Difference between revisions of "Proofs without words"
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<!-- | <!-- | ||
− | + | Links: | |
− | + | http://usamts.org/Gallery/G_Gallery.php | |
− | + | http://mathoverflow.net/questions/8846/proofs-without-words | |
− | + | http://gurmeet.net/computer-science/mathematical-recreations-proofs-without-words/ | |
− | |||
hockey stick? | hockey stick? | ||
+ | Catalan numbers / reflection principle | ||
+ | geometric series: http://mahalanobis.twoday.net/stories/3472911/ | ||
+ | Fibonacci: sum_{i=1}^{2n-1} F_iF_{i+1} = F_{2n}^2 | ||
+ | sum of squares, http://legacy.lclark.edu/~mathsci/invitation.pdf | ||
+ | weighted sum of triangle numbers: http://legacy.lclark.edu/~mathsci/weightedsum.pdf | ||
+ | pentagonal/octagonal numbers: http://legacy.lclark.edu/~mathsci/octnos.pdf | ||
geometry: | geometry: | ||
Pythagorean Theorem (multiple ways - 5?) | Pythagorean Theorem (multiple ways - 5?) | ||
− | + | smallest triangle in angle (Engel 315 5) | |
− | smallest triangle in angle | ||
− | |||
equilateral triangle in three lines | equilateral triangle in three lines | ||
− | |||
Fermat point problem | Fermat point problem | ||
− | |||
Fagnano problem | Fagnano problem | ||
− | |||
− | |||
− | |||
regular heptagon identity? | regular heptagon identity? | ||
− | |||
− | |||
− | |||
− | |||
− | |||
Law of Cosines | Law of Cosines | ||
− | |||
Law of Sines | Law of Sines | ||
− | |||
radical axes/homotheties? | radical axes/homotheties? | ||
+ | Engel 313 E12, 316 6, 11, 323 4, 326/7 | ||
+ | arbelos/salinon: http://legacy.lclark.edu/~mathsci/arbelos.pdf, http://legacy.lclark.edu/~mathsci/salinon.pdf | ||
+ | abc/4R: http://legacy.lclark.edu/~mathsci/euler.pdf | ||
+ | area of triangle xy: http://legacy.lclark.edu/~mathsci/area-rt-tri.pdf | ||
misc. | misc. | ||
3D QM-GM | 3D QM-GM | ||
− | + | AM-GM from http://jwilson.coe.uga.edu/emt725/AMGM/AMGM.5.html | |
− | |||
− | |||
Minkowski's inequality | Minkowski's inequality | ||
− | |||
rt(a^2 - ab + b^2) rt(b^2 - bc + c^2) < rt(a^2 + ac + c^2) | rt(a^2 - ab + b^2) rt(b^2 - bc + c^2) < rt(a^2 + ac + c^2) | ||
− | |||
− | |||
− | |||
− | |||
cardinality of interval with R | cardinality of interval with R | ||
− | + | domino tiling, per mathoverflow | |
+ | hexagon tiling with cube (aops logo) | ||
--> | --> | ||
== Summations == | == Summations == | ||
Line 55: | Line 44: | ||
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 ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } |
/* triangle */ | /* triangle */ | ||
draw((-r,0)--(-r,-n+1)^^(r,-n+1)--(r,0),linetype("4 4")); | draw((-r,0)--(-r,-n+1)^^(r,-n+1)--(r,0),linetype("4 4")); | ||
− | for(int i = 0; i < n; ++i) | + | for(int i = 0; i < n; ++i) |
draw((-i,-i)--(i,-i)); | draw((-i,-i)--(i,-i)); | ||
