My parents (now in their eighties) were taught the square root by hand method in grade 4. I learned it in my 40s when reading a book on abacus calculations. I wasn't quite sure why the algorithm worked and meant to derive a proof eventually. What you presented here is essentially the same algorithm. So now I don't have to figure it out. THANKS!
Here is another way of looking at the process. Think of the graph of y = x^2. It has a point on the curve (x, 17) and x is of course the square root of 17. We know that (4, 16) is pretty close. Mentally draw a tangent line at that point. The derivative of Y= x^2 is 2x, so the slope at our initial guess is 2*4=8. Delta x = delta Y / slope = 1/8. So the 2nd approximation is 4 + (17-16)/8.
This works, but accuracy increases in single iterations of newton's method for larger numbers, so you would get a more accurate calculation if you use larger numbers, and nearby perfect squares. For example: √999 ≈ 32 + (999 - 1024) / 64 To simplify, I would say do something similar to what you suggest. (999-1024) / 64 = 25/64 Multiply by 10. 250 / 64 = 256 / 64 - 3/32 The last term is pretty close to 1/10, so: 250 / 64 ≈ 3.9 Then divide by 10. 25 / 64 ≈ 0.39 So we're looking for 32 - 0.39 = (3200 - 39)/100, which yields 31.61. This squares to 999.19. The method you describe is great if you don't know a nearby perfect square, though.
This is the direct application of Newton's method to find the zero of a function: Let f(x)=x²-a, so that we want to find x such that f(x)=0 x²=a we then have x_n+1 = x_n - f(x_n)/f'(x_n) It becomes x_n+1 = x_n - (x_n²-a)/(2*x_n), which corresponds to the algorithm described in the video
@@voicutudor7331 me too. A simpler summary of this formula is that your next guess is the mean of x and a/x. Hence, x_n+1=(x_n+a/x_n)/2. If for sqrt(2) you know a good start is 7/5, then next one is 7/10+5/7=99/70 which is good for 4 decimal places.
Thanks for showing the table at 0:33. It shows clearly that the numbers from 1 to 59 were represented as two digits, where the leftmost digit shows the number of tens and the rightmost digit shows the number of ones. I.e., these numbers are represented very similar to the way we do, except that they didn't use a symbol for zero, they used different sets of digits for the tens and for the ones, and their maximum number of tens was 5.
Each of those is a single digit. The tens and units within a digit are not separate digits, they're subparts of the same digit. It's a common misconception about cuneiform numbers.
Babylonians did not have zero. In the 60 based system that was not so much of a problem as it rarely occurred. They did develop a place holder to express it but they only used it in the middle of numbers.
Many have pointed out that this is what you get if you use Newton's method (a.k.a. Taylor approximation of order one). It is also the result of the following: Let x be a first approximation of √s. Then we would very much like to calculate the geometric mean of x and s/x, as that is exactly √s. But we can't. What we CAN do is to calculate the arithmetic mean. And that yields exactly this method. As an added bonus, both Taylor approximation and the mean method have natural ways to bound their respective errors. Taylor has Taylor's theorem. And we know the geometric mean lies between the arithmetic and harmonic means.
When I saw the title, “How to calculate any square root”, I was initially excited-as I thought Presh had done a video about Square Root Long Division, a means to calculate, EXACTLY, any square root. It’s well worth looking up, and it would be a great subject for one of Presh’s videos. I actually used the technique on a standardized test-and was later accused of cheating by using a calculator, as I “couldn’t possibly have calculated that root” without a calculator. This video isn’t about how to CALCULATE a square root, but to ESTIMATE one. It should really be renamed.
Since the method in the video can be repeated, as many times as you wish, and the result is guaranteed to be much better for each repetition, there is no distinction between what this method accomplishes and what your long division method accomplishes. They are both repeatable methods that for each repetition gives you more correct digits. They also both never terminate in the case of irrational roots. So I see no objective reason why you would call one exact and the other an approximation.
He said that you can repeat the process as many times as necessary, meaning you can calculate the sqrt to any arbitary amount of precision that you want. And to be completely technical, any algorithm that we could use to calculate sqrts will always only be an approximation because we can only calculate up to a finite amount of precision regardless of how effecient it is
... but it uses the same logic. Notice the factor of 2. Repeated iterations yield lower and lower lines as one continues the square root by hand calls. The big difference is that this presentation allows picking a higher number square, whereas doing it long-divisionish style we always choose a smaller number.
