# Difference between revisions of "Proofs without words"

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int n = 6; pair shiftR = ((n+2),0); real r = 0.3; | int n = 6; pair shiftR = ((n+2),0); real r = 0.3; | ||

pen colors(int i){ return rgb(i/n,0.4+i/(2n),1-i/n); } /* shading */ | pen colors(int i){ return rgb(i/n,0.4+i/(2n),1-i/n); } /* shading */ | ||

− | void htick(pair A, pair B,pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } | + | void htick(pair A, pair B, pair C, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } |

/* triangle */ | /* triangle */ | ||

Line 75: | Line 75: | ||

draw(A-ticklength--A+ticklength); | draw(A-ticklength--A+ticklength); | ||

draw(B-ticklength--B+ticklength); | draw(B-ticklength--B+ticklength); | ||

+ | draw(A--C--B); | ||

} | } | ||

## Revision as of 13:56, 2 November 2018

The following demonstrate proofs of various identities and theorems using pictures, inspired from this gallery.

## Summations

The sum of the first odd natural numbers is .

defaultpen(linewidth(0.7)); unitsize(15); int n = 6; pair shiftR = ((n+2),0); real r = 0.3; pen colors(int i){ return rgb(0.4+i/(2n),i/n,1-i/n); } /* shading */ void htick(pair A, pair B,pair ticklength = (0.15,0)){ draw(A--B); draw(A-ticklength--A+ticklength); draw(B-ticklength--B+ticklength); draw(A--C--B); } /* triangle */ draw((0.5,0)--(n-0.5,-n+1),linetype("4 4")); for(int i = 0; i < n; ++i) draw((0,-i)--(i,-i)); for(int i = 0; i < n; ++i) for(int j = 0; j <= i; ++j) filldraw(CR((j,-i),r),colors(i)); /* arc arrow */ draw( arc((n,-n+1)/2, (1.5,-1.5), (n-1.5,-1.5), CW) ); fill((n-1.5,-1.5) -- (n-1.5,-1.5)+r*expi(5.2*pi/6) -- (n-1.5,-1.5)+r*expi(3.3*pi/6) -- cycle); /* manual arrowhead? avoid resizing */ /* square */ draw(shiftR+(0.5,0)--shiftR+(n-0.5,-n+1),linetype("4 4")); for(int i = 0; i < n; ++i) draw(shiftR+(0,-i)--shiftR+(i,-i)^^shiftR+(n,-n+1)-(0,-i)--shiftR+(n,-n+1)-(i,-i)); for(int i = 0; i < n; ++i) for(int j = 0; j < n+1; ++j) filldraw(CR((j,-i)+shiftR,r),colors((j <= i) ? i : n-1-i)); /* labeling and ticks */ htick(shiftR+(-1,r),shiftR+(-1,-n+1-r)); label("$n$",shiftR+(-1,(-n+1)/2),W,fontsize(10)); htick(shiftR+(-r,-n),shiftR+(n+r-1,-n),(0,0.15)); label("$n$",shiftR+((n-1)/2,-n),S,fontsize(10)); htick(shiftR+(n-r,-n),shiftR+(n+r,-n),(0,0.15)); label("$1$",shiftR+(n,-n),S,fontsize(10)); (Error compiling LaTeX. a798cc87c0a2b0e6c7c397693c8279de7918840b.asy: 10.9: no matching variable 'C')

The sum of the first positive integers is .

The sum of the first positive integers is .^{[1]}

The alternating sum of the first odd natural numbers is . (Source)

Nichomauss' Theorem: can be written as the sum of consecutive integers, and consequently that .

Here, we use the same re-arrangement as the first proof on this page (the sum of first odd integers is a square). Here's another re-arrangement to see this:

This also suggests the following alternative proof:

An animated version of this proof can be found in this gallery.

The th pentagonal number is the sum of and three times the th triangular number.

If denotes the th pentagonal number, then .

The identity , where is the th Fibonacci number.

### Geometric Series

The infinite geometric series .

The infinite geometric series .

The infinite geometric series .

Another proof of the identity .

The infinite geometric series .

The arithmetic-geometric series , also known as Gabriel's staircase.

^{[2]}

## Geometry

The Pythagorean Theorem (first of many proofs): the left diagram shows that , and the right diagram shows a second proof by re-arranging the first diagram (the area of the shaded part is equal to , but it is also the re-arranged version of the oblique square, which has area ).

^{[3]}

Another proof of the Pythagorean Theorem (animated version).

Another proof of the Pythagorean Theorem; the left-hand diagram suggests the identity , and the right-hand diagram offers another re-arrangement proof.

A dissection proof of the Pythagorean Theorem.

^{[6]}(Cut-the-knot)

COMING: The last proof of the Pythagorean Theorem we shall present on this page, this one by dissection.

^{[10]}

(Comment: we do not need to re-arrange the triangles to a trapezoid to see this, but this re-arrangement works due to alternate interior angles/angle bisector properties of the incenter.)

The area of a dodecagon is , where is the circumradius.

The smallest distance necessary to travel between , the x-axis, and then for is given by .

^{[4]}

In trapezoid with , then .

Varignon's theorem: the area of the outer parallelogram is twice the area of the quadrilateral and four times the area of the midpoint parallelogram, so the midpoint parallelogram of a (convex) quadrilateral has area of the quadrilateral.

Proof for Volume of a Cone: http://www.mathematische-basteleien.de/wuerfel16.gif

## Miscellaneous

from . (Source)

(Source)

The Root-Mean Square-Arithmetic Mean-Geometric Mean inequality, .

The Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality.^{[5]}

Fermat's Little Theorem: for (above ).

There exists a homeomorphism, the stereographic projection, between the punctured hypersphere and for .

Sum of arctangents formula:

## References

- ^ MathOverflow
- ^ Wolfram MathWorld
- ^ Attributed to the Chinese text Zhou Bi Suan Jing.
- ^ This is more of a proof without words of the AM-GM inequality ; though the lengths of the segments labeled RMS and HM can easily be verified to have values of , respectively, it might not be obvious from the diagram. It still serves as a useful graphical demonstration of the inequality.