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Complex number generalizations

  • Thread starter 314Jason
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Homework Statement



Use de Movire to find solutions for the following:

z^5 = i
z^4 = i
z^3 = i

Find generalization for z^n = x+iy,
where modulus of x+iy is 1

Explore when |x+iy| is not equal to 1


Homework Equations



[rcis(theta)] = (r^n)cis(theta*n)

r = /sqrt(y^2 + x^2)

theta = arctan(y/x)

Finding the roots of complex numbers: (r^1/n)cis(2kpi/n), where k = 0,1,...n-1


The Attempt at a Solution



I tried using moivre's theorem to find the roots of z, but I ended up with theta=arctan(1/0), so it is undefined. Yes, and now I'm stuck. Could the answer possibly be undefined/no solution?
 

Answers and Replies

  • #2
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Welcome to PF, 314Jason! :smile:

Homework Statement



Use de Movire to find solutions for the following:

z^5 = i
z^4 = i
z^3 = i

Find generalization for z^n = x+iy,
where modulus of x+iy is 1

Explore when |x+iy| is not equal to 1


Homework Equations



[rcis(theta)] = (r^n)cis(theta*n)

r = /sqrt(y^2 + x^2)

theta = arctan(y/x)

Finding the roots of complex numbers: (r^1/n)cis(2kpi/n), where k = 0,1,...n-1


The Attempt at a Solution



I tried using moivre's theorem to find the roots of z, but I ended up with theta=arctan(1/0), so it is undefined. Yes, and now I'm stuck. Could the answer possibly be undefined/no solution?

arctan(1/0) is indeed undefined, but that only means that theta is either pi/2 or -pi/2.
In your case, for the complex number "i", the corresponding theta is pi/2.

Your problem is arctan. It is not quite the right function to find theta.
What you actually need is atan2(x, y) or atan2(y, x) to find theta.
See http://en.wikipedia.org/wiki/Atan2 for details.
Or otherwise, you could write:
[tex]\theta = \textrm{Arg}(x+iy)[/tex]
See http://en.wikipedia.org/wiki/Argument_(complex_analysis)" [Broken] for details.
 
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  • #3
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Oh okay, so atan2(y, x) is the angle between the point (x,y) and the real axis. And when y>0 and x=0, theta = pi/2. YAY!

I've done the all z^3, z^4 and z^5 = i. Turns out, the conjecture for the distance between two adjacent roots (when z^n = i) is the same as when z^n = 1. Which is 2|sin(pi/n)|

Now I have to explore when |x+iy| ≠ 1. On the graph, this the modulus of z only adds to the radius (when mod(z) > 1) or subtracts to the radius (when mod(z) < 1).

In de moivre's theorem, z = r^(1/n) cis(ø + 2kpi)
r = mod(z)
ø = arg(z)
n = number of roots / exponent of z
k = 0, 1, ..., n-1
r^(1/n) is equal to the radius of the circle: x^2 + y^2 = r^(2/n)

THEN I wanted to find the conjecture interms of r, ø, n and k

z = r^(1/n)cis(ø+2kπ) <------------ de moirve
where r = mod(z) = √(x^2 + y^2)
where ø = arg(z) = atan2(x,y) = arctan (y/x)
where n = number of roots / exponents of z
where k = 0,1,...,n-1
distance formula = √(x2+y2)
Point 1(r^(1/n)cos(ø), r^(1/n)sin(ø))
Point 2(r^(1/n)cos(ø+2pi), r^(1/n)sin(ø+2pi))

I plugged the two points into the distance formula and got an answer of 0. I even put it through Wolframalpha (without the r^(1/n)) and got 0.

http://www.wolframalpha.com/input/?i=2cos^2(ø+π)sin^2(π)+-+2sin^2(ø+π)sin^2(π)

^ this is after I applied trig's sum-to-product formulas and extracted r^(1/n) since they're common.

THE DISTANCE CANNOT BE ZERO! What's wrong!? :(
 
  • #4
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I think you've got the wrong distance between 2 adjacent roots.
It's not 2|sin(pi/n)|.
It should be |cis(2pi/n)-1|.