for(int i = 0; i < n; ++i) | for(int i = 0; i < n; ++i) | ||
Line 149: | Line 138: | ||
The sum of the first <math>n</math> positive integers is <math>{n+1 \choose 2}</math>.{{ref|1}}<br><br> | The sum of the first <math>n</math> positive integers is <math>{n+1 \choose 2}</math>.{{ref|1}}<br><br> | ||
+ | </center> | ||
+ | |||
+ | <center><asy>unitsize(15); defaultpen(linewidth(0.7)); | ||
+ | int n = 5; pair shiftR = (2*n + 4, 1); real r = 0.35; pen sm = fontsize(10); | ||
+ | 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); } | ||
+ | |||
+ | /* draw plus/minus circle at A */ | ||
+ | void plus (pair A) { filldraw(CR(A,r),rgb(0.5,1,0.5)); MP("+",A,(0,0), sm); } | ||
+ | void minus (pair A) { filldraw(CR(A,r),rgb(1,0.5,0.5)); MP("-",A,(0,0), sm); } | ||
+ | |||
+ | /* triangle 1 */ | ||
+ | path p1 = (-1,-1/2) -- (1,-1/2); | ||
+ | draw((-n+1-r,-n+1-2*r)--(n-1+r,-n+1-2*r)); | ||
+ | for(int i = 0; i < n; ++i) { | ||
+ | for(int j = 0; j < 2*i+1; ++j) { | ||
+ | if (i % 2 == 0) | ||
+ | plus((j-i,-i)); | ||
+ | else | ||
+ | minus((j-i,-i)); | ||
+ | } | ||
+ | if (i % 2 != 0) | ||
+ | p1 = (-i-1/2,-i-3/2)--(-i-1/2,-i+1/2)--p1--(i+1/2,-i+1/2)--(i+1/2,-i-3/2); | ||
+ | plus((2*i - n + 1, -n + 1 - 4*r)); | ||
+ | } | ||
+ | p1 = p1 -- cycle; | ||
+ | draw(p1, rgb(0.85,0.85,0.85)); | ||
+ | |||
+ | /* triangle 2 */ | ||
+ | n = n+1; | ||
+ | path p2 = (-0.1,1/2) -- (0.1,1/2); | ||
+ | draw(shiftR+ (-n+1-r,-n+1-2*r)-- shiftR+ (n-1+r,-n+1-2*r)); | ||
+ | for(int i = 0; i < n; ++i) { | ||
+ | for(int j = 0; j < 2*i+1; ++j) { | ||
+ | if (i % 2 != 0) | ||
+ | plus(shiftR+ (j-i,-i)); | ||
+ | else | ||
+ | minus(shiftR+ (j-i,-i)); | ||
+ | } | ||
+ | if (i % 2 == 0) | ||
+ | p2 = (-i-1/2,-i-3/2)--(-i-1/2,-i+1/2)--p2--(i+1/2,-i+1/2)--(i+1/2,-i-3/2); | ||
+ | plus(shiftR+ (2*i - n + 1, -n + 1 - 4*r)); | ||
+ | } | ||
+ | p2 = p2 -- cycle; | ||
+ | draw(shift(shiftR)*p2, rgb(0.85,0.85,0.85)); | ||
+ | htick(shiftR+(-n,r),shiftR+(-n,-n+1-r)); label("$n$",shiftR+(-n,(-n+1)/2),W,fontsize(10)); | ||
+ | </asy> <br> | ||
+ | |||
+ | The alternating sum of the first <math>n</math> odd natural numbers is <math>\sum_{k=1}^n (-1)^{n-k}(2k-1) = n</math>. ([http://www.math.hmc.edu/~benjamin/papers/OddPWW.pdf Source])<br><br> | ||
</center> | </center> | ||
Line 184: | Line 222: | ||
</asy><br> | </asy><br> | ||
− | [[Nichomauss' Theorem]]: <math>n | + | [[Nichomauss' Theorem]]: the sum of the first <math>n</math> cubes can be written as the square of the sum of the first <math>n</math> integers, a statement that can be written as <math>1^3+2^3+...+n^3=(1+2+...+n)^2</math>. <br><br> |
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: | 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: | ||
