Construct a line segment of length n-th root of a given length b

Discussion in 'Physics & Math' started by DNA100, Sep 2, 2015.

  1. DNA100 Registered Senior Member

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    A line segment of length b is given, how can we construct a line of length n-th root of b?
    I know how to construct the length square root of b, but...
     
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  3. Dinosaur Rational Skeptic Valued Senior Member

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    4th root is easy.
    Construct a square with the line as a diagonal.
    Construct a square with the side of the previous square as the diagonal.
    It seems that the above can be the basis for 8th, 16th, et cetera roots.

    Cube root could be constructed by constructing a cube with the line as the diagonal, but I do not think plane geometry allows such a construction.

    As for other roots, my guess is that it cannot be done.
     
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  5. krash661 [MK6] transitioning scifi to reality Valued Senior Member

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    this maybe off topic but IMO, it will help. elaborate this meaning and possibly give a value.
    IN---> IN[subscript]f---> f(#) ---> OUT[subscript]f = IN[subscript]f-1 --->f^-1(#)--->OUT[subscript]f-1=IN[subscript]f--->out
     
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  7. rpenner Fully Wired Valued Senior Member

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    4,833
    Incorrect.

    The ratio between the edge of a cube and it's interior corner-to-corner diagonal is \(1 \!: \!\sqrt{3}\) . You can see that because for any three sides that meet at a point, they are all of length x and at 90° to each other. So any two form a square (with equal diagonals) and the diagonal from the point between any two is a diagonal of length \(\sqrt{x^2 + x^2} = \sqrt{2 x^2} = x \sqrt{2}\) which is at right angles to the third side of length x. The theorem of Pythagoras gives the diagonal of the of the cube as \(\sqrt{ x^2 + \left( x \sqrt{2} \right)^2} = \sqrt{x^2 + 2x^2} = \sqrt{3 x^2} = x \sqrt{3}.\) Therefore the ratio between side and interior diagonal for any cube is \(x\!:\!x \sqrt{3}\, :: \, 1 \!: \! \sqrt{3}\). QED.

    An ancient problem, not solvable by the methods of geometric construction, is to construct the side of a cube with a volume twice that of a given cube. This construction is the construction of the ratio \(1 \!: \!\sqrt[3]{2}\) is impossible with compass and unmarked straightedge.

    Therefore the only roots accesible with compass and unmarked straightedge are those in the form \(2^n\); 2, 4, 8, 16, etc.

    https://en.wikipedia.org/wiki/Doubling_the_cube
     
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  8. rpenner Fully Wired Valued Senior Member

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    4,833
    Also,
    That does not result in a 4th root. That results in a new segment half the size of the original.
    The correct procedure to construct a square root requires two segments, a segment of length 1 and a segment of length x.
    Construct a segment of length 1+x marked so that AB has length 1 and BC had length x. Find the midpoint of the segment D. Construct a semicircle from A to C centered at D with radius AD. Construct at B a line perpendicular to AC and extend through the semicircle at point E. Then ABE and CBE are similar right triangles. Therefore AB:BE :: BE:CB and BE has length equal to the geometric mean of AB and BC. Therefore BE has length \(\sqrt{x}\). QED.

    Numeric example for x = 9/4. Then AC = x + 1 = 13/4. Then AD = DC = DE= \(\frac{x + 1}{2}\) = 13/8. Then BD = \(\frac{ | x - 1 | }{2}\) = 5/8. Since DBE is also a right triangle then it follows that BE = \(\sqrt{ DE^2 - BD^2} = \sqrt{ \left( \frac{x + 1}{2} \right)^2 - \left( \frac{x - 1}{2} \right)^2 } = \sqrt{ \frac{( x^2 + 2 x + 1) - ( x^2 - 2x + 1) }{4} } = \sqrt{\frac{4x}{4}} = \sqrt{x} = \frac{3}{2}.\)
    With the same length 1, this process can be repeated, getting \(\sqrt[4]{x}, \sqrt[8]{x}, \sqrt[16]{x}, \dots\)
     
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  9. DNA100 Registered Senior Member

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    259
    Thanks.
    So, it can't be constructed using compass and ruler(straight-edge).
    However, can it be constructed in any other way?
    Because what I really want is a practical and intuitive way to understand that the number \(\sqrt[3]{2}\) or more generally \(\sqrt[n]{2}\) actually exists in real world and not just a hypothesis.
    That is to say that such a distance exists.