As for the distance when the mod is not 1.
Note that the distances between successive adjacent roots will not be the same.


What you have is the first 2 roots, but you've got their argument (angle) wrong.
Apparently you forgot to divide the angle by n.
It should be:
Point 1(r^(1/n)cos(ø/n), r^(1/n)sin(ø/n))
Point 2(r^(1/n)cos((ø+2pi)/n), r^(1/n)sin((ø+2pi)/n))
 
  • #5
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I think you've got the wrong distance between 2 adjacent roots.
It's not 2|sin(pi/n)|.
It should be |cis(2pi/n)-1|.
I don't understand how :cry:

Here's how I got the generalization:

When z^n = 1, z = cis(2kπ/n), where k = 0,...,n-1

distance formula = √(x^2+y^2)
P1(1, 0)
P2(cos(2pi/n), sin(2pi/n)

d = √[ (cos(2pi/n)-1)^2 + (sin(2pi/n)-0)^2 ]
= √[ cos^2(2pi/n) - 2cos(2pi/n) + 1 + sin^2(2pi/n) ]
= √[-2cos(2pi/n)+2]
= √[-2 ( 1-2sin^2(pi/n) ]+2]v
= √[-2+4sin^2(pi/n)]+2]
= √[4sin^2(pi/n)]
= ± 2[sin(pi/n)] <---- CONJECTURE
Since distance cannot be negative,d = 2|sin(pi/n)|

When z^n = i, z =cis[(π + 4kπ)/2n], where k = 0,...,n-1

distance formula = √(x2+y2)
P1(cos(pi/2n), sin(pi/2n))
P2(cos(5pi/2n), sin(5pi/2n))

d = √[ (sin(5pi/2n)-sin(pi/2n))^2 + (cos(5pi/2n)-cos(pi/2n))^2 ]

Using trig's sum-to-product formula...

= √[ (2cos(3pi/n)sin(2pi/n))^2 + (-2sin(3pi/n)sin(2pi/n))^2 ]
= √[ 4cos^2(3pi/n)sin^2(2pi/n) + 4sin^2(3pi/n)sin^2(2pi/n) ]
= √[ 4(1-sin^2(3pi/n))(sin^2(2pi/n))+ 4sin^2(3pi/n)sin^2(2pi/n) ]
= 2 √ [ (sin^2(2pi/n)-sin^2(2pi/n)sin^2(3pi/n) + sin^2(2pi/n)sin^2(3pi/n) ]
= 2 √ [sin^2(2pi/n)]
= ± 2sin(2pi/n)
Since distance cannot be negative,d = 2|sin(2pi/n)|

As for the distance when the mod is not 1.
Note that the distances between successive adjacent roots will not be the same.
Oh you're right! As n increases, the circle formed from the polygon becomes smaller as well. So the distance is decreasing faster than when |x+iy| = 1. How do I use this information?

What you have is the first 2 roots, but you've got their argument (angle) wrong.
Apparently you forgot to divide the angle by n.
It should be:
Point 1(r^(1/n)cos(ø/n), r^(1/n)sin(ø/n))
Point 2(r^(1/n)cos((ø+2pi)/n), r^(1/n)sin((ø+2pi)/n))
Okay, here's the fix:

d^2 = (r^(1/n)cos((ø+2pi)/n)-r^(1/n)cos(ø/n)^2 + (r^(1/n)sin((ø+2pi)/n-r^(1/n)sin(ø/n))^2