<br><asy>defaultpen(linewidth(0.7)); unitsize(15); pen heavy = linewidth(2); | <br><asy>defaultpen(linewidth(0.7)); unitsize(15); pen heavy = linewidth(2); | ||
Line 261: | Line 299: | ||
real h = 0.6; // scale factor of diagram | real h = 0.6; // scale factor of diagram | ||
pair shiftR1 = (n*h+1,0), // middle diagram shift offset | pair shiftR1 = (n*h+1,0), // middle diagram shift offset | ||
− | shiftR2 = shiftR1 + (n*h+1, | + | shiftR2 = shiftR1 + (n*h+1,-1); // right diagram shift offset |
int lvl(int i) { return ceil(((8*i+9)^.5-1)/2); } /* return level of square i */ | int lvl(int i) { return ceil(((8*i+9)^.5-1)/2); } /* return level of square i */ | ||
Line 418: | Line 456: | ||
<center>[[#toc|Back to Top]]</center> | <center>[[#toc|Back to Top]]</center> | ||
− | === Geometric | + | === Geometric Series === |
<center><asy>defaultpen(linewidth(0.7)); unitsize(15); | <center><asy>defaultpen(linewidth(0.7)); unitsize(15); | ||
int n = 10; /* # of iterations */ | int n = 10; /* # of iterations */ | ||
Line 496: | Line 534: | ||
</asy><br> | </asy><br> | ||
− | Another proof of the identity <math>\frac 14 + \frac {1}{4^2} + \frac {1}{4^3} + \cdots = \frac 13</math>. <br><br> | + | Another proof of the identity <math>\frac 14 + \frac {1}{4^2} + \frac {1}{4^3} + \frac {1}{4^4}+\cdots = \frac 13</math>. <br><br> |
</center> | </center> | ||
Line 551: | Line 589: | ||
} | } | ||
</asy><br><br> | </asy><br><br> | ||
− | The [[arithmetic-geometric series]] <math>\sum_{n=1}^{\infty} nr^n = \sum_{n=1}^{\infty} \sum_{i=n}^{\infty} r^i = \sum_{n=1}^{\infty} \frac{r^{ | + | The [[arithmetic-geometric series]] <math>\sum_{n=1}^{\infty} nr^n = \sum_{n=1}^{\infty} \sum_{i=n}^{\infty} r^i = \sum_{n=1}^{\infty} \frac{r^{n}}{1-r} = \frac{r}{(1-r)^2}</math>, also known as Gabriel's staircase.{{ref|2}}<br><br></center> |
<center>[[#toc|Back to Top]]</center> | <center>[[#toc|Back to Top]]</center> | ||
Line 809: | Line 847: | ||
In trapezoid <math>ABCD</math> with <math>\overline{AB} \parallel \overline{CD}</math>, then <math>\overline{AC} \perp \overline{BD} \Longleftrightarrow AC^2 + BD^2 = (AB + CD)^2</math>.<br><br></center> | In trapezoid <math>ABCD</math> with <math>\overline{AB} \parallel \overline{CD}</math>, then <math>\overline{AC} \perp \overline{BD} \Longleftrightarrow AC^2 + BD^2 = (AB + CD)^2</math>.<br><br></center> | ||
+ | <center><asy> | ||
+ | // feel free to change these four points! | ||
+ | pair A = (0,0), B = (3, -2), C = (5,1), D = (1,3); | ||
+ | |||
+ | // | ||
+ | // Rest of code | ||
+ | // | ||
+ | size(200); defaultpen(linewidth(0.9)); pen lightgreen = rgb(0.6,1,0.6), lightred = rgb(1,0.6,0.6), smdash = linewidth(0.7)+linetype("2 2"); | ||
+ | |||
+ | pair E = IP(A--C,B--D), AB = (A+B)/2, BC = (B+C)/2, CD = (C+D)/2, DA = (D+A)/2, ABE = 2*AB-E, BCE = 2*BC-E, CDE = 2*CD-E, DAE = 2*DA-E; | ||