@@ericfielding668 While I see your point, it’s not really the same logic, as with this technique, you start with a table of squares. This is great if you’re estimating the square root of a two digit number-but is that practical if you’re starting with, for example, an 8-digit number? Do you truly know what square is closest to that? With square root long division, there’s no need to know that-you derive it as you go.
Isn't 111 closer to 11² = 121 than 10²=100? at ~5:54; So sqrt(111) ~~11 - 10/22 = 11 - 5 / 11 = 10.545454... which is obviously closer to 10.536... (sqrt(111)) than 10.55 (the approximation given in the video by starting with 10²)
The closer you are to the squared value, the more accurate the adjustment value will be it seems. Since 111 is closer to 11² then using 11 + adjustment will be more accurate than 10 + adjustment indeed
I take exception that you are choosing to round some values and not round other values to make it seem like they were more accurate than they actually were. For example, at 5:09, you show 8.312 and 8.307. You did not round the first number up, but you rounded the second number up. It should be 8.3125 and 8.3066, and if you rounded them both, would be 8.313 and 8.307...
Calculating square root by hand is derived from this method. Of further interest is applying this scheme to binary numbers. It dramatically simplifies to appending -01 to the right of the existing answer to test against the remainder. If it "fits", append 1 to the right of the answer. If not, append 0 & discard the result of the subtraction.
9:05 where did you get the 2? Is it just a 2, or is it because the original value we're square-rooting is a 2? The Babylonians were clearing using base 60. Does that have an effect on how many terms are needed in base 10 to get the same accuracy? On 2nd thought, I'd say no, or we'd use base 60. But it is interesting.
The method I learned in grade school is to take the average of a and s/a. This gives the same result as Presh's "a + (s - a^2)/(2a)", but I find "(a + s/a) / 2" to be more intuitive. If "a" is a bit low, then s/a will be a bit high (and vice versa), so halfway between them will be closer than either of them.
This is Newton's iteration for the square root. Suppose you want to find the square root of "a". You start with an initial approximation x as in the video. The next approximation follows the formula 1/2(x + a/x) which is equivalent to the given formula. Simply repeat the formula starting with the new closer approximation. The number of correct digits approximately *doubles* with each iteration. So if one approximation has two correct digits, the number of correct digits in succeeding iterations will be approximately 4, then 8, then 16, then 32 and so on. Using newton's iteration for the square root of 2 as in the last example, after only 7 iterations we have 60 correct digits. I didn't know that the Babylonians had previously figured this out.
They didn't. As explained in the video, one can arrive at the method using geomtric means without knowing calculus. It would be possible, using similar reasoning, to arrive at the iterative approximation S≈a³+(S-a³)/3a² by splitting the difference over the three square faces of a cube, with sides 'a' to get a new approximation.
Another way to get a better value to approximate root(2) is to observe that root(98) = root(49)*root(2) = 7*root(2). Using the method, root(98) = approx 10 - 2/20 = 99/10. Dividing by 7 gives root(2) = approx 99/70 = 1.4142857... which is already a lot better than 1.5
Interesting that they can approximate square roots so well because it's easy to check the answer, whereas they couldn't compute pi nearly as well because there was no easy way to check the answer
Our past ancestors were so smart! The ways they went about solving problems show that thinking outside the box is key, not following some rigid process.
I was taught in 8th grade a system whereby you do long division taking two digits at a time in the numerator and the divisor. I don't remember it now but it was cool.
I am undergrade maths students . I want to now , how to make maths vidoe like you , I mean which software you use to demonstrate the mathematical equation and all this
(1+24/60+51/60²+10/60³)² =1+(24+24)/60+(24²+51+51)/60²+... =1+48/60+678/60²+... =1+48/60+11/60+18/60²+... =1+59/60+18/60²+... ~2 (I didn't do the ² for terms in 1/60³ and beyond)
Another way to get a more accurate approximation of sqr(2) would be to calculate sqr(200) and then divide the result by 10. Using this method, sqr(200) would be approximately 14 + 4/28, or 14.14286, so sqr(2) would be approximately 1.41429.