_______________________________________
u + v / 2
(ø+2π + ø)/2n = 2ø+2π)/2n = (ø+π)/n

u - v / 2
(ø+2π - ø)/2n = π/n
_______________________________________


d^2 = (r^(1/n) * 2cos((ø+π)/n)sin(π/n))^2 + (-r^(1/n) * 2sin(ø+π)sin(π))^2
= 4r^(2/n) [ cos^2((ø+π)/n)sin^2(π/n) - sin^2((ø+π)/n)sin^2(π/n) ]
= 4r^(2/n) [ (1-sin^2((ø+π)/n))sin^2(π/n) - sin^2((ø+π)/n)sin^2(π/n) ]
= 4r^(2/n) [ sin^2(π/n) - sin^2((ø+π)/n)sin^2(π/n) - sin^2((ø+π)/n)sin^2(π/n) ]
= 4r^(2/n) [ sin^2(π/n) - 2sin^2((ø+π)/n)sin^2(π/n) ]
= 4r^(2/n) [ sin^2(π/n) (1 - 2sin^2((ø+π)/n) ]
= 4r^(2/n) [ sin^2(π/n) (cos^2((2ø+2π)/n) ]
= 4r^(2/n) [ sin^2(π/n) (cos^2((2ø+2π)/n) ]
d = 2r(1/n)|sin(π/n)cos((2ø+2π)/n)|

Is this right? And how do I prove if its true?? :confused:

Btw, THANK SO MUCH for all this, I would have gotten this one wrong if you hadn't pointed it out.
 
  • #6
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My bad.
Your distance of 2|sin(pi/n)| is right.


I don't understand how :cry:

Here's how I got the generalization:

When z^n = 1, z = cis(2kπ/n), where k = 0,...,n-1

distance formula = √(x^2+y^2)
P1(1, 0)
P2(cos(2pi/n), sin(2pi/n)

d = √[ (cos(2pi/n)-1)^2 + (sin(2pi/n)-0)^2 ]
There it is: d = |cis(2pi/n)-1|
But nevermind. Your distance is right.


= √[ cos^2(2pi/n) - 2cos(2pi/n) + 1 + sin^2(2pi/n) ]
= √[-2cos(2pi/n)+2]
= √[-2 ( 1-2sin^2(pi/n) ]+2]v
= √[-2+4sin^2(pi/n)]+2]
= √[4sin^2(pi/n)]
= ± 2[sin(pi/n)] <---- CONJECTURE
Since distance cannot be negative,d = 2|sin(pi/n)|

When z^n = i, z =cis[(π + 4kπ)/2n], where k = 0,...,n-1

distance formula = √(x2+y2)
P1(cos(pi/2n), sin(pi/2n))
P2(cos(5pi/2n), sin(5pi/2n))

d = √[ (sin(5pi/2n)-sin(pi/2n))^2 + (cos(5pi/2n)-cos(pi/2n))^2 ]

Using trig's sum-to-product formula...

= √[ (2cos(3pi/n)sin(2pi/n))^2 + (-2sin(3pi/n)sin(2pi/n))^2 ]
= √[ 4cos^2(3pi/n)sin^2(2pi/n) + 4sin^2(3pi/n)sin^2(2pi/n) ]
= √[ 4(1-sin^2(3pi/n))(sin^2(2pi/n))+ 4sin^2(3pi/n)sin^2(2pi/n) ]
= 2 √ [ (sin^2(2pi/n)-sin^2(2pi/n)sin^2(3pi/n) + sin^2(2pi/n)sin^2(3pi/n) ]
= 2 √ [sin^2(2pi/n)]
= ± 2sin(2pi/n)
Since distance cannot be negative,d = 2|sin(2pi/n)|
Good!
And nice work!

Btw, there's also a geometrical interpretation and proof.
If you draw the roots in the complex plane, it should become obvious that the distance between the roots is the same for z^n=1 and for z^n=i.


Oh you're right! As n increases, the circle formed from the polygon becomes smaller as well. So the distance is decreasing faster than when |x+iy| = 1. How do I use this information?
Never mind.
The distance between the roots is equal after all.

FYI, the distance between root k and (k+1) is:
|r^(1/n)\cis((ø+2(k+1) pi)/n) - r^(1/n)cis((ø+2k pi)/n)|
But if you work it out, you'll see it's the same distance.


