How to find a square root 

Tripping over a root while discussing roots... priceless 0:26

The dedication to do an opening bit in the woods just so you could trip on a root at the right time is an admirable commitment to comedy.

Start = root End = square You have a great sense of humour Mr Parker.

0:27 goes to the middle of the forest just ro make a root pun. 0.0

Ahh, a Parker Square Root

Oh so it’s just what I watched 2 days ago but again. *watches again anyways*

The root joke was levels of comedic genius never before witnessed by the human race

I had a Dynamics professor, Dr. Penrod, at Auburn Univ. who could work out roots and powers simultaneously (e.g. 2.68^2/3) and give an answer to about 3 or 4 decimal places. It was absolutely amazing.

0:27 - Looks like you found a different type of root :)

Did you stumble (0:27 s) over the root on purpose? Maybe because you end the video on the city square. ;)

in high school i once overestimated a really simple square root on a test, and couldn't figure it out. so to make sure i never got stuck on any root again, i learned an iterative method by heart which only uses simple operators (so i could do it on paper, albeit very slowly). naturally, we never got any square roots on tests after that

He tripped over a root when talking about square roots

Keep it up Matt. Always excited when I see that you uploaded a new video.

When I find a square root, I sketch a big parabola and then estimate distances using an elaborate system of paperclips and hourglasses.

00:26 “Find the square ROOT (trips over tree root, notices that he accidentally made a Parker pun, and smiles)

I predict that the frequency at which Matt's hair is going to change will not be random but rather contain some sort of coded message.

0:27 when he was saying “root” and literally tripped over the root.

thanks for the explanation I've often wondered, and had studens asking me, this seems like a good way to explain what is going on with root squares.

n ** 0.5

I wouldn't mind if my hair made an hilarious reappearance.

For the summer, I'd gotten a pretty bleak hair cut as well (if that's the way to use the word 'bleak' ) But after watching your video, I am pondering over the option of going full blown shiny head

Really like this, wanted to know how to do this for a long time!

Newton's method for √N: Next term = xₙ/2 + N/2xₙ

Pi-head was the best one. thx for brightening up my day with the reflection of the sun on your perfectly round head. ^^ Big thumbs up

"Find the ROOT-" trips on actual tree root. Love it.

Your method reminds me of using log tables (not logs for square roots before anyone says anything, but my log table book which had all the lists of numbers to work out things like cos, sin, sq rt, sq, logs, nat logs etc etc etc). I still have my book - used it for my O levels. My little sister (she's 50 soon lol) was in the first year to be allowed to use calculators. My year and the one between us were absolutely horrified that they'd be _given_ the answers! Where was the working out?! Ah, time flies, eh? Btw, hubby's got the same haircut. Had it for years. Much neater than a comb over lol!

I looked and immediately wondered whether what you did or finding the solutions to x^2+2xy+y2=10005 is easier. After attempting the latter it's still a bit of guess and check but I was able to place some restraints on the possible value of y that were helpful in increasing the accuracy by any amount desired.

If you do it geometrically and algebraically with the binomial theorem I think it is quite straight forward. The advantage is that you can see whether the number is a perfect square or not. But at the same time you have to do all the sub-calculations increasing with the growing answer 😅 just giving you a single digit every time. Therefor Herons method remains my favorite.

so ez yet so helpfull i love it thank you!

The method I learned to do square roots looks a lot like long division but it is done two digits at a time.