Strangely enough, I had come up with basically the same method earlier this year without knowing the Babylonian technique. My idea was that sqr(y) is some whole value a plus a fractional value x, so that y = a^2 + 2ax + x^2. With x being less than 1, x^2 can be ignored and x would be about (y - a^2) / 2a.
Very simple techniques, you can come up with your own approximations pretty easily, here are some other ways: (sqrt(p_2) - sqrt(p_1))(x - p_1) / (p_2 - p_2) + sqrt(p_1) (p_1 + sqrt(p_1) + x) / (2sqrt(p_1) + 1) (p_2 - sqrt(p_2) + x) / (2sqrt(p_2) - 1) Where x is the number you are finding the square root of, p_1 is the closest perfect square smaller than x, and p_2 is the closest perfect square greater than x. These actually approximate the square root of x more accurately when x is a larger number.
There is a faster algorithm which computers use, at least any computers that still calculate square roots using software. The algorithm use doubles the number of significant digits at each iteration, as opposed to the method in this video which produces additional significant digits linearly in the number of iterations. If you only want 4 or 5 digits, it doesn't really matter, but if you want 10 or 20 digits, an algorithm that doubles the number of digits each time is superior. On the other hand most modern computers do square roots in hardware with a single instruction. I have no idea how this is done, when even regular division is done weird (whence the Pentium division bug of yore)
Is there any reason you even need to start from the closest square other than minimizing how many times you go through the cycle? Shouldn't this keep getting closer regardless of what number you start with?
I would like to know,how they pick up 1,24,51,10 numbers. And kunoor represents only upto 60 numbers then how 60^2 and 60^3 they got and division also exists? How many such mathematical clay tablets ( or other forms) exists
Here's a recent rant I had on November 15: You might find it interesting as it has to do with root numbers, also. How would we calculate an ever finite to the root, IE 10 decimals? 1040496217739/2500000000000 or 0.4161984870956 : what would work for all units. 6/15 fist unit is 0.4... {(23)(24)(550) | 529/550/576, 21 | 49} this method is too large. hmm. To refine your estimate of the square root of 55, you can use a method called the Newton-Raphson method. It's an iterative numerical technique that gives progressively closer approximations. Here's a simplified version of how you can use it: (7.4 + 55/7.4) / 2 = 7.416216(216rp) this method is fairly close, but you can do this forever, basically going to add the last faction into the original 55, and it will pretty much push the 0000000 in the decimal place, I call it chasing the zero..., kind of like pushing a crease out of rug ,you can make your calculation more and more hwole. Whats going on is your making a fraction of your fraction, doubling and halfing it which hsould get you closer and closer to your target, either 99999 or 0000. in this case were always larger, going smaller, towards 55.. You can keep doing this method, to refine your answer.... 55.000xxxx I would then keep. 7.416 and refine the next set of zeros. so (7.416 + 55/7.416) / 2. 7.4161984897518878101402373247033 sq 55.000-000-0-3939818160448442913017 So I would count the 0's and take the decimals and repeat further pushing the zero. Your constantly fractioning your fraction in half to get closer to 55. 7.4161984 = 000,000,0 so then you go, (7.4161984 + 55/7.4161984) / 2 = 7.416198487095663460136125808069 ||= 0.4161984870956 ^^=0.4161984870956 How close did that make our square root? THis is an fundamental teaching for math that is often not taught. Seems like its really accurate, we can do this forever, or until its whole, but they typically because of how math is designed in a language the root of 55, is never resolved. like most non squares, since, its liek another way to look at it griding down forever, in a scale, there is information lost, more often the proper solution would be to multply teh desire root of 55, into a gridable whole number, to its origin, this works in reality if yo uwant a root 55 wall, you simply multiply it around into a whole number, or change the scale, since you can measure in shoelaces if you wanted to. So math is misleading, as all numbers are whole, decimals can be useful, but typically most peopel adjust unresovables to non fractions for a means of working with, which provides better accuracy of hte main subject whatever it is. IE temperature guage for specific elements verses water works to convey functional relation information. Anyways you would repeat this forever, the nubmer squared is: 55.000-000-000-000-00-758565459356129 14th decimal place 7.416-198-487-095-66-3-460136125808069 7.41619848709566((repeat function)) 7.4161984870956629487113974408013 55.000-000-000-000-000-000-000-000-000-00-9 The next one we take the 29 decimal place and repeat the same method... and get a more precise answer. This method although, 6/15 helps, from the interrelated powers, you just need a number close to it, and then you always put your number (target55)/ divided and added to itself since your rounding the square or flattening the carpet, lol You could start wit hseven and it would end up self sorting itself out by dividing, as long as you do itself, This also works in cubes and more exponenets, just increases the adding of the self, to refine teh decimal place of a cube. -.- This is good for art, spacing and your giving nonsense numbers, or whatever.