Okay, here's the fix:

d^2 = (r^(1/n)cos((ø+2pi)/n)-r^(1/n)cos(ø/n)^2 + (r^(1/n)sin((ø+2pi)/n-r^(1/n)sin(ø/n))^2

_______________________________________
u + v / 2
(ø+2π + ø)/2n = 2ø+2π)/2n = (ø+π)/n

u - v / 2
(ø+2π - ø)/2n = π/n
_______________________________________


d^2 = (r^(1/n) * 2cos((ø+π)/n)sin(π/n))^2 + (-r^(1/n) * 2sin(ø+π)sin(π))^2
I'm afraid you have the wrong trig identities here.
You should use:

cos u - cos v = -2sin[(u+v)/2] sin[(u-v)/2]
sin u - sin v = 2cos[(u+v)/2] sin[(u-v)/2]


= 4r^(2/n) [ cos^2((ø+π)/n)sin^2(π/n) - sin^2((ø+π)/n)sin^2(π/n) ]
= 4r^(2/n) [ (1-sin^2((ø+π)/n))sin^2(π/n) - sin^2((ø+π)/n)sin^2(π/n) ]
= 4r^(2/n) [ sin^2(π/n) - sin^2((ø+π)/n)sin^2(π/n) - sin^2((ø+π)/n)sin^2(π/n) ]
= 4r^(2/n) [ sin^2(π/n) - 2sin^2((ø+π)/n)sin^2(π/n) ]
= 4r^(2/n) [ sin^2(π/n) (1 - 2sin^2((ø+π)/n) ]
= 4r^(2/n) [ sin^2(π/n) (cos^2((2ø+2π)/n) ]
= 4r^(2/n) [ sin^2(π/n) (cos^2((2ø+2π)/n) ]
d = 2r(1/n)|sin(π/n)cos((2ø+2π)/n)|

Is this right? And how do I prove if its true?? :confused:
What you have is a proof.
What you need is a way to verify independently there are no calculation mistakes.

Geometrically (that is, if you draw it), you can tell that the distance must be:
2r^(1/n)|sin(π/n)|


Btw, THANK SO MUCH for all this, I would have gotten this one wrong if you hadn't pointed it out.
No problem! :smile:
 
  • #7
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I don't understand how :cry:

When z^n = i, z =cis[(π + 4kπ)/2n], where k = 0,...,n-1

distance formula = √(x2+y2)
P1(cos(pi/2n), sin(pi/2n))
P2(cos(5pi/2n), sin(5pi/2n))

d = √[ (sin(5pi/2n)-sin(pi/2n))^2 + (cos(5pi/2n)-cos(pi/2n))^2 ]

Using trig's sum-to-product formula...

= √[ (2cos(3pi/n)sin(2pi/n))^2 + (-2sin(3pi/n)sin(2pi/n))^2 ]
= √[ 4cos^2(3pi/n)sin^2(2pi/n) + 4sin^2(3pi/n)sin^2(2pi/n) ]
= √[ 4(1-sin^2(3pi/n))(sin^2(2pi/n))+ 4sin^2(3pi/n)sin^2(2pi/n) ]
= 2 √ [ (sin^2(2pi/n)-sin^2(2pi/n)sin^2(3pi/n) + sin^2(2pi/n)sin^2(3pi/n) ]
= 2 √ [sin^2(2pi/n)]
= ± 2sin(2pi/n)
Since distance cannot be negative,d = 2|sin([STRIKE]2[/STRIKE]pi/n)|
I noticed a mistake. For z^n = i, I got 2|sin(2pi/n)| but for z^n = 1, I got 2|sin(pi/n)|. I'm sure the later is correct. I've tried and tried again, but I just can't find the mistake. And I KNOW it must be the same length in either casses.


Also, for z^n = x+iy, when |x+iy| ≠ 1, I got d = 2r^(1/n)|sin(pi/n)|. But, I don't know how to show my generalization on a graph. I don't know what to do with the "r" value :( ...Is there any other way I can prove this is true? I need to explore a variety of different methods to make sure.
 
  • #8
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I noticed a mistake. For z^n = i, I got 2|sin(2pi/n)| but for z^n = 1, I got 2|sin(pi/n)|. I'm sure the later is correct. I've tried and tried again, but I just can't find the mistake. And I KNOW it must be the same length in either casses.

d = √[ (sin(5pi/2n)-sin(pi/2n))^2 + (cos(5pi/2n)-cos(pi/2n))^2 ]

Using trig's sum-to-product formula...

= √[ (2cos(3pi/n)sin(2pi/n))^2 + (-2sin(3pi/n)sin(2pi/n))^2 ]
The mistake is right here.
You forgot to divide (u+v) and (u-v) by 2.