+ | path midpts = AB--BC--CD--DA--cycle; | ||
+ | filldraw(shift(-E)*scale(2)*midpts,lightgreen); filldraw(midpts,lightred); | ||
+ | draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw((A+ABE)/2--(C+BCE)/2,smdash); draw((B+ABE)/2--(D+DAE)/2,smdash); draw((B+BCE)/2--(D+CDE)/2,smdash); draw((A+DAE)/2--(C+CDE)/2,smdash); | ||
+ | |||
+ | dot(A); dot(B); dot(C); dot(D); label("$A$",A,W);label("$B$",B,S);label("$C$",C,E);label("$D$",D,N); | ||
+ | </asy><br> | ||
+ | |||
+ | [[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 <math>1/2</math> of the quadrilateral. | ||
+ | </center> | ||
+ | |||
+ | Proof for Volume of a Cone: http://www.mathematische-basteleien.de/wuerfel16.gif | ||
<center>[[#toc|Back to Top]]</center> | <center>[[#toc|Back to Top]]</center> | ||
Line 828: | Line 887: | ||
draw((0,1)--(pi/2,1),linewidth(1)); | draw((0,1)--(pi/2,1),linewidth(1)); | ||
</asy> | </asy> | ||
− | <math>\int_0^{\pi/2} \sin^2 x = \int_0^{\pi/2} \cos^2 x = \frac {\pi}{4}</math> from <math> | + | <math>\int_0^{\pi/2} \sin^2 x \, dx = \int_0^{\pi/2} \cos^2 x \, dx = \frac {\pi}{4}</math> from <math> |
<br><br></center> | <br><br></center> | ||
+ | <center><asy> | ||
+ | import graph; size(170); defaultpen(linewidth(0.7)); Label k; k.p=fontsize(8); | ||
+ | |||
+ | real xmax = 1.35, xmin = -0.35, ymax = 1.35, ymin = -0.35, lblpt = 1/2 + 0.15, alpha = 2, epsilon = 0.015; | ||
+ | |||
+ | real f(real x) { return x^alpha; } | ||
+ | |||
+ | xaxis(xmin,xmax,Ticks(k, 0.25, NoZero),Arrows(6)); yaxis(ymin,ymax,Ticks(k, 0.25, NoZero),Arrows(6)); | ||
+ | filldraw(graph(f,0,1)--(1,0)--(0,0)--cycle,gray(0.7),linewidth(1)); | ||
+ | draw(graph(f,1,1.1),linewidth(0.7),EndArrow(4)); | ||
+ | draw((lblpt-epsilon,f(lblpt))--(0+epsilon,f(lblpt)),Arrows(4)); draw((lblpt,f(lblpt)-epsilon)--(lblpt,0+epsilon),Arrows(4)); | ||
+ | label("$x = y^{1/\alpha}$",(lblpt/2,f(lblpt)),N,fontsize(10)); | ||
+ | label("$y = x^{\alpha}$",(lblpt,f(lblpt)/2),E,fontsize(10)); | ||
+ | draw((0,1)--(1,1),linewidth(1)); | ||
+ | </asy> | ||
+ | <math>\alpha > 0 \Longrightarrow \int_0^{1} \left(x^\alpha + x^{1/\alpha}\right) \, dx = 1.</math> ([http://www.math.ubc.ca/~morey/talk/proofwowords.html Source]) | ||
+ | <br><br></center> | ||
<center><asy>unitsize(15); defaultpen(linewidth(0.7)); | <center><asy>unitsize(15); defaultpen(linewidth(0.7)); | ||
Line 918: | Line 994: | ||
[[Fermat's Little Theorem]]: <math>a^p \equiv a \pmod{p}</math> for <math>\text{gcd}\,(a,p) = 1</math> (above <math>a=2,p=3</math>).<br><br> | [[Fermat's Little Theorem]]: <math>a^p \equiv a \pmod{p}</math> for <math>\text{gcd}\,(a,p) = 1</math> (above <math>a=2,p=3</math>).<br><br> | ||
</center> | </center> | ||
+ | <center><asy> | ||
+ | defaultpen(linewidth(0.7)); unitsize(30); | ||