That was the system they used. The 1.5 would be 1 30. Then 17/12 would be 1 25. If you square it you get 2 0 25. Subtract 2 from it and you get just 0 0 25. Divide it by 2 50 and you get 0 0 8 49. Now subtract that from 1 25 and you get 1 24 51 11. Well they got 10 as the last. Note they would not use zeroes so 0 0 25 would be just 25. They would somehow keep track of the magnitude like one does with a slide rule. The precision is way more than used in any engineering so they were interested in pure math.
Thought the same at first too, but in this case it is about to use the next *smaller* square number so that the difference between the two stays positive. This way you can already say in advance that the result will be 10,... (in this case). It would also be legitimate to take 121 though since in the next example (6:00) for sqrt(23) it is shown that it works for negative differences as well (since it's more intuitive to take 25 as next square number to 23 instead of 16).
The problem with IQ tests, is that they are predictable. If you have no idea what types of questions they will ask, you will not do so great on your first attempt. But if you take a few example exams in advance so that you know what types of questions will be on it, you will at least double your score on the next attempt. But if it is supposed to measure a constant intelligence value, then how could you suddenly become twice as smart just by preparing a little in advance? The first time I took one, I had no idea what types of questions were going to be on it, and I got a certain score. Then I just took it again, with all new questions, but I knew what to expect and my score was 30% better, at least. Did I suddenly become 30% smarter after taking the first test?
My parents (now in their eighties) were taught the square root by hand method in grade 4. I learned it in my 40s when reading a book on abacus calculations. I wasn't quite sure why the algorithm worked and meant to derive a proof eventually. What you presented here is essentially the same algorithm. So now I don't have to figure it out. THANKS!
Here is another way of looking at the process. Think of the graph of y = x^2. It has a point on the curve (x, 17) and x is of course the square root of 17. We know that (4, 16) is pretty close. Mentally draw a tangent line at that point. The derivative of Y= x^2 is 2x, so the slope at our initial guess is 2*4=8. Delta x = delta Y / slope = 1/8. So the 2nd approximation is 4 + (17-16)/8.
So Newton's method
For large numbers, divide by 100 before and multiply 10 after.
√999?
√9.99 ≈ 3 + .99/6
√999 ≈ 30 + 9.9/6 = 31 + 3.9/6 = 31.65
Actual value 31.6069
This works, but accuracy increases in single iterations of newton's method for larger numbers, so you would get a more accurate calculation if you use larger numbers, and nearby perfect squares.
For example:
√999 ≈ 32 + (999 - 1024) / 64
To simplify, I would say do something similar to what you suggest.
(999-1024) / 64 = 25/64
Multiply by 10.
250 / 64 = 256 / 64 - 3/32
The last term is pretty close to 1/10, so:
250 / 64 ≈ 3.9
Then divide by 10.
25 / 64 ≈ 0.39
So we're looking for 32 - 0.39 = (3200 - 39)/100, which yields 31.61. This squares to 999.19.
The method you describe is great if you don't know a nearby perfect square, though.
This is the direct application of Newton's method to find the zero of a function:
Let f(x)=x²-a, so that we want to find x such that f(x)=0 x²=a
we then have x_n+1 = x_n - f(x_n)/f'(x_n)
It becomes x_n+1 = x_n - (x_n²-a)/(2*x_n), which corresponds to the algorithm described in the video
Fascinating
i was searching for a comment that realised this 😂
@@voicutudor7331 me too. A simpler summary of this formula is that your next guess is the mean of x and a/x. Hence, x_n+1=(x_n+a/x_n)/2. If for sqrt(2) you know a good start is 7/5, then next one is 7/10+5/7=99/70 which is good for 4 decimal places.