Also, for z^n = x+iy, when |x+iy| ≠ 1, I got d = 2r^(1/n)|sin(pi/n)|. But, I don't know how to show my generalization on a graph. I don't know what to do with the "r" value :( ...Is there any other way I can prove this is true? I need to explore a variety of different methods to make sure.
If you draw a graph for z^n=1, you find a regular polygon.
For z^n=i, you find the same polygon, only rotated by 90 degrees.

For z^n=x+iy, again you find the same polygon, but rotated over Arg(x+iy) and scaled by r^(1/n).






And if you're interested, here's another way to show your result.
It depends on Euler's formula, which is the generalization of De Moivre's theorem.

Euler's formula is:
[tex]e^{i \phi} = \cos \phi + i \sin \phi[/tex]
Which can be rewritten as:
[tex]\cos \phi = {1 \over 2}(e^{i \phi} + e^{i \phi})[/tex][tex]
\sin \phi = {1 \over 2i}(e^{i \phi} - e^{i \phi})[/tex]


Starting with your problem, you have:
z^n = i
The solution as you basically already know is:
[tex]z = (e^{i ({\pi \over 2} + 2k\pi)})^{1 \over n} = e^{i ({\pi \over 2} + 2k\pi) {1 \over n}}[/tex]

The distance between the first and second root is:

[tex]d = |e^{i ({\pi \over 2} + 2\pi) {1 \over n}} - e^{i ({\pi \over 2}) {1 \over n}}|[/tex]
[tex]d = |e^{i ({\pi \over 2} + \pi) {1 \over n}}(e^{i {\pi \over n}} - e^{-i {\pi \over n}})|[/tex]
[tex]d = 1 \cdot |(e^{i {\pi \over n}} - e^{-i {\pi \over n}})|[/tex]
[tex]d = |2i \sin{\pi \over n}|[/tex]
[tex]d = 2 |\sin{\pi \over n}|[/tex]
 
  • #9
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If you draw a graph for z^n=1, you find a regular polygon.
For z^n=i, you find the same polygon, only rotated by 90 degrees.

For z^n=x+iy, again you find the same polygon, but rotated over Arg(x+iy) and scaled by r^(1/n).
Oooo. Is there a way to prove this? I drew the graphs and they same to be rotating all over the place.

If I were to make this the "original polygon" (z^n = 1 with ø = 0)
Then this (z^n = i with ø = π/2) would rotate it 90/n degrees anti-clockwise.

E.g. When z^3 = x+iy, the "original polygon" would rotate 90/3 = 30 degrees anti-clockwise.

I feel like I need to adress this pattern ("rotated over Arg(x+iy) and scaled by r^(1/n)") but I don't know how I can show it or prove it mathematically! :frown:


And if you're interested, here's another way to show your result.
It depends on Euler's formula, which is the generalization of De Moivre's theorem.

Euler's formula is:
[tex]e^{i \phi} = \cos \phi + i \sin \phi[/tex]
Which can be rewritten as:
[tex]\cos \phi = {1 \over 2}(e^{i \phi} + e^{i \phi})[/tex][tex]
\sin \phi = {1 \over 2i}(e^{i \phi} - e^{i \phi})[/tex]


Starting with your problem, you have:
z^n = i
The solution as you basically already know is:
[tex]z = (e^{i ({\pi \over 2} + 2k\pi)})^{1 \over n} = e^{i ({\pi \over 2} + 2k\pi) {1 \over n}}[/tex]

The distance between the first and second root is:

[tex]d = |e^{i ({\pi \over 2} + 2\pi) {1 \over n}} - e^{i ({\pi \over 2}) {1 \over n}}|[/tex]
[tex]d = |e^{i ({\pi \over 2} + \pi) {1 \over n}}(e^{i {\pi \over n}} - e^{-i {\pi \over n}})|[/tex]
[tex]d = 1 \cdot |(e^{i {\pi \over n}} - e^{-i {\pi \over n}})|[/tex]
[tex]d = |2i \sin{\pi \over n}|[/tex]
[tex]d = 2 |\sin{\pi \over n}|[/tex]
Wow this is great! Thanks!

