Approximate Square Roots Visually!

[@MathVisualProofs]

When I say we shade 6/25 of this diagram, I meant to say “we shade an extra 6/25 in the L-shaped diagram”

[@dougr.2398]

You can call it a gnomon, seriously. Make them look it up if they don’t know the term?!? You, too!! 😂

[@robertveith6383]

@MathVisualProofs -- That should be "... shade *an* extra ..."

[@starsun2347]

I know that already bro

[@notcring_h3re2012.ieatrocks]

256 < 267 < 289 16 < sqr. 267 16 + 1/3

[@BananaPelted]

I was SO close. I calculated 267 - 256 as 21. So i rounded 21/33 to ⅔ to get 16.666.... I guess I wasn't so close

[@RishavSingla-wb8fe]

Easy way to approximate is 267= 256 + 11 + (11÷32)^2 - (11÷32)^2=(16+11÷32)^2 we can neglect - (11÷32)^2 as it is so small that it will not effect result that badly so final answer of √267=16 + 11÷32 and it is more accurate

[@davidjonathan4445]

a < b < c sqrt(a)

[@fdasfafs]

sqrt(b) depends on which sqrt of other number is closer, but usually, yeah, it is the smaller one

[@xismailsix]

It's (c + a)

[@thehexagon_yt]

@@xismailsix it's (sqrt(c) + sqrt(a)) or (c - a). In example from video, a is 144, c is 169. c - a is 169 - 144, which is 25. sqrt(c) + sqrt(a) is 13 + 12, which is also 25.

[@xismailsix]

@@thehexagon_yt sure, my mistake. Sorry 😬

[@NWRIBronco6]

If we let "a" be "n^2" then c = (n+1)^2 and we can do some adjusting to think in terms of the sqrts rather than the sqrs. Sqrt(b) = n + (b-n^2)/(2n+1) Not really any different except calculating the denominator may be a bit faster.

[@StratosFair]

It's nice but you need to know the two perfect squares which sandwich your number. Also i'd be curious to know how bad the error of this approximation can get !

[@madb69a]

Since every number has a perfect square, the error would be in the fractional part of the result. So, the larger the number, the smaller the error.

[@Aladin-tp1]

The smaller is the number the larger Is the error for exemple sqrt(2) with thecnique equal 1.333..... But the real value is 1.4142.... so it's pretty big error Also this thecnique work only with natural number you can't approximate the others

[@StratosFair]

@@madb69a sounds like the error decreases like 1/sqrt(2n) asymptotically, right ? it's basically the reciprocal of the square root of (n+1)^2 - n^2

[@Umbra451]

If you look at the closest square number rather than round down, the error will be at most 0.25.

[@simonelflaco10]

Why do you use 25 as the denominator of the fraction?

[@KKPMONSTAofficial]

Underrated channel

[@MathVisualProofs]

Hey, thanks! :)

[@KKPMONSTAofficial]

@MathVisualProofs Bro Saw it ❤️❤️ 10min after comment W

[@lanevalhalla1225]

​@MathVisualProofs for some reason you sound like J perm ❤

[@user-rv9uo7xp2u]

You’re linearly interpolating the square root function here. It could be better to use a tangent line approximation rather than a secant line like this one if you’re close to a known point

[@MathVisualProofs]

Yes, that's the follow up video! :)

[@lonko173]

I developed this technique myself when I was 12 😋 been using it all the time

[@sylvesterogbolu-otutu1498]

The square root of 267 is 16+1/3. 16^2 = 256 17^2 = 289 The range of 267 is: 256 < 267 < 289 Therefore, the square root of 267 will be at least 16 and less than 17. 289 - 256 = 33 267 - 256 = 11 The square root of 267 = 16 + 11/33 => 16 + 1/3.

[@hebermoreno7963]

I'm mesmerized!

