@Mr H: On the last example, instead of 100-97, you could use 97-100 (as same as in the first two examples) and this would lead to -3 on the numerator. So the method works the same way for the upper as well as the lower approach.
@AizenSosukesama
19 күн бұрын
Thats very helpful,thank you!!
@incendiohawk1725
19 күн бұрын
This can be generalised into nth root of (x^n + a)^(1/n) ≈ x + a/(nx^(n-1)) e.g. cube root of 70 = (4^3+6)^(1/3) ≈ 4 + 6/((3)(4²)) = 4.125. The actual value is 4.121 to 3 decimal places
@rogerphelps9939
12 күн бұрын
This is just Newton's method for successive approximaion of a square root.
@StereoSpace
18 күн бұрын
Awesome teaching, as always with Mr H.
@MeMadeIt
19 күн бұрын
Thx! 👍 I had long forgotten how to do this. 😄 Since school, I've mostly just guesstimated the difference between the radicand and its closest perfect squares. 😄
@hywell_
19 күн бұрын
Just learned linear approximation recently so it finally makes a bit more sense to me now why this method works.
@homosapien6031
19 күн бұрын
Just curious. Is that linear algebra or…?
@homosapien6031
19 күн бұрын
Asking cuz I’ve taken up to Calc 2 so far and haven’t really gotten there ywt
@hywell_
19 күн бұрын
@@homosapien6031 For me this is Calc 1 but our curriculums may be different - it's basically drawing a tangent line at a close known value and using that tangent line to approximate the answer (therefore, linear approximation). I've seen this method before from another channel (maybe it was bprp) but only now do I realize that this method follows the formula used for linear approximation.
@homosapien6031
19 күн бұрын
@@hywell_ ohh I vaguely remember that thanks. I took AP Calc BC a few years ago in truth. That covers up to Calc 2 so that’s why I said that, but it’s been a few years, so I forgot I learned that
@t.b.4923
18 күн бұрын
this is the first term of the taylor expansion of sqrt(1+x). Assume we have sqrt(b) if you take the closest square out, lets call it a then the rest will be sqrt( a^2 + (b - a^2)) = sqrt(a^2)*sqrt( 1 + (b-a^2)/a^2). Now we have in in form of the taylor expansion. (basically approximating a function with a polynomial that have the same y value , derivative, second derivative ect. ) The taylor exppansion of sqrt(1+x) is 1 + x/2 - x^2/8 ... substituting the formula derived earlier we have (and only using the first term because the series converges rapidly) 1 + b-a^2/2a^2, which is the formula of the video.
@dannmann17
15 күн бұрын
You are an awesome teacher👍🏻🇺🇸
@smalin
16 күн бұрын
I'd like to see a geometric explanation of why this approximation works.
@bjorncedervall5291
15 күн бұрын
Check out Newton Raphson method (my 72th birthday today - still remember the essence - made many iterations & approximations since I was young and before I learned more rational approaches - my teachers never understood what I did...).
@anotherelvis
14 күн бұрын
Alternatively start with the equation of a tangent line (or Taylor expansion) t(x)=f(x0) + f'(x0)*(x-c0) Insert f(x0)=sqrt(x0), and f'(x0)=1/(2*sqrt(x0)) to get t(x)=sqrt(x0) + (x-x0)/(2*sqrt(x0))
@bjorncedervall5291
14 күн бұрын
@@anotherelvis I like Taylor expansions - such an approach was actually the basis for a short radiation biology publication I and a British scientist wrote in 2002*. It was a relatively simple thing - the key was just to realize that Taylor expansion could be used to make an important point (about fragmentation of irradiated DNA depending on chromosomal sizes and related to the species). * The Fraction of DNA Released on Pulsed-Field Gel Electrophoresis Gels may Differ Significantly between Genomes at Low Levels of Double-Strand Breaks. Radiation Research, Vol. 158(2), pp. 247-249.
@Loots1
13 күн бұрын
@@bjorncedervall5291 cool!
@kaijabyenkya
17 күн бұрын
This is really cool!