+ | |||
+ | real r = 0.2; | ||
+ | pair endPt1 = (-2.5,-1), endPt2 = (2.5,-1); // endpoints of R^1 line | ||
+ | real projections[] = {-1.5, -0.75, 1, 2.2}; // x-coordinates of steoreographic projections on y=-1 | ||
+ | |||
+ | void makeshiftarrow(pair A, real dir, real arrowlength = r){ /* Arrow option resizes, so draw makeshift arrows */ | ||
+ | fill(A--A+arrowlength*expi(dir+pi/8)--A+arrowlength*expi(dir-pi/8)--cycle); | ||
+ | } | ||
+ | |||
+ | draw(endPt1 -- endPt2); | ||
+ | draw(unitcircle); | ||
+ | for(int i = 0; i < projections.length; ++i) { | ||
+ | draw((0,1) -- (projections[i],-1), linetype("2 2")); | ||
+ | dot((projections[i],-1), Fill(red)); | ||
+ | dot(OP((0,1) -- (projections[i],-1),unitcircle), Fill(yellow)); | ||
+ | } | ||
+ | dot((0,1), Fill(green)); | ||
+ | |||
+ | makeshiftarrow(endPt1, 0); makeshiftarrow(endPt2, pi); | ||
+ | </asy><asy> | ||
+ | import three; defaultpen(linewidth(0.7)); unitsize(20); | ||
+ | |||
+ | currentprojection = orthographic(0.3,-2,0.6); | ||
+ | |||
+ | pen gridpen = dotted; | ||
+ | int gridmin = -2, gridmax = 2, gridlines = 6; | ||
+ | pair projections[] = {(1,-1),(1,-2),(2,-2),(2,-1)}; | ||
+ | triple IPs[] = new triple[4]; | ||
+ | draw((1,-1,-1)--(1,-2,-1)--(2,-2,-1)--(2,-1,-1)--cycle); | ||
+ | |||
+ | for(int i = 0; i < gridlines; ++i) { | ||
+ | draw((gridmin,gridmin+(gridmax-gridmin)*(i+1)/(gridlines+1),-1)--(gridmax,gridmin+(gridmax-gridmin)*(i+1)/(gridlines+1),-1), gridpen); | ||
+ | draw((gridmin+(gridmax-gridmin)*(i+1)/(gridlines+1),gridmin,-1)--(gridmin+(gridmax-gridmin)*(i+1)/(gridlines+1),gridmax,-1), gridpen); | ||
+ | } | ||
+ | |||
+ | draw(unitsphere,white); | ||
+ | draw(arc((0,0,0), (1,0,0), (-1,0,0), (0,0,1)), linetype("2 2")); | ||
+ | draw(arc((0,0,0), (-1,0,0), (1,0,0), (0,0,1))); | ||
+ | // draw(circle((0,0,0), 1, (0,1,0))); | ||
+ | |||
+ | // draw projection points | ||
+ | for(int i = 0; i < projections.length; ++i) { | ||
+ | real px = projections[i].x/2, py = projections[i].y/2, pxy = 1 + px*px + py*py; | ||
+ | draw((0,0,1)--(2*px, 2*py, -1), linetype("2 2")); | ||
+ | triple IP = (2*px/pxy, 2*py/pxy, (pxy-2)/pxy), OP = (2*px, 2*py, -1); | ||
+ | dot(IP, yellow); | ||
+ | draw(circle(IP, 0.06, (0,1,0))); | ||
+ | IPs[i] = IP; | ||
+ | dot(OP, red); | ||
+ | draw(circle(OP, 0.06, (0,1,0))); | ||
+ | if(i != 0) | ||
+ | draw(IPs[i] -- IPs[i-1], dotted); | ||
+ | } | ||
+ | draw(IPs[0]--IPs[3], dotted); | ||
+ | |||
+ | dot((0,0,1),green); | ||
+ | draw(circle((0,0,1), 0.06, (0,1,0))); | ||
+ | </asy> | ||
+ | There exists a [[homeomorphism]], the [[stereographic projection]], between the punctured hypersphere <math>S^n \setminus \{(1,\underbrace{0, \ldots, 0}_{n-1\text{ zeroes}})\}</math> and <math>\mathbb{R}^n</math> for <math>n = 1,2</math>. | ||