    Even more generally what I want to understand is that a distance \(b^a\) where b>=0 and real
    and a is real is a an actually a real distance than can occur in our universe.
     
    Last edited: Sep 3, 2015
  10. rpenner Fully Wired Valued Senior Member

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    4,833
    That's why I linked the Wikipedia page.
    Math isn't about the real world. We model space as a continuous manifold because we are ignorant if it is continuous or just really fine-grained beyond our ability to measure.
    The same is true about continuous actions like rotation. If something turns a full circle in 1 second, we assume it turns 1° in 1/360 second and 0.00000036° in 1 ns, but there is no practical way to measure such a small angle. Unlike geometry every physical measurement has associated error, every manufactured object is only correct within certain tolerances.

    So there is no way to prove space is a continuous manifold and that possible distances correspond exactly to the mathematical real numbers. That's just an assumption made in ignorance, for the convenience of physical theories. Physics isn't about the nature of reality -- Physics is about the behavior of phenomena in reality. Because that's all we can do is poke reality and see how it jumps -- that type of inquiry will never tell you what reality is, only how it behaves. We have very good mathematical models of how it will behave in all sorts of circumstances, but the map is not the territory.
     
  11. DNA100 Registered Senior Member

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    259
    So \(\sqrt[3]{2}\) is possible with neusis construction.
    What about \(\sqrt[n]{2}\) or \(\sqrt[n]{a}\)?
    Are these possible to constructy in any way?

    OK, so space can be discrete.
    But at least in the euclidean system, given that we can construct lengths a X b or a / b, it means we can construct any decimal number distance to any desired degree of accuracy.


    I want to ask an even more fundamental question
    what does \(b^a\) even mean when a and b are positive real numbers?

    And why is that definition intuitively meaningful?
    And what is the minimum set of assumptions needed to prove that a unique such (real) number exists.

    Can ordered field axioms prove it?
     
    Last edited: Sep 5, 2015
  12. rpenner Fully Wired Valued Senior Member

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    4,833
    I'm not saying it is -- just that we are ignorant if it really is fundamentally like a mathematical manifold at the tiniest length scales.


    This construction is the construction of the ratio 1:∛2 is impossible with compass and unmarked straightedge. Therefore not every nth root is constructible. It follows that not every distance corresponding to a real number is constructible using the rules of Euclidean plane constructions.

    However the real numbers, by construction, have exactly the property that every bounded subset has a least upper bound, the supremum. The supremum of all rational numbers whose cube is less than 2 includes an infinite subset of approximates, { 1, 5/4, 34/27, 286/227, 635/504, … }, and this set is obviously bounded from above (by say 3), so it has a supremum in the real numbers and that real number has the name ∛2. Similarly, every (primary) nth root of a non-negative real number exists in the non-negative reals. And when b is positive and a is any real number, b^a determines a specific real number. How do we know that number?

    So from the field axioms of complex numbers, we can define when \(z \ne 0\), \(z^n\) as a finite product of z or it's reciprocal. From this we get the laws of exponents:
    the most important of which is \(z^{n+m} = z^n \, z^m\). Using the concept of the supremum, we can establish the behavior of the limit of an infinite sum (when it exists) and define \(\textrm{exp}(z) = \sum_{k=0}^{\infty} \frac{z^k}{k!}\). Then we can prove that this exponential function is one-to-one from the reals onto the positive reals. We can also prove it is an isomorphism between the operation of addition on real numbers and multiplication on complex numbers. And that it has (over the positive reals) a well-defined inverse function, log. Skipping over the notion of the log of a complex or negative real number, we can then define a complex exponentiation operation:

    \( w^z = \left\{ { \begin{array}{lll} 1 & \quad \quad \quad & \textrm{if} \; w = 0 \; \textrm{and} \; z = 0 \\ 0 & & \textrm{if} \; w = 0 \; \textrm{and} \; \Re(z) > 0 \\ \textrm{exp}( z \, \textrm{log} (w) ) & & \textrm{if} \; w \ne 0 \end{array} } \right.\)

    \(0^i\) is undefined because then \((0^i)^i\) would have to be a complex number, but \((0^i)^i = 0^{i\times i} = 0^{-1} = \frac{1}{0}\) is also undefined.