There is a sort of long division method to find square roots that works in binary very easily: 1. write the number in digit pairs left and right of the point, adding a zero before the first digit if odd number of digits before point, appending a zero after the ladt digit idmf an odd number of digits after the point. (extra pairs of zeros can be added after the nunber after rhe point.) 2. Do a long division taking the pairs of digits of the nunber (dividend) as a unit and creating a new divisor at each step: 2.1. double the answer so far and multiply by 10 (the base). 2.2.. Now find a digit to put in the ones place to create a divisor that when multiplied by this digit is the largest result that does not exceed the "working" digits (of the division) 2.3. Put this digit over the pair of digits that form the lat pair of digits of the "working" number. 2.4. Multiply the divisor by this digit and subtract (like long division) 2.5. Bring down next pair of digits to continue the division. 3. Continue the "division" until all pairs of digits have been used and result of last subtraction is zero, or until enough precision has been reached. In binary the double is a shift left 1 bit, and multiply by 10 (the base - in decimal this is also 2) shift left anothe bit; only possible values for the units bit are 1 and 0, so try 1 - put a 1 in the units place and if divisor is less than "working" number can subtract and a 1 goes in the answer; otherwise a 0 goes in answer and next 2 bits of dividend are appended to the end of the working number. - highly efficient for a digital computer. For 10005 the division is: 10005 -> 01 00 05 . 00 00 00 00 Put a decimal point above the point in the "dividend" first pair is 01; answer do far is 0 -> 0_ so need units digit 1; 1 in answer above 01, 01×1=01 subtract to get 0 and bring down next pair 00 to make the "working" number 0 00 As this is zero can put a 0 above the 00 and bring down the next pair 05 to make the working number 0 00 05 Double answer and multiply by 10 gives 20_ which is greater than 5 so put 0 above 05. Bring down next pair 00 to make working number [0 00] 05 00 Double etc answer -> 200_ again 0 in answer; bring down next pair for 05 00 00 Double etc -> 2000_ this time can put a 2 in the units place -> 20002, multiply by that 2 to get 40004 and subtract to get 99 96, bring down 00 Double etc -> 20 00 4_ this time a 4 to give 20 00 44 × 4 = 80 01 76 to subtract = 19 94 24; bring down 00 Double etc -> 2 00 04 8_ units = 9 to give 2 00 04 89 x 9 = 18 00 44 01 to subtract = 1 93 79 99 and repeat as required. (The "division" is easier to see (and do) when written out as a long division with the new divisor written down each time along side each "working" number, extending the vertical line of the divisiom "bus stop shelter" down the page.)

4k maths video, just what I needed XD

There is an issue in your correction at 2:55 . When you subtract the area of x-square. You first cut off 100+x times y. With this cut you leaf not enough area for the second cut. So, your new estimate is also overvalued and it’s overvalued by y square, because you can’t cut this square of. In the next step you need to subtract the new estimate by y square. And so on, and so on.

Hey man great video and you look way better short/bald like that!

I found a video not so long ago telling how to algebraicly (spelling no?) calculate the square root, digit by digit, without approximation. 1. Start off by having your number, in this case 10005. 2. Now write down the closest square root integer that's closest to your number without going over. In this case 100. 3. Next up, write down 100 at the start of the result line, and add a period. While also subtracting the square (10000) from 10005 and remembering that. In this case it's 5. 4. Next up, double the result, but without the decimal place, and remember the integer, in this case 200. 5. Now if you have decimal places of your starting number, bring down 2 of them and place them after your subtracted number. In this case we have 500. 6. We now do the same as we did on #2, but we take the closest integer x without going over where we can multiply 200_ * _ where the _ gets replaced by the digit x. In this case we can't so it's 0. 7. Once again we add 0 to the result, and we subtract from the number above so we still have 500. 8. Just like at #4, we double the result without the decimal place, and remember it, in this case 2000. 9. We now bring down 2 additional decimal places just like #5 and we end up with 50000. 10. Now repeat from #6 over and over and we will end up calculating digit by digit what the exact square root of the number is. To show an example, I will continue doing a few of the ones following. We need 2000_ * _ < 50000. In this case it's 2, since 20002 * 2 < 50000. The result is 40004. We subtract 40004 from 50000 and get 9996 while we also add 2 to the result. We turn 100.02 from the result into 10002 and then 20004, and we also carry down 2 decimals to have 999600. We do 20004_ * _ < 999600 and we get that it's 4 since 200044 * 4 = 800176 which is just as far as we can go. While the trade off of using this method is that the numbers used in the formula gets insanely huge quickly, it does still mean that if you're able to perform them, the answer will always be accurate, compared to the more used techniques which indeed is easier but they're not exact. Credits to RU-vid user Davidson1956 who made a video on this 10 years ago.