[@abhiramvartak4149]

Another clever approximation is as follows: Sqrt(x) = (x+y)/(2(y^0.5)) where y is the nearest whole root to the required number x. So in case x = 150, then y would be 144 (viz 12 squared) The neat thing is that you can just plug in the values and the answers are really accurate for larger numbers.

[@davidmurphy563]

UnboundLocalError: local variable 'y' referenced before assignment

[@ITinkerWithTech]

he actually explained this method in his video on how to do this with calculus😊

[@alfadog67]

@@davidmurphy563 Did you reboot? Did you clear your cache?

[@abhiramvartak4149]

@@ITinkerWithTech oh, didn't know that. Had seen this method in Twitter

[@juliavixen176]

This is the first two terms of the Taylor Series for the square root. You're basically doing a first order linear aproximation.

[@santiagohull3303]

I plotted this on desmos, and it's just a square root with linear interpolation between each perfect square

[@conanedojawa4538]

on a grid of 1×1 squares suppose you have a big square n×n , inside it a square m×m such that n=m+1 , n,m are positive integers , Like in this video we will shade "a" number of 1×1 squares from the "L" shape to get "k" number of 1×1 squares, now we have √(k) =~ m+a/(n+m) , but k= m×m+a=m^2+a , n=m+1 , "a" is small enough, then we get √(m^2+a)=~ m+a/(2m+1) , if you let a=1 , for big m you will get a very good approximation, because if you try to graph this function: y=√(x^2+1)-x-1/(2x+1) you will notice that as x approach to infinity, y approach to 0 , fun fact: if you integrate this function from x=0 to infinity you will get nicely exactly 1/4 , Maths is fun and a world of an amazing curiosity

[@ALOK_KUMAWAT_4]

Yes It is very useful method , (for heigher values ~150≤ ) also variance become less than 0.01 for values ≥156.25 😮

[@arifyesehehehehhewahahahah3445]

A classical way for physicists to approximate values is by using Taylor series: (1+x)ⁿ ≈ 1+nx for small nx (nx

[@secdtan2650]

This method works by approximating f(x) = √x for a

[@sylvesterogbolu-otutu1498]

You are using a method from the Calculus. The problems simply called for the use of a technique from Algebra.

[@genio2509]

Algebraically rather than geometrically you can do the following. Let k be your number and n² be the preceding square The distance between n² and (n+1)² is 2n+1, that's the blank space. The filled blank is then k-n². So √k≈ n + (k-n²)/(2n+1) √267 ≈ 16 + (267-256)/(32+1) = 16+11/3 = 16 1/3

[@ТимурГубаев-ж8ы]

Tried to understand geometrically why one adds 6/25. Here’s what I came up with. The actual square that we’re looking for can be obtained by morphing the rest of the area (6 yellow squares) into an L-shape such that its area is also 6 and it perfectly complements the red square. One way to obtain an approximation of such L-shape is to take a rectangle of length 25 (because it should be aligned with the cyan L-shape) and thickness 6/25 (because its area = length * thickness), cut it into two equal rectangles with height 6/25 and one square with side length of 6/25, and assemble an L-shape using two smaller rectangles as its arms, and a square to fill in the corner. Finally the square that we’re looking for is the red square plus the L-shape that we’ve just constructed. The square root in question is the side length of the square, i.e., side length of the red square plus the thickness of the obtained L-shape. Now, where is the source of the error? The lengths of the arms of our L-shape will be (25 - 6/25) / 2 which is clearly slightly more than 12. Going back and increasing the thickness should make the arms shorter and get them closer to 12 - that indicates that the error comes from inaccurate estimation of the thickness, and this method of approximating the square root will always yield results less than or equal to the actual square root.