    http://us.metamath.org/mpegif/efeul.html ⊢ (A∈ℂ → (exp ‘A) = ((exp ‘(ℜ ‘A)) · ((cos ‘(ℑ ‘A)) + (i · (sin ‘(ℑ ‘A)))))) Eulerian representation of the complex exponential. This explains how the complex exponential function relates to the real exponential function and the cosine and sine functions (in radians!).
    http://us.metamath.org/mpegif/efzval.html ⊢ (N∈ℤ → (exp ‘N) = (e↑N)) The exponential function at integer values is precisely equal to e raised to that integer power.
    http://us.metamath.org/mpegif/reeff1o.html ⊢ (exp ↾ℝ):ℝ–1-1-onto→ℝ+ The exponential function maps real arguments one-to-one onto positive reals. Therefore it has an inverse on the positive reals.
    http://us.metamath.org/mpegif/cxpexp.html ⊢ ((A∈ℂ ∧ B∈ℕ0) → (AcB) = (AB)) The complex exponential definition gives the same answer as the definition based on finite multiplication.
    http://us.metamath.org/mpegif/cxpadd.html ⊢ (((A∈ℂ ∧ A ≠ 0) ∧ B∈ℂ ∧ C∈ℂ) → (Ac(B + C)) = ((AcB) · (AcC))) Sum of exponents law for complex exponentiation.
    http://us.metamath.org/mpegif/mulcxp.html ⊢ (((A∈ℝ ∧ 0 ≤ A) ∧ (B∈ℝ ∧ 0 ≤ B) ∧ C∈ℂ) → ((A · B)↑cC) = ((AcC) · (BcC))) Complex exponentiation of a product.
    http://us.metamath.org/mpegif/cxpmul.html ⊢ ((A∈ℝ+∧B∈ℝ∧C∈ℂ) → (Ac(B · C)) = ((AcB)↑cC)) Product of exponents law for complex exponentiation.
    http://us.metamath.org/mpegif/cxpmul2.html ⊢ ((A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℕ0) → (Ac(B · C)) = ((AcB)↑C)) Product of exponents law for complex exponentiation with more general conditions on A and B when C is an integer.
    http://us.metamath.org/mpegif/rpcxpcl.html ⊢ ((A∈ℝ+ ∧ B∈ℝ) → (AcB) ∈ℝ+) Positive real closure of the complex power function.

    So for two real positive numbers, a & b, \(b^a = \textrm{exp}( a \textrm{log} (b) )\) where \(\textrm{log} (b)\) is the solution to \( \textrm{exp}( x ) = b\), and when a & b are positive real numbers, so is \(b^a\).
     
    Last edited: Sep 5, 2015
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  13. Dinosaur Rational Skeptic Valued Senior Member

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    R Penner: Mea Culpa! My construction was incorrect.

    At least let me claim credit for being one of the very few Posters who admit to errors.
     
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  14. DNA100 Registered Senior Member

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    Thanks.
    But actually that definition using complex numbers with complex functions exp and log is precisely what I wanted to avoid.

    I think intuitively complex numbers can be understood in terms of real numbers(as distances) and trigonometry and geometric rotations. But for that we have to first understand real numbers.
    And as much as I know, it is not necessary to use complex numbers to construct real numbers.

    I wanted to understand real numbers, and more specifically positive real numbers and the meaning of \(a^b\), from an intuitive point of view.

    My personal view point is that while being mathematically rigorous is good, we must not throw away our intuitive understanding because all great discoveries ultimately rely upon intuitive understanding.
     
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  15. rpenner Fully Wired Valued Senior Member

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    4,833
    Pbbbt. There is an isomorphism between positive reals (x,y) using multiplication and reals using addition (log x, log y).

    So for any positive real x and any integer n, \(x^n\) corresponds to \(n\; \textrm{log} \, x\). THAT's the intuitive picture. You're just using the number line with different labels.

    3 log x and 2 log x are separated by the same distance as 101 log x and 100 log x because these ratios are equal: \(\frac{x^3}{x^2} = \frac{x^{101}}{x^{100}}\).

    The difference between \(x^2, x^3, x^4, \dots\) and \(y^2, y^3, y^4, \dots\) is just the magnitude of x and y. If \(x < 1 < y\) then they are also oriented differently.