Nice pun with that root

Your sense of humor is amazing. Heisenberg haircut looks badass.

Damn, I thought the video will be about botanical research...

"How I was able to find the root..." *Trips on a root*

The way I was shown to do square roots by hand is using this formula (x+(r/x))/2 where r is the number you want the square root of and x is a number close to √r. If you plug the answer you get back into the formula as x, the result you get gets closer and closer to √r. ex. (2+(5/2))/2=(9/4)=2.25 ((9/4)+(5/(9/4)))/2=161/72≈2.2361111111... ((161/72)+(5/(161/72)))/2=51841/23184≈2.2360679779... which is a good approximation for √5, what I was trying to find: 2.2360679775...

Here I was beginning to think I was weird for using an estimation algorithm I learned a long time ago from a website about bit manipulation in C-like programming languages. I usually carry out the estimation in my head. Turns out it's actually very similar to this estimation technique you demonstrate here. Just adjusted for binary since I work just as well in binary as I do in decimal and conversion between them is simple enough. That and it's how I first learned of the algorithm before learning what it was approximating.

Yay, he's back! Math dad.

0:24 "there's a whole section on how I was able to find the roo...*trips on root*...t"

It is basically just the taylor series expansion (1+x)^m=1+mx+m(m-1)/2 x^2... For the special case of m=1/2, x=0.0005 and a scaling factor of 100 (10005)^(1/2)=100* (1+.0005)^(1/2) But this is a nice intuitive geometric illustration! Great job!

I like the new doo too. Matt's eyebrows now take center stage and rule the school.

At the end you were filming in Trafalgar SQUARE. Haha nice!

That ROOT totally made my day

"find the root" he says as he trips over a root. Great pun. A+

You could also do a taylor expansion from the closest perfect square!

“How I was able to find the root” Trips on a tree root.

3:25 Aside from the whole nicely approximating errors thing, it's nice to know that I'm not the only person who does the weird hovering raptor arm thing.

I remember when I ran across the technique of finding a square root by hand. I was fascinated. I didn't understand why it worked (and still don't, I'm a humanities guy), but it was fascinating.

4:12 "Another overlycomplicated VDt5Dvision....." Masterpiece

You could use Newton's Method to approximate the root of the polynomial x^2-n. Successive iterations of the method would yield better and better approximations of root n.

This is awesome

Subtle root placement; I appreciate that cheeky pun delivered through physical prop comedy. Thank you for that.

Very interesting!!!!

Cue-ball look is a good look.

I remember i used the same method to square numbers 10-20 with a simmilar method. I was like 7and somehow found this out.

Calculating a square root is a fairly simple process. It is similar to long division except you pair the dividend and do some additional calculations on each subsequent subtrahend. I learned it in primary school back when calculators were the size of buildings. We were still faster.

It’s a bit of a Parker sphere, though, tbh. (This video is going into my pocket as a demo of successive approximation techniques. Thanks for making it! In my historical math elective in undergrad, I did a small survey essay on ancient square root methods, which gave me a solid appreciation for this sort of thing. If you haven’t read The Crest of the Peacock, you really should!)

Very nice graphical method and a beautiful illustration of what (I think???) amounts to a Taylor expansion without resorting to calculus! Another nice method might be the Newton-Raphson method but again, needs calculus.

I appreciate the pun. You're at Piccadilly circus, which is actually a square and you speak about interest in finding a square, while being at circus that's actually a square. That's a good one

Nailing that grothendieck look!

Could we do a little binomial expansion? Factorials and fancy brackets would look good on a whiteboard.

Back in the day when desktop PCs executed 16-bit software at about 20MHz, without hardware floating point, where integer multiplication was slow and division was ridiculously slow, I implemented efficient code to do square roots. I used Newton's Method of approximations. If you have not covered that, you might consider it, especially with a focus on using only simple operations (add or subtract, multiply or divide by 2, bitwise logical operations like AND and OR, compare). In my application (Lambert shading) the value would not change much from one pixel to the next, so the last result is used as an approximation to start with. For successive pixels being rasterized, it could adjust it with a single iteration of Newton's method, and skip the overhead of the looping.

