Tiggsy

Hi Melody.

My method is pretty much the same as yours, though I did arrive at a different result, and I do have a suggestion for taking it further.

\(\displaystyle \text{If } z^{7}=-1=1.\text{cis}(\pi), \text{then} \\ z = \{1.\text{cis}(\pi+2k\pi)\}^{1/7}=\text{cis}(\pi/7+2k\pi/7), \;k=0,1,2,\dots,6.\)

Now, if

 \(\displaystyle z=\cos(\theta)+i\sin(\theta), \text{ then}\\ 1-z=1-\cos(\theta)-i\sin(\theta) \text{ so}\\ \mid1-z\mid^{2}=(1-\cos(\theta))^{2}+\sin^{2}(\theta)=2-2\cos(\theta) =2(1-\cos(\theta)).\)

Then, using the trig identity \(\displaystyle \cos2A = 1-2\sin^{2}A,\)

\(\displaystyle \mid1-z\mid^{2}=4\sin^{2}(\theta/2).\)

So,

\(\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}=\frac{1}{4}\left\{\frac{1}{\sin^{2}(\pi/14)} +\frac{1}{\sin^{2}(3\pi/14)}+\dots +\frac{1}{\sin^{2}(13\pi/14)}\right\}.\)

Summing the terms inside the curly bracket gets you the number 49.

That's just too much of a coincidence, 49 being 7 squared.

I checked it out for z^3 = -1 and z^5 = -1 and the results were 9 and 25.

So, it would seem that this is a standard result, one the I don't recall seeing before, and for the moment don't see how to prove.

I'll see if I can take it further

Tiggsy

It's 1 - z, not 1 + z isn't it ? That removes the undefined term.

Just substitute for x, y, z, expand the brackets and cancel down.

You're just left with the number 3.

Put the number into polar form \(\displaystyle (1/2,\pi/2), \)  (abbreviated).

We're finding a root so the number has to be put into its multivalued form, \(\displaystyle (1/2,\pi/2+2k\pi), k=0,1,2,...\)

\(\displaystyle z=(1/2,\pi/2+2k\pi)^{1/2},k=0,1,\\=(1/\sqrt{2},\pi/4+k\pi),k=0,1,\\ =(k=0),(1/\sqrt{2},\pi/4)=(1/2+i/2),\\ \text{or }(k=1),(1/\sqrt{2},5\pi/4)=(-1/2-i/2).\)

If you're looking for the smallest positive b, then b = 18 gets you 99 squared.

Answer #1 is not correct.

Best is to differentiate wrt x, turning the integral equation into a differential equation.

\(\displaystyle \frac{d}{dx}f(x)= -\{1.[f(x)]^{2} - 0.[f(0)]^{2} +\int^{x}_{0}\frac{\partial}{\partial x}[f(t)]^{2}dt , \)

\(\displaystyle \frac{d}{dx}f(x)= - [f(x)]^{2}.\)

Now integrate that.

There will be a constant of integration, so substitute back into the original equation to determine its value.

Write (1) as     \(\displaystyle \frac{f ' (x)}{f(x)}=\frac{f ''(x)}{f '(x)}\), and integrate.

\(\displaystyle \log_{e}f(x)=\log_{e}f '(x) + \log_{e}C=\log_{e}\{Cf '(x)\}\) ,

so

\(\displaystyle f(x)=Cf '(x).\)

Now integrate a second time, (same sort of technique), to get an expression for f(x).

Suppose that the common roots are a and b and that the other, individual roots, are m and n.

We then have (Veita),

a + b + m = -5, .................(1)

a + b + n  =  5, .................(2)

ab + bm + ma = p, ...........(3)

ab + bn + na = p, .............(4).

So, 

(2) - (1):  n - m = 10, ........(5)

(4) - (3):  b(n - m) + a(n - m) = 0, .....(6).

from which

b + a = 0

and then, from (1) and (2),

m = -5 and n = 5.

\(\)

Suppose that the roots are

\(\displaystyle a-d, a, a+d,\)

then the equation will be

\(\displaystyle (x-a+d)(x-a)(x-a-d) \\ =x^{3}-3ax^{2}+(3a^{2}-d)x-a^{3}+ad^{2} =0, \)

so equating coefficients,

\(\displaystyle a = 5,\\ d = 1, \\ c = -120.\)

Extend AB to B' and AC to C'.

Call angle CBO alpha, then angle OBB' = alpha, so angle ABC = 180 - 2(alpha).

Call angle BCO beta, then angle ACB = 180 - 2(beta).

From triangle ABC, (alpha + beta) = 115 deg, and then from triangle BOC, x = 65 deg.