[@CharlieG143]

Since (16)^2 = 256 and (17)^2 =289 And, 256

[@mattheoprannac]

For some reason I was thinking of exactly this method for approximating square roots yesterday night, and first thing I see when I open youtube is this, how did that happen lmaoo

[@diya2910]

There's the [squeezing theorem] in calc1 wich is pretty cool and a simple concept

[@JennyBlaze253]

For sqrt(267), 267 is between 256 and 289, 16 and 17 squared respectively. 267-256 is 11, and 289-256 is 33, so the square root is approximately 16 1/3.

[@unkown9685]

√256 < √267 < √289 i.e 16 < √267 < 17 Therefore, 16+ 11/33 16 + 1/3 16 + 0.33 Therefore √267 approximately equal to 16.33.

[@ALOK_KUMAWAT_4]

You can plot it's graph in desmos ,as I did : f(x)= x½ or = √x g(x)=floor(√x) + (x-(floor(√x))²)/{(floor(√x)+1)² - (floor(√x))²} or g(x)= [√x] + (x - [√x]²)/{ ([√x] +1)² - [√x]² } where [.] =floor(x) , stand for greatest integer function For variance do f(x) - g(x)

[@anirudhgovada2753]

You can also use linearisation/differentials

[@ArthurWLittle]

When I was in school, we learned how to calculate square roots manually, similar to doing long division. You can calculate as many digits as you want.

[@physjim]

Brilliant

[@Sandipan_Mandal_YT]

Always remember, shifting to swords is way more faster than reloading your pistol.

[@RishiSingh-kn7tr]

Totally works

[@NHWY25]

It really works √ 160 =12.64 Thank you ❤❤❤❤❤

[@cl0p38]

√256 < √267 < √289 16 < √267 < 17 √267 ≈ 16 + (267-256)/(289-256) ≈ 16 + 11/33 √267 ≈ 16,33

[@Polaris12295]

I like that you used the word investigate. "Let's see an algebraic way to approximate square roots by investigating the square root of 150." I don't know why, but I like that.

[@MathVisualProofs]

😀

[@ThePeterDislikeShow]

Isn't this just linear interpolatio

[@MathVisualProofs]

Essentially

[@Rg-nk3rc]

Proof by drawing each boxes out

[@MohamadrezaGanjkhani]

There's another way thats more precise. sqrt(a²+b) ~ a+(b/2a) which in general is: nth root of a²+b ~ a+(b/n.a^(n-1))

[@vishwanathmali4338]

It works!!

[@EfremJackson]

Sqrt 267: 256 < 267 < 289 267 - 256 = 11 289 - 256 = 33 Sqrt 256 = 16 16 + (11 / 33) 16 + (1 / 3) Sqrt 267 is approximately 16.333333... Sqrt 267 is actually 16.34013... Difference is approxemately 0.0068...

[@danielwinter441]

I got 16+ 11/33 or 16.3continued Real root is 16.3401

[@LewisBrandt]

This was in Jordan Ellenberg’s Wonderfull book, Shape. It supposedly makes a mean party trick.

[@Martinlarh]

Hey please show how to calculate cube root

[@RK-tf8pq]

So the fraction is (150-144)/(169-144) or (150-144)/(12+13).

[@Ymir._]

Another approximation would be where [b≈a^.5] (b+a/b)/2

[@adamgrimsley2900]

Nice

[@JBOboe720]

My favorite is to do a continued fraction. Set your number to be in the form x=n²+m, so 50=7²+1, 150=12²+6, etc. 1. Take 2n 2. Take the reciprocal 3. Multiply by m 4. Add 2n Repeat steps 2-3-4 as many times as desired. 5. Subtract n This number will approach sqrt(x).

[@1337w0n]

If n^2

[@dougr.2398]

How about continuing with another square and gnomonic method to obtain either a more accurate last decimal place or a better overall next approximation?

[@NerdBoiVR]

One way I’ve done before is to is you take the sqrt of 6 for example, and since 6 is in between the perfect squares 4 and 9, you draw a number line with the square roots of them, so you draw a number line from 2 to 3. Next you take the number 2.5 and square it. Since it is bigger than six you now know that sqrt of 6 is in between 2 and 2.5. Keep doing this and you’ll realize that sqrt of 6 to the tenths place is 4. You can keep doing this for the hundredths, thousandths, and so on.