You should mention that is exactly what you get if you apply Newtons' method.
Thanks for showing the table at 0:33. It shows clearly that the numbers from 1 to 59 were represented as two digits, where the leftmost digit shows the number of tens and the rightmost digit shows the number of ones. I.e., these numbers are represented very similar to the way we do, except that they didn't use a symbol for zero, they used different sets of digits for the tens and for the ones, and their maximum number of tens was 5.
Not exactly as they did have a symbol for 10 which could not be expressed only with 1..9.
Each of those is a single digit. The tens and units within a digit are not separate digits, they're subparts of the same digit.
It's a common misconception about cuneiform numbers.
Babylonians did not have zero. In the 60 based system that was not so much of a problem as it rarely occurred. They did develop a place holder to express it but they only used it in the middle of numbers.
Why did you choose the number “69”?
Nice
There's an old joke (that only works when spoken aloud) "What's the square root of 69? 8 something."
"What's the square-root of '69'?" "Ate something!"
Because he's nice.
Very nice! This reminds me of high school days! Enjoyed very much haha
Many have pointed out that this is what you get if you use Newton's method (a.k.a. Taylor approximation of order one). It is also the result of the following: Let x be a first approximation of √s. Then we would very much like to calculate the geometric mean of x and s/x, as that is exactly √s. But we can't. What we CAN do is to calculate the arithmetic mean. And that yields exactly this method.
As an added bonus, both Taylor approximation and the mean method have natural ways to bound their respective errors. Taylor has Taylor's theorem. And we know the geometric mean lies between the arithmetic and harmonic means.
I'm sure the Ancient Babylonians knew this Taylor guy intimately
I have enjoyed this video thoroughly. Thank you very much for sharing.
When I saw the title, “How to calculate any square root”, I was initially excited-as I thought Presh had done a video about Square Root Long Division, a means to calculate, EXACTLY, any square root. It’s well worth looking up, and it would be a great subject for one of Presh’s videos. I actually used the technique on a standardized test-and was later accused of cheating by using a calculator, as I “couldn’t possibly have calculated that root” without a calculator.
This video isn’t about how to CALCULATE a square root, but to ESTIMATE one. It should really be renamed.
Since the method in the video can be repeated, as many times as you wish, and the result is guaranteed to be much better for each repetition, there is no distinction between what this method accomplishes and what your long division method accomplishes.
They are both repeatable methods that for each repetition gives you more correct digits. They also both never terminate in the case of irrational roots. So I see no objective reason why you would call one exact and the other an approximation.
He said that you can repeat the process as many times as necessary, meaning you can calculate the sqrt to any arbitary amount of precision that you want.
And to be completely technical, any algorithm that we could use to calculate sqrts will always only be an approximation because we can only calculate up to a finite amount of precision regardless of how effecient it is
... but it uses the same logic. Notice the factor of 2. Repeated iterations yield lower and lower lines as one continues the square root by hand calls. The big difference is that this presentation allows picking a higher number square, whereas doing it long-divisionish style we always choose a smaller number.
@@ericfielding668 While I see your point, it’s not really the same logic, as with this technique, you start with a table of squares. This is great if you’re estimating the square root of a two digit number-but is that practical if you’re starting with, for example, an 8-digit number? Do you truly know what square is closest to that? With square root long division, there’s no need to know that-you derive it as you go.
This is also essentially the same as the Newton's method. The 2 in the bottom can be explained by the fact that the derivative of x² is 2x.
Isn't 111 closer to 11² = 121 than 10²=100? at ~5:54; So sqrt(111) ~~11 - 10/22 = 11 - 5 / 11 = 10.545454... which is obviously closer to 10.536... (sqrt(111)) than 10.55 (the approximation given in the video by starting with 10²)
Yes, but he didn't have 11 on his table
The closer you are to the squared value, the more accurate the adjustment value will be it seems. Since 111 is closer to 11² then using 11 + adjustment will be more accurate than 10 + adjustment indeed
I noticed that right away also.