    The average of n logarithms corresponds to the geometric mean. \( \frac{1}{n} \sum_{k=1}^n \textrm{log} \, x_k \leftrightarrow \left( \prod_{k=1}^n x_k \right)^{\frac{1}{n}}\). So any nth root can be found by dividing the logarithm of x. So any rational number q can be sensibly used as an exponent. And since the relationship is 1-to-1, onto, monotonic and continuous, then you can use the fact that the reals are a complete ordered field to show that when x is a positive real number and y is any real number, that

    \(x^y = \left\{ { \begin{array}{lll} \textrm{sup} \left\{ x^q | q \in \mathbb{Q} \; \textrm{and} \; q < y \right\} & \quad \quad \quad & \textrm{if} \; x \geq 1 \\ \textrm{sup} \left\{ x^q | q \in \mathbb{Q} \; \textrm{and} \; q > y \right\} & & \textrm{otherwise} \end{array} } \right.\)

    Until your intuition is trained to reason with the supremum, you aren't mentally using the real numbers.

    \(\phi^{\pi} = \textrm{exp} ( \pi \; \textrm{log} \, \phi ) = \textrm{sup} \left\{ \phi^q | q \in \mathbb{Q} \; \textrm{and} \; q < \pi \right\} = \textrm{sup} \left\{ \phi^{3}, \; \phi^{\frac{333}{106}}, \; \phi^{\frac{103993}{33102}}, \; \phi^{\frac{208341}{66317}}, \dots \right\} \approx \frac{477247408977619887286}{105242109338451627749}\)
     
    Last edited: Sep 6, 2015
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  16. DNA100 Registered Senior Member

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    Thanks.
     
  17. danshawen Valued Senior Member

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    A geometrical interpretation of taking the root of a number is, as you mentioned, would be a shrinking scale on a slide rule.

    To put it another way, using log (common or natural) tables, you would find the root by taking the log, dividing the value of the log by n (the root you want), and find the antilog. The answer is your root. Log tables tended to lose digits of resolution quickly, however.

    In this manner, it was common for most slide rules to provide extra log scales specialized for taking square and cube roots, but there is no reason in principle this process could not be extended indefinitely. Apps exist on some smartphones to implement infinite resolution slide rules, which is über cool. I haven't seen any with root scales yet. Too bad.

    The spirals of Archimedes and of Theodorus in particular is an interesting way to visualize and construct square roots, but does not have relationships to the golden mean or counterparts in nature that the logarithmic spiral has.
     
    Last edited: Sep 9, 2015
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  18. DNA100 Registered Senior Member

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    Thanks.
    Is there a way to geometrically construct log x?
     
  19. rpenner Fully Wired Valued Senior Member

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    Do you mean with straightedge and compass? No.

    Since for \( x \geq 1 \), \(\log x = \int_{t=1}^{x} \frac{1}{t} \, dt \geq 0\) you can think of log(x) as an area under a hyperbola.

    Then \(\log(x \times y) \; = \; \log x \, + \, \log y\) is graphically \(\int_{t=1}^{x \times y} \frac{1}{t} \, dt \; = \; \int_{t=1}^{x} \frac{1}{t} \, dt \; + \; \int_{t=1}^{y} \frac{1}{t} \, dt\). So, it follows that \(\int_{t=x}^{x \times y} \frac{1}{t} \, dt = \int_{t=1}^{y} \frac{1}{t} \, dt\) which is an interesting graphical relation.
     
  20. danshawen Valued Senior Member

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  21. rpenner Fully Wired Valued Senior Member

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    4,833
    Just as ratios of consecutive Fibonacci numbers do not give the Golden Ratio, so the construction given by danshawen is not a true golden spiral.
    Since the above-mentioned spiral consists of piecewise segments which are arcs of circles of constant radius, it exhibits discontinuities in curvature and is not as smooth as a true logarithmic spiral.

    Let O=(0,0), A=(0,1), B=(Φ, 0), and C=(Φ, 1), D=(1,0), E=(1,1), F=(1, 2 - Φ), G=(Φ, 2 - Φ). Then rectangle OACB is a golden rectangle which can be partitioned into square OAED and golden rectangle DECB. Golden rectangle DECB can be partitioned into square FECG and golden rectangle DFGB. Unlike the rectangle generated by the Fibonacci sequence, this cutting off square off golden rectangles can proceed forever.