[@leopartanen8752]

And you can get the amout of non-common squares (25) by adding 12 + 13.

[@itsphoenixingtime]

≈ 16 1/3 or 16.333 Actual value is 16.340

[@thechocolatemonster3392]

Question - great explanation but why is there an error? For some reason, I thought there would not be an error.

[@MathVisualProofs]

This is assuming square root is linear, which it is not. But it is close to linear so this gives an approximation. Check follow up short for a related explanation.

[@thegomillionairemindset6719]

You could also approximate it using linearization if the visuals are throwing you off I got 12.25

[@MathVisualProofs]

Heck next short :)

[@LakeA-pz1fm]

this is the motivation i needed to learn my perfect square roots

[@tharunb7282]

Schools never teach these things

[@happyjobkebobkebrawl-stars9116]

Why would they

[@aluminatestrontium7298]

They do in primary school, but it's too early for children to pay attention. At higher levels, there is barely time for more complex stuff and you have access to calculators anyway. Note: I clearly remember thinking "why would I ever need to remember this stuff?", but this is not a Mandela effect, I recently opened my old school books (early 2000's) and many of these visual proofs are there at various levels (this one in primary/middle school books, other videos related to pi in high school books, etc.). At least they teach this stuff here in Europe.

[@danielevilone]

Using Taylor at first order I got √150 = √(144+6) = 12 + 6/(2*12) = 12.25, which is an even better approximation.

[@MathVisualProofs]

Yep. That’s the follow up short :)

[@FundamSrijan]

But why did you add 6/25 in this video , can u explain me pls 🙏​@@MathVisualProofs

[@wanlim9548]

​@FundamSrijan it is because of the non-shaded region. He took at first 12x12 squere cells, then the rest of them were 25 in total, that goes to denominator. And as 150 is larger than 144 with 6 cells, he added 6 on top of the 25. Therefore we have 12 + 6/25

[@FundamSrijan]

@@wanlim9548 y did 25 go in denominator

[@wanlim9548]

@FundamSrijan that's the whole trick of it. When we took 12x12 out of 13x13, there were left 25. So 25 would be our total for 6

[@robmbez]

I don't understand why 6?

[@justfaqih]

150-144

[@robmbez]

@@justfaqih thanks

[@matei_woold_wewu]

Integral method B)

[@colinthomasson3948]

I'd say you arrived at that approximation by way of geometry

[@MathVisualProofs]

Yes. I think of it as algebra since it is related to the fact that this diagram shows the algebra of (n+1)^2 = n^2 +2n+1 :)

[@borjecarlsson4860]

Newton gives 12.25, if starting with 12, next one is 43209/3528=12 +873/3528

[@LamSH1109]

What part of this is an algebraic solution?

[@MathVisualProofs]

It’s related to the fact that (n+1)^2 = n^2 + 2n+1, which is an algebraic fact demonstrated by this image. But I agree that might be a stretch (just how I think about it)

[@ShinjiCarlos]

I just used a Taylor series approximation to the first order to get about 12.25.

[@thegomillionairemindset6719]

Is that different from using linearization because I got 12.25 as well

[@MathVisualProofs]

First order Taylor series is linearization.

[@ZainAhmed456]

My guess was 12.3 so i was SO close lol

[@topquark22]

There's a way to do this using calculus. It's just the first iteration of the Newtom-Raphson method.

[@LauraEdgar-xv1bs]

COOL 😊

[@Ndahiya-n6u]

Why did you only shade 6 squares in the end ?