But I don't quite understand what you did here:

The distance between the first and second root is:
[tex]d = |e^{i ({\pi \over 2} + 2\pi) {1 \over n}} - e^{i ({\pi \over 2}) {1 \over n}}|[/tex]

Is there a rule to find the distance between the two points with Euler? I haven't learned Euler.

Also, I happen to study Mathematical Induction in class, and that's pretty much the only proving I know how to do. But it doesn't seem to work here! Since you must have a lot more experience with this, from the methods listed in the link below, what do you think I can use?

http://en.wikipedia.org/wiki/Mathematical_proof#Direct_proof
 
  • #10
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Oooo. Is there a way to prove this? I drew the graphs and they same to be rotating all over the place.

If I were to make this the "original polygon" (z^n = 1 with ø = 0)
Then this (z^n = i with ø = π/2) would rotate it 90/n degrees anti-clockwise.

E.g. When z^3 = x+iy, the "original polygon" would rotate 90/3 = 30 degrees anti-clockwise.

I feel like I need to adress this pattern ("rotated over Arg(x+iy) and scaled by r^(1/n)") but I don't know how I can show it or prove it mathematically! :frown:
The "original" polygon has its points on the unit circle.
The angle at the origin between successive points is 2pi/n.
Therefore the polygon is regular, meaning all its edges have the same length.

When x+iy has an angle ø, the first root is at ø/n, and on the circle with radius r^(1/n).
Each successive root is on the same circle and has an angle 2pi/n further along.
Therefore the polygon is regular, meaning all its edges have the same length.

I believe this would suffice as a proof.



Wow this is great! Thanks!

But I don't quite understand what you did here:

The distance between the first and second root is:
[tex]d = |e^{i ({\pi \over 2} + 2\pi) {1 \over n}} - e^{i ({\pi \over 2}) {1 \over n}}|[/tex]

Is there a rule to find the distance between the two points with Euler? I haven't learned Euler.
A complex number represents a point in the complex plane, meaning it has an x-coordinate (the real part) and an y-coordinate (the imaginary part).
The distance between 2 of these points is simply the absolute value of the difference of the 2 complex numbers.

The exponential expression is simply another way to write a complex number, using its distance to the origin and the angle with the real axis.
The exponential expression is also another way to express De Moivre's theorem.



Also, I happen to study Mathematical Induction in class, and that's pretty much the only proving I know how to do. But it doesn't seem to work here! Since you must have a lot more experience with this, from the methods listed in the link below, what do you think I can use?

http://en.wikipedia.org/wiki/Mathematical_proof#Direct_proof
In practice I basically use only 3 proofs: forward proof, proof by contradiction, and full induction.
Actually I rarely see full induction except in a math context that is dedicated to it.
I guess I do use the others too, but I'm not really aware that I'm doing that.
I haven't studied how you would call the specific ways of proving things.
 
Last edited:
  • #11
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Hi! Sorry for the late reply, I've been busy completing a ton of assignments.

I talked to my teacher about proving the conjecture and he said he expects me to use mathematical induction, and that its the "best, valid proof".

I just have no idea how I could apply that here. According to this: http://people.richland.edu/james/lecture/m116/sequences/induction.html

I need to make the left hand side equal the right hand side. But my conjecture is (the distance between 2 adjacent roots of z^n=1) = 2|sin(pi/2)|

From the examples I found online, its generally listing out the values (in this case the distances) and seeing the pattern, then forming a conjecture from the values observed. And proving it by making the expressions equal.

For example, I have to find the distances of n=3, n=4, n=5, etc.

n distance
3 √3
4 √2
5 1.755
6 1
...

And then see from the values above that its 2|sin(pi/2)|. THEN I'll prove it by simplifying the distance formula. But the thing is, I can't form a conjecture just by looking at those values!
 
  • #12
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Uum... didn't you already proof this for the general case with Direct proof?

I'm not sure what else there is to do.
I do not see how to proof this by full induction.

Listing specific cases for a sequence of n-values makes the conjecture likely.
I guess this is also a form of induction, but it is no proof.
 

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