@MohammedShayaanX
19 күн бұрын
Thank you sir
@whatzause
12 күн бұрын
Very good and will be occasionally useful. I enjoyed your presentation. What kind of audio system are you using? I found the audio somewhat muffled due to an apparent deficiency in the high frequency range. Therefore, it would have been better if I could have understood what you said better.
@shabbirghulam5454
18 күн бұрын
Thank you so much Sir ❤
@gregwolter1536
18 күн бұрын
Thank you, sir.
@Phylaetra
17 күн бұрын
It is important to keep in mind - you are not actually finding the square root - you are approximating the square root. The approximation it the worst what the number is between two squares; for example 156 is 12 more than 144 or 13 less than 169, giving you approximations of 12.500 either way; but the actual root is 12.490. There are two methods that you can use to calculate a square root - one is Newton's method (using a method of his from Calculus, though the actual method was known in Babylonian times) and another is called the 'long-division method', so named because it looks kind of like long division. Newton's method is very fast, and pretty easy to do. Guess what the root of your number n is (say r_1), then find the average of r_1 and n/r_1; call this number r_2. Repeat - find the average of r_2 and n/r_2. Continue until you are as close as you need to be. For 150, we can start with 12: average 12 and 150/12: 1/2 * (12 + 150/12) = 1/2 * (144/12 + 150/12) = 1/2 * (294/12) = 147/12 = 12.250 But if we do that again we get: 43209/3527 = 12.24744898, the actual value being 12.24744871 -- off in only the last two decimal places! You need to work it out as fractions - but you double the number of accurate decimal places with each time you do it. The 'long division' method is a bit more difficult to show in a comment like this, but it has the advantage of stopping if you have a perfect square, and everything stays as a decimal. It adds one decimal point with each row, basically, but - each additional decimal point takes a little more work.
@paulstudier5706
16 күн бұрын
Use a slide rule to get the square root to 2 or 3 decimal places, then apply this method. This saves an iteration.
@Phylaetra
16 күн бұрын
@@paulstudier5706 Assuming you have a slide rule (which _I_ do, but not everyone does...)
@samueldeandrade8535
16 күн бұрын
Hahahahaha. You saying "you are not actually finding the square root - you are approximating the square root" is so funny. Hahahahahahahahahaha.
@brnmcc01
16 күн бұрын
@@paulstudier5706 Using a slide rule what you are actually doing though is taking the log of the number, and dividing it in half. Then raising the base to the power of the answer. For example, log to the base 10 of 156 is 2.19312 and 10 to the power of half of that is 12.49. 12.49*12.49=156.0001 Close enough
@charlespartrick528
19 күн бұрын
Fantastic - another great method I can teach my students.
@-.-4
17 күн бұрын
Cool, I haven’t done square roots since school in the early 1970”s.😊 and we didn’t have calculators just paper ,pencil , and slide rule. I didn’t get a calculator til I was in college. Part of me wants to learn advanced math again. Now that I’m older, things make more sense 😂 I’ve subscribed.
@rivenoak
15 күн бұрын
woah, that was not discussed in my math lessons. but definitely should be in curriculum !
@nopaandriananugraha9473
19 күн бұрын
very nice
@robhill5732
13 күн бұрын
He is using the derivative of X^0.5 to find the fractional part added to the perfect square. Easy!
@Assassin4174
16 күн бұрын
Thank you Sir.
@naderhumood1199
17 күн бұрын
Great apprauch thanks v much
@bowlineobama
2 күн бұрын
I have never seen this method before, but you still have to deal with fractions which you may still have to use a calculator to take care of the fraction portion. I recommend using the binomial expansion method.
@bearohan
19 күн бұрын
chad math teacher❤
@user-in8hu9yt2h
19 күн бұрын
Wow thank you
@tvesaatamannamohanty5948
18 күн бұрын
Thank you so much 😭🙏
@Aeyo
19 күн бұрын
Sir, I have a question. In the denominator of decimal part it is 2 multiplied by the given number. Had this been the case of cube root finding, then would we use 3 in place of 2 in the decimal part?