[@MathVisualProofs]

That got me from 144 up to 150

[@Deficard]

i perferred not to draw a square and instead use my brain. i approximated √269 this way: I first find the square. so it is 16 and 17. i choose 16. now we make an adjustment. so im going to subtract 17² by 16². this gives me 33. i will call this x for algebraic purpose. then i will subtract the number by 16², or simply 269-256. this gives me 13. so it is approximately equal to 16+13/33. using a calculator, it equaled to 16.3939393939. whereas the actual value, it is 16.401. thats actually very close! if i round the approximate, it turns out to be 16.4, which matches. cool!

[@alfadog67]

A table of Squares and their sqrt accuracy using this method: 110: 99.773% Accurate 2550: 99.99% accurate 25122 : 99.999% accurate. The accuracy grows with the square, and has valleys between the "sandwich" squares. The numbers above are in the valleys.

[@davidmurphy563]

But what's the algorithm? def sqrt(x): return what? Do you want me to loop through all the squares and find the one before and after the number? That will be an order of magnitude more expensive. Imagine it was a large number? This isn't practical. I guess you could calculate the derivative of the slope but you'd still have to loop.

[@dhalsam]

you took “all mathematical proofs are technically infinite” too seriously

[@mr.wretchfromthebackrooms2468]

I automatically rounded up to 12.25, so i got exact

[@honeyb1scuit]

49/3

[@carlplz00001]

Exactly what I thought 👌🏽

[@jaja47_coolness]

I wish I remembered the extension of this method that the Babylonians used it was basically this but again and again but I forgot how

[@jaja47_coolness]

I'd like to point out that this is typically what calculators do to find square roots

[@heroman3259]

Ye ill just keep using taylor for such problems

[@Mhrn.Bzrafkn]

Why 6??

[@MathVisualProofs]

6 more to get from 144 to 150

[@Mhrn.Bzrafkn]

@MathVisualProofs Ahaa, thanks. 🙋🏻‍♂️

[@dougr.2398]

I don’t see a clear reason to use six of the 25 remaining gnomonic squares, other than that in hindsight, it approximates the correct answer. Backwards solutions should provide a rationale

[@MathVisualProofs]

144+6 is 150. So 25 out of 25 gets you to 13. And if we use the fractions of 25 we get approximations for all numbers from 145-169.

[@Георгий.Цыфаркин]

лучше брать не 25, а 12*2=24, так даже тоснее будет

[@strodion2105]

For westerners it’s like a high math probably 😂

[@Manocheher.A]

I use something called calculator instead of drawing squares.

[@joeindia1647]

Don’t insult algebra. This is not algebra.

[@officialalphabet]

16.333333333...

[@calabiyauisdeadmeow]

This is not an algebraic method. Not to mention this is overtly tedious and takes much an unnecessary long time. Definitely not recommended to use in exams. Newton’s method of approximation is honestly infinitely better.

[@themodestai9603]

not everything in math is about exams. its interesting

[@Rifat91804]

16.33333333

[@joshuajoseph8849]

kzbin.infooTXZSah48hQ?si=ZCEM7Gbx7PloXdrl

[@TestAccount-l2y]

First😊

[@carlplz00001]

That’s what she said!!! 😂

[@GourangaPL]

You can show it's between 12 +6/25 and 12+6/26

[@MathVisualProofs]

I think between 12+6/24 and 12+6/25

[@GourangaPL]

@@MathVisualProofs of course, my bad, that's what i meant

[@MathVisualProofs]

@@GourangaPL :) no worries. I'm always making minor errors (like my misleading use of the word diagram to mean two different diagrams in this video :) ).

[@charl10439]

💩

[@Wawacat4436]

267 + 256 / 2x16 = 523 / 32 = 16 11/32 = 16.34375 Actual value 16.34013

[@robertveith6383]

What you wrote is wrong because you are missing needed grouping symbols: (267 + 256)/(2*16) = 523/32

[@Wawacat4436]

@@robertveith6383 thanks bro but i don't remember asking nor caring

[@JAYDELROSARIO-l5t]

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