I take exception that you are choosing to round some values and not round other values to make it seem like they were more accurate than they actually were.
For example, at 5:09, you show 8.312 and 8.307. You did not round the first number up, but you rounded the second number up. It should be 8.3125 and 8.3066, and if you rounded them both, would be 8.313 and 8.307...
Calculating square root by hand is derived from this method. Of further interest is applying this scheme to binary numbers. It dramatically simplifies to appending -01 to the right of the existing answer to test against the remainder. If it "fits", append 1 to the right of the answer. If not, append 0 & discard the result of the subtraction.
9:05 where did you get the 2? Is it just a 2, or is it because the original value we're square-rooting is a 2?
The Babylonians were clearing using base 60. Does that have an effect on how many terms are needed in base 10 to get the same accuracy? On 2nd thought, I'd say no, or we'd use base 60. But it is interesting.
The method I learned in grade school is to take the average of a and s/a. This gives the same result as Presh's "a + (s - a^2)/(2a)", but I find "(a + s/a) / 2" to be more intuitive. If "a" is a bit low, then s/a will be a bit high (and vice versa), so halfway between them will be closer than either of them.
This is Newton's iteration for the square root. Suppose you want to find the square root of "a". You start with an initial approximation x as in the video. The next approximation follows the formula 1/2(x + a/x) which is equivalent to the given formula. Simply repeat the formula starting with the new closer approximation. The number of correct digits approximately *doubles* with each iteration. So if one approximation has two correct digits, the number of correct digits in succeeding iterations will be approximately 4, then 8, then 16, then 32 and so on. Using newton's iteration for the square root of 2 as in the last example, after only 7 iterations we have 60 correct digits.
I didn't know that the Babylonians had previously figured this out.
Now we know newton took this from the Babylonians 😂
Yes I noticed this its very very similar to the newtons method because it is an iterative method
They didn't. As explained in the video, one can arrive at the method using geomtric means without knowing calculus.
It would be possible, using similar reasoning, to arrive at the iterative approximation S≈a³+(S-a³)/3a² by splitting the difference over the three square faces of a cube, with sides 'a' to get a new approximation.
@@patelk464 calculus originates from approximation of smooth curves by triangles. So one can say calculus is one of the many geometric means.
Another way to get a better value to approximate root(2) is to observe that root(98) = root(49)*root(2) = 7*root(2).
Using the method, root(98) = approx 10 - 2/20 = 99/10. Dividing by 7 gives root(2) = approx 99/70 = 1.4142857... which is already a lot better than 1.5
How to APPROXIMATE any square root. Misleading title :(
What an irrational comment 🤭
@@ciscou omg hahaha you didn't!
Wow amazing......what a fantastic way of Babylonian
Interesting that they can approximate square roots so well because it's easy to check the answer, whereas they couldn't compute pi nearly as well because there was no easy way to check the answer
F(X)=√X
Taylor series:
F(X+e)=F(X)+F'(X)*e+F''(X)/2*e²+O(e³)
√(X+e)~√X+e/(2√X)-e²/(4X√X)
Very interesting!
And clever cleaver!
you can also just use point-slope form using the next closest whole number square root where the slope is equal to the derivative at that point
Our past ancestors were so smart! The ways they went about solving problems show that thinking outside the box is key, not following some rigid process.
And then this "outside the box" becomes rigid process
@Noobish_Monk Ah! The staight-jacket of the mind. Must guard against!
this is quite wrong
"thousands of years ago, before LLMs and the declared onside kick..."
The automatic captions call you "Press Tow Walker"
Just basically did this with Linear Approximation for the first half in my AP Calculus Class.
I was taught in 8th grade a system whereby you do long division taking two digits at a time in the numerator and the divisor. I don't remember it now but it was cool.
I am undergrade maths students . I want to now , how to make maths vidoe like you , I mean which software you use to demonstrate the mathematical equation and all this
Really interesting, thank you
(1+24/60+51/60²+10/60³)²
=1+(24+24)/60+(24²+51+51)/60²+...
=1+48/60+678/60²+...
=1+48/60+11/60+18/60²+...
=1+59/60+18/60²+...
