joke)

[scottleung]

Nice - I knew I was doing it wrong when I got an answer that wasn't elegant enough.

[damirdukic]

The general formulas for the first method go like this:
If "n" is a positive integer, then:
z^(2n) + 1/z^(2n) = (z^n + 1/z^n)^2 - 2
z^(2n+1) + 1/z^(2n+1) = (z^(n+1) + 1/z^(n+1)) (z^n + 1/z^n) - (z + 1/z)
These can be applied to any value of "z + 1/z".

[MrLidless]

7:10 You could also have sin a = 0 and should investigate that branch. You don’t yet know z is complex, other than the variable is “z”.

[alejrandom]

I just used the quadratic formula and recognized (sqrt2+isqrt2)/2 is exp(+-i*pi/4) lol after that it's just powers of unitary number

[DrQuatsch]

I did this the long way, for some reason. I squared both sides to get z^2 + 1/z^2 = 0. Then I took the third power of that equation, getting that z^6 + 1/z^6 = 0. I then multiplied z + 1/z with that to get the result that z^7 + 1/z^7 = -(z^5 + 1/z^5). So I had to back engineer z^5 + 1/z^5. So I calculated (z^2 + 1/z^2)^2, which gave me z^4 + 1/z^4 = -2. I multiplied that with z + 1/z to get that z^5 + 1/z^5 + z^3 + 1/z^3 = -2sqrt(2). So now I needed z^3 + 1/z^3 as well. Which I got by multiplying z + 1/z with z^2 + 1/z^2. That gave me z^3 + 1/z^3 = -sqrt(2). Which I could use to find that z^5 + 1/z^5 = -sqrt(2), which resulted in z^7 + 1/z^7 = sqrt(2). So yeah, in the end I had a whole list of all of them from exponent 1 to 7.

[sohampine]

How did 1/cos(x)+isin(x) become cos(-x) + isin(-x)??

[VinTheFox]

I solved for z first using the quadratic formula and plotted it. Since its radius is 1, z^7 is just (+/-)45° * 7 which is still (+/-)45°. z^7=1/z and vise versa. Both sums are the same, root 2

[abdu]

ന മ്മ ൾ മലയാളിയാകുന്നു അയാളുടെ ഇംഗ്ലീഷ് ക്ലിയർ ആ കുന്തില്ല.