Awesome.

@ 0:27... "...able to find the *ugh, root"... As he trips over actual root... hahahahahaha Fell over laughing.

Shaven head looks good!

0:27 This root almost got you ! ;)

Don't assume your head is a sphere.

My favorite algorithm is a recursive one. It involves arrangement of the equation into the form f(x) = x, then plugging the estimate for x into f(x) and solving for a new estimate of x. From your intitial estimate of .025 I was able to come up with .024996875 relatively quickly compared to the amount of work you put in. However, the next interaction didn't buy me much for a very long hand multiplication. I suspect another iteration would be required to make the work pay off.

"...a whole section about how I was able to find the root..." Right as you trip over a root. I'm not sure if that was intentional or accidentally perfect. Either way, I loved it.

Another iterative technique that requires less thinking to do right is to take whatever estimate x that you have, and calculate (x+10005/x)/2. For a first guess of 100, it gives 100.025, and starting at 100.025 it gives too many correct digits for my phone calculator to tell the difference. (You still have to divide by 100.025, so calculation wise it may not be easier, but there are fewer things to get wrong). The idea is that what you're really after is the geometric mean of x and 10005/x, so you use the arithmetic mean instead, which is a little bit bigger. But with a good starting point, x and 10005/x will be quite close, and then the arithmetic mean and geometric mean are really close to one another.

Haircut's looking great.

I do like the "coincidence" of saying "finding the root" and almost stumbling on a root!

I like the Left Raptor arm between 3:22 and 3:38 lol

Liking the new haircut, Matt! :) Suits you

0:26 I see what you did there clever cookie.

Why don't you cover the method for determining an integer root by hand some time? The one where you break it up into groups of digits that match the base (2 for square, 3 for cube, 4, 5, etc.) and work it sort of like long division but with what gets subtracted being determined with the Pascal 's Triangle numbers.

"route", with a "root", to a "square". masterful

Nice touch ending in a square.

For next pi day, can we get a video of you manually approximating the pi root of pi?

I watched your show last Wednesday in Soho theatre (didn't meet you after the show. My non-nerd friend was in a hurry). Anyway, that Piendulum music was intense. Where can I find it?

What if you cannot factorize the number which you want to find the square root of to two distinct numbers one of which is already a perfect square? Can you do the same with square root 2. I wonder how it goes with this method ...

0:27 Ha! Ever the master of timing! You couldn't have made that up if you'd tried!

0:26 "I was able to find the *ROOT*" as he trips on a root lol

While this explanation works well for a square, I think that the general approach of Newton's method. In this particular case, you're doing exactly the same computations, but the use of newton's method is more versatile when a geometric interpretation is not so clear. In fact, for working out by hand, this newton's method approach gives you fewer digits in your division problem ( 625 vs 3125 and 20005 vs 100025). y(x):=x^2- A

It was your approximation of sqrt(10 005) that limited the precision of your approximation of pi. Using the value you got here for the Chudnovski algorithm first approximation would have given you about the same precission as you got for the second approximation - with a lot less work.

0:54 Classic parker square

0:27 How many takes did it take for you to have that joke go the way you wanted? Guh, it's so awful... and it's why I'm subscribed. :)

I saw a video of different way of approximating sqrt using the function y=x^2-c. Sqrt (c) is a solution. You start with a guess. And you calculate the slope at your guess and make a line from your guess with that slope. Find the 0 of that line. This is your new guess. This method doubles the number of digits at every step. Seems like your method also doubles the digits at every step.

0:28 "I was able to find the root" *trips over root*

A high pitched Mark Strong!

Do it with cubes!

I love how he tripped on a root while filming this

There is a simpler method which involves no estimating or rounding. Search "afjarvis square root" and there's a pdf file. It can even be done mentally, since there are only two variables to keep track of which change with every step of the algorithm.

I very much enjoy all your videos. Please consider eliminating the distracting "musical" accompaniment. It simply adds noise.

"...how I was able to find the roOT!" Falls over root.

4:24 oOo0o0o000ooo0OOoOooof LOL