~2
(I didn't do the ² for terms in 1/60³ and beyond)
for 111 you should subtract from 121 as it is closer to 111
Another way to get a more accurate approximation of sqr(2) would be to calculate sqr(200) and then divide the result by 10.
Using this method, sqr(200) would be approximately 14 + 4/28, or 14.14286, so sqr(2) would be approximately 1.41429.
Strangely enough, I had come up with basically the same method earlier this year without knowing the Babylonian technique. My idea was that sqr(y) is some whole value a plus a fractional value x, so that y = a^2 + 2ax + x^2. With x being less than 1, x^2 can be ignored and x would be about (y - a^2) / 2a.
You need to write *"sqrt."* That is short for "square root," not "sqr."
@@jhouck1969 -- 2a in the denominator in must be inside grouping symbols, such as (2a).
Imagine how far the Babylonians could have gone if their writing system wasn’t so ill-equipped..
wow. drake can finally have his question answered
May be they knew, may be they didn't but looks like they were doing "d/dx of x^2"
Very simple techniques, you can come up with your own approximations pretty easily, here are some other ways:
(sqrt(p_2) - sqrt(p_1))(x - p_1) / (p_2 - p_2) + sqrt(p_1)
(p_1 + sqrt(p_1) + x) / (2sqrt(p_1) + 1)
(p_2 - sqrt(p_2) + x) / (2sqrt(p_2) - 1)
Where x is the number you are finding the square root of, p_1 is the closest perfect square smaller than x, and p_2 is the closest perfect square greater than x. These actually approximate the square root of x more accurately when x is a larger number.
Thanks!
"Geometry is God's path to true knowledge"
- A pitagorean student 😮
There is a faster algorithm which computers use, at least any computers that still calculate square roots using software. The algorithm use doubles the number of significant digits at each iteration, as opposed to the method in this video which produces additional significant digits linearly in the number of iterations. If you only want 4 or 5 digits, it doesn't really matter, but if you want 10 or 20 digits, an algorithm that doubles the number of digits each time is superior.
On the other hand most modern computers do square roots in hardware with a single instruction. I have no idea how this is done, when even regular division is done weird (whence the Pentium division bug of yore)
Walk like an Egyptian, and root like a Babylonian. 💀💀💀
We want JEE level question based on logarithm.Please❤❤❤❤
Is there any reason you even need to start from the closest square other than minimizing how many times you go through the cycle? Shouldn't this keep getting closer regardless of what number you start with?
looks a lot like newton's method. impressive
Please try to make for calculating the value sin thetha without any use of logarithmic table or any calculator
Put the angle in grouping symbols: sin(theta).
I would like to know,how they pick up 1,24,51,10 numbers. And kunoor represents only upto 60 numbers then how 60^2 and 60^3 they got and division also exists? How many such mathematical clay tablets ( or other forms) exists
wait. 111 is closest to 11 squared. did you mean “ closest to but not greater than?”
Very smart ! ....
Why they didn't take an ipad instead of a clay tablet ?
lmao
lmao
The iPad didn't have a calculator app back then.
Good question. Apples definitely existed in Babylon. Maybe they were too expensive?
Here's a recent rant I had on November 15: You might find it interesting as it has to do with root numbers, also.