    The true golden spiral is the smooth logarithmic spiral that passes through diagonal points of successive lopped off squares. So, it would pass though O, E, G, etc.
    Since square OAED has image ECGF under rotation, scaling and translation, it follows that we can use matrix methods to describe this rotation of 90° through the origin, shrinking by a factor of Φ⁻¹ and then a translation of the origin O to point E as:
    \( M = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \Phi^{-1} & 0 & 0 \\ 0 & \Phi^{-1} & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & \Phi^{-1} & 1 \\ - \Phi^{-1} & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix}\)

    The transform \( \begin{pmatrix} x' \\ y' \\ 1 \end{pmatrix} = M \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} \) then maps H = \(\left ( \frac{5 + 3 \sqrt{5}}{10}, \, \frac{5 - \sqrt{5}}{10} \right)\) to itself. H is the "center" of our golden spiral.

    What's the relation between r and θ as measured from H? \(r \propto \Phi^{ \frac{ 2 }{\pi} \theta }\)

    \(\begin{array}{l|ll} \textrm{Point} & r & \theta \\ \hline \\ O - H & \sqrt{1+\frac{1}{\sqrt{5}}} & \pi - \tan^{-1} \left( 2 - \sqrt{5} \right) \\ E - H & \sqrt{1-\frac{1}{\sqrt{5}}} & \pi - \tan^{-1} \left( 2 + \sqrt{5} \right) \\ G - H & \sqrt{2-\frac{4}{\sqrt{5}}} & - \tan^{-1} \left( 2 - \sqrt{5} \right) \end{array} \)
    Every quarter turn of the circle, the radius shrinks by a factor of the golden ratio.
     
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  22. DNA100 Registered Senior Member

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    no, no.
    When I say "construct " by no means I have only ruler and compass in mind.
    It is absolutely anything that has a geometric existence.
    For example, \(\pi\) is constructible number to me because it is the length of a circle of radius 1/2(which can be easily constructed).


    Thanks.
     
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  23. rpenner Fully Wired Valued Senior Member

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    For \(M= \begin{pmatrix} 0 & x^{-1} & 1 \\ - x^{-1} & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix}\) always leaves the point \(H = \left( \frac{x^2 + x}{x^2 + 1}, \; \frac{x^2 - x}{x^2 + 1} \right)\) unchanged.

    When x equals the golden ratio, \(\Phi^2 = 1 + \Phi\) can be used to simplify \(H = \left( \frac{\Phi^2 + \Phi}{\Phi^2 + 1}, \; \frac{\Phi^2 - \Phi}{\Phi^2 + 1} \right) = \left( \frac{2\Phi + 1}{\Phi + 2}, \; \frac{1}{\Phi + 2} \right)\)

    \(\begin{array}{l|ll} \textrm{Point} & r & \theta & \cos \theta & \sin \theta \\ \hline \\ O - H & \sqrt{1+\frac{1}{\sqrt{5}}} & \pi - \tan^{-1} \left( 2 - \sqrt{5} \right) & -\sqrt{\frac{1}{2} + \frac{1}{\sqrt{5}}} & -\sqrt{\frac{1}{2} - \frac{1}{\sqrt{5}}} \\ E - H & \sqrt{1-\frac{1}{\sqrt{5}}} & \pi - \tan^{-1} \left( 2 + \sqrt{5} \right) & -\sqrt{\frac{1}{2} - \frac{1}{\sqrt{5}}} & \sqrt{\frac{1}{2} + \frac{1}{\sqrt{5}}} \\ G - H & \sqrt{2-\frac{4}{\sqrt{5}}} & - \tan^{-1} \left( 2 - \sqrt{5} \right) & \sqrt{\frac{1}{2} + \frac{1}{\sqrt{5}}} & \sqrt{\frac{1}{2} - \frac{1}{\sqrt{5}}} \end{array} \)

    And the curve is:
    \( x = \frac{2\Phi + 1}{\Phi + 2} + \sqrt{2 - \frac{4}{\sqrt{5}}} e^{\frac{2 \ln \Phi}{\pi} \, \tan^{-1}( 2 - \sqrt{5} )} e^{ \frac{ 2 \ln \Phi}{\pi} \theta } \cos \theta \\ y = \frac{ 1}{\Phi + 2} + \sqrt{2 - \frac{4}{\sqrt{5}}} e^{\frac{2 \ln \Phi}{\pi}\, \tan^{-1}( 2 - \sqrt{5} )} e^{ \frac{ 2 \ln \Phi}{\pi} \theta } \sin \theta \)
    Plotted for \(\theta \in \left( -\infty, \pi - \tan^{-1} \left( 2 - \sqrt{5} \right) \right]\)
     

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