+ | <br><br></center> | ||
+ | |||
+ | Sum of arctangents formula: | ||
+ | |||
+ | <asy> | ||
+ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | ||
+ | import graph; usepackage("amsmath"); size(13cm); | ||
+ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
+ | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ | ||
+ | pen dotstyle = black; /* point style */ | ||
+ | real xmin = -0.4717093412177357, xmax = 7.405441345585962, ymin = -1.1854534297865673, ymax = 7.342957746870971; /* image dimensions */ | ||
+ | pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); | ||
+ | |||
+ | draw((0.,0.)--(0.,1.)--(1.,1.)--cycle); | ||
+ | draw((1.,1.)--(0.,3.)--(0.,1.)--cycle, uququq); | ||
+ | draw((0.,3.)--(6.,6.)--(1.,1.)--cycle, aqaqaq); | ||
+ | draw(arc((1.,1.),0.3101240427875472,180.,225.)--(1.,1.)--cycle, qqwuqq); | ||
+ | draw(arc((1.,1.),0.3101240427875472,116.56505117707799,180.)--(1.,1.)--cycle, qqwuqq); | ||
+ | draw(arc((1.,1.),0.3101240427875472,45.,116.56505117707799)--(1.,1.)--cycle, qqwuqq); | ||
+ | /* draw grid of horizontal/vertical lines */ | ||
+ | pen gridstyle = linewidth(0.7) + cqcqcq; real gridx = 1., gridy = 1.; /* grid intervals */ | ||
+ | for(real i = ceil(xmin/gridx)*gridx; i <= floor(xmax/gridx)*gridx; i += gridx) | ||
+ | draw((i,ymin)--(i,ymax), gridstyle); | ||
+ | for(real i = ceil(ymin/gridy)*gridy; i <= floor(ymax/gridy)*gridy; i += gridy) | ||
+ | draw((xmin,i)--(xmax,i), gridstyle); | ||
+ | /* end grid */ | ||
+ | |||
+ | /* draw figures */ | ||
+ | draw((0.,0.)--(0.,1.)); | ||
+ | draw((0.,1.)--(1.,1.)); | ||
+ | draw((1.,1.)--(0.,0.)); | ||
+ | draw((1.,1.)--(0.,3.), uququq); | ||
+ | draw((0.,3.)--(0.,1.), uququq); | ||
+ | draw((0.,1.)--(1.,1.), uququq); | ||
+ | draw((0.,3.)--(6.,6.), aqaqaq); | ||
+ | draw((6.,6.)--(1.,1.), aqaqaq); | ||
+ | draw((1.,1.)--(0.,3.), aqaqaq); | ||
+ | label("$\arctan 1 + \arctan 2 + \arctan 3 = \pi$",(1.544096936901321,0.5925910821953678),SE*labelscalefactor,fontsize(10)); | ||
+ | /* dots and labels */ | ||
+ | dot((0.,0.),linewidth(3.pt) + dotstyle); | ||
+ | dot((0.,1.),dotstyle); | ||
+ | dot((1.,1.),dotstyle); | ||
+ | dot((0.,3.),dotstyle); | ||
+ | dot((6.,6.),dotstyle); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
+ | /* end of picture */ | ||
+ | </asy> | ||
<center>[[#toc|Back to Top]]</center> | <center>[[#toc|Back to Top]]</center> |
Latest revision as of 15:07, 16 September 2024
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 .
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: the sum of the first cubes can be written as the square of the sum of the first integers, a statement that can be written as .
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.
(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.