How would we calculate an ever finite to the root, IE 10 decimals? 1040496217739/2500000000000 or 0.4161984870956 : what would work for all units. 6/15 fist unit is 0.4... {(23)(24)(550) | 529/550/576, 21 | 49} this method is too large. hmm. To refine your estimate of the square root of 55, you can use a method called the Newton-Raphson method. It's an iterative numerical technique that gives progressively closer approximations. Here's a simplified version of how you can use it: (7.4 + 55/7.4) / 2 = 7.416216(216rp) this method is fairly close, but you can do this forever, basically going to add the last faction into the original 55, and it will pretty much push the 0000000 in the decimal place, I call it chasing the zero..., kind of like pushing a crease out of rug ,you can make your calculation more and more hwole. Whats going on is your making a fraction of your fraction, doubling and halfing it which hsould get you closer and closer to your target, either 99999 or 0000. in this case were always larger, going smaller, towards 55.. You can keep doing this method, to refine your answer.... 55.000xxxx I would then keep. 7.416 and refine the next set of zeros. so (7.416 + 55/7.416) / 2. 7.4161984897518878101402373247033 sq 55.000-000-0-3939818160448442913017 So I would count the 0's and take the decimals and repeat further pushing the zero. Your constantly fractioning your fraction in half to get closer to 55. 7.4161984 = 000,000,0 so then you go, (7.4161984 + 55/7.4161984) / 2 = 7.416198487095663460136125808069 ||= 0.4161984870956 ^^=0.4161984870956 How close did that make our square root? THis is an fundamental teaching for math that is often not taught. Seems like its really accurate, we can do this forever, or until its whole, but they typically because of how math is designed in a language the root of 55, is never resolved. like most non squares, since, its liek another way to look at it griding down forever, in a scale, there is information lost, more often the proper solution would be to multply teh desire root of 55, into a gridable whole number, to its origin, this works in reality if yo uwant a root 55 wall, you simply multiply it around into a whole number, or change the scale, since you can measure in shoelaces if you wanted to. So math is misleading, as all numbers are whole, decimals can be useful, but typically most peopel adjust unresovables to non fractions for a means of working with, which provides better accuracy of hte main subject whatever it is. IE temperature guage for specific elements verses water works to convey functional relation information. Anyways you would repeat this forever, the nubmer squared is: 55.000-000-000-000-00-758565459356129 14th decimal place 7.416-198-487-095-66-3-460136125808069 7.41619848709566((repeat function)) 7.4161984870956629487113974408013 55.000-000-000-000-000-000-000-000-000-00-9 The next one we take the 29 decimal place and repeat the same method... and get a more precise answer. This method although, 6/15 helps, from the interrelated powers, you just need a number close to it, and then you always put your number (target55)/ divided and added to itself since your rounding the square or flattening the carpet, lol You could start wit hseven and it would end up self sorting itself out by dividing, as long as you do itself, This also works in cubes and more exponenets, just increases the adding of the self, to refine teh decimal place of a cube. -.- This is good for art, spacing and your giving nonsense numbers, or whatever.
Why on the tablet were the denominators always a power of 60. This is not part of the taught method.
That was the system they used. The 1.5 would be 1 30. Then 17/12 would be 1 25. If you square it you get 2 0 25. Subtract 2 from it and you get just 0 0 25. Divide it by 2 50 and you get 0 0 8 49. Now subtract that from 1 25 and you get 1 24 51 11. Well they got 10 as the last. Note they would not use zeroes so 0 0 25 would be just 25. They would somehow keep track of the magnitude like one does with a slide rule.
The precision is way more than used in any engineering so they were interested in pure math.
Isn't 111 closer to 121 (11 x 11) than to 100?
Thought the same at first too, but in this case it is about to use the next *smaller* square number so that the difference between the two stays positive. This way you can already say in advance that the result will be 10,... (in this case). It would also be legitimate to take 121 though since in the next example (6:00) for sqrt(23) it is shown that it works for negative differences as well (since it's more intuitive to take 25 as next square number to 23 instead of 16).
My way is better but not gonna tell, thanks for teaching be a plan B
I'm extremely curious Presh, what is your IQ?
The problem with IQ tests, is that they are predictable. If you have no idea what types of questions they will ask, you will not do so great on your first attempt. But if you take a few example exams in advance so that you know what types of questions will be on it, you will at least double your score on the next attempt. But if it is supposed to measure a constant intelligence value, then how could you suddenly become twice as smart just by preparing a little in advance?
The first time I took one, I had no idea what types of questions were going to be on it, and I got a certain score. Then I just took it again, with all new questions, but I knew what to expect and my score was 30% better, at least. Did I suddenly become 30% smarter after taking the first test?
I did an IQ test once and it says my IQ was squirrel
You could do a binomial approximation of sqrt(a^2 + b), which the first order should give, to my knowledge a + b/2a.
You need grouping symbols: a + b/(2a).
1st
The whole idea loses its sense at the moment you start squaring at 9:02.
No explanation as to why we multiply by 2
Because √s = a + b/(2a) . The proof was towards the end if the video
Your videos always leave a mark in my heart and mind. Thank you for your creativity and passion!🪀🔞🦖
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