@carultch
12 күн бұрын
Yes indeed. In a general sense for finding the nth root with a method similar to this, we rewrite this nth root as a power of 1/n. nth root of x = x^(1/n) Then, we use the power rule to take the derivative: d/dx x^p = p*x^(p - 1) Apply for p = 1/n: d/dx nth root of x = 1/n*x^(1/n - 1) So as you can see, this is where the 1/2 comes form for square roots, and where 1/3 would come from for cubic roots. Plug in x = x0, for the reference point. Call this k. k = d/dx x^(1/n) at x = x0: k = 1/n*x0^(1/n - 1) The tangent line L(x) to the original x^(1/n) function, will therefore be: L(x) = k*(x - x0) + x0^(1/n)
@kingbeauregard
18 күн бұрын
Here is a reason to do this even if you have a calculator: if you are, for whatever reason, trying to find the difference between sqrt(64) and sqrt(64.000001). Your calculator may have trouble calculating that, or at least displaying it with enough precision.
@CommDao
19 күн бұрын
I'm not smart, but this made me feel like I have the potential to be smart. 🙏
@sharmota2760
19 күн бұрын
Thank you
@njd2342
17 күн бұрын
Square numbers until you get close.
@claireli88
19 күн бұрын
This method is derived from the approximation of the first two terms of the binomial expansion (a+x)½ where a is the required perfect square.
@okaro6595
19 күн бұрын
It can also be seen as use of the Newton's method for x²-65=0. Just the signs are slightly different. That also explains why the constant is 2. In the Newton's method it would be 8 - (64-65) / (2*8)
@Thauan7020
19 күн бұрын
Incrivel. Calculei como seria o 81 eu posso fazer isso? Ou é só o número mais proximo? Incredible. I calculated what 81 would be like. Can I do this? Or is it just the closest number?
@joshi1q2w3e
19 күн бұрын
So, why does this actually work? I understand how but not why…
@crix_h3eadshotgg992
18 күн бұрын
I think it’s a the first term of the expansion of its Taylor series. What does this actually mean? Well, each function, like sqrt(x) can be represented as an infinite series. Each term of the series is a fraction, and each successive term of the fraction is smaller and smaller in value. So if you just use the first term, you’ll already be close enough to have a good approximation. I don’t actually know how that works (still haven’t had Taylor series, but did have the series of e^x, sin(x) and cos(x), and another commenter might’ve put the thought in my mind. You can look up “Taylor Series” on Wikipedia and try to change the language to “simple English” for a better understanding. Hope this helps!
@kevinchen9389
17 күн бұрын
In sqrt(x), he made a tangent line on the point where x = 64, then he took the derivative of sqrt(x), which is 1/2sqrt(x), then he plugged it in, 1/2sqrt(64), which means the tangent line is x/16, and since 65-64 = 1, you do 1/16 * 1, and added to 8, and you get 8 1/16. This also means that the larger the number is, the better it approximates it.
@nafnist
16 күн бұрын
This is not finding the squareroot, but doing an approximation.
@roland3et
6 күн бұрын
Nice trick**, but (@MrH) you better should have called it _estimate_ instead of _finding_ the sqrt, don't you think? 🙂👻 ** more a clever idea using the derivative of the sqrt-function for a pretty good estimate rather than a 'trick'.
@mahmoudibra5822
17 күн бұрын
Subtle Way...thanks
@cejII
16 күн бұрын
Why is the constant always 2 in the denominator? And would this metbod work with cube root?
@samueldeandrade8535
16 күн бұрын
Because (√x)' = 1/(2√x)
@carultch
11 күн бұрын
This comes from finding the derivative (i.e. slope of the tangent line) of y=sqrt(x), and then using it to construct a tangent line called linearization, to approximate it in the neighborhood of a known square root. For a general power function, y=x^p, the derivative is: d/dx x^p = p*x^(p - 1) For square roots, p=-1/2. For roots in general, the power p = 1/n. Plug in p=1/n and get: d/dx x^(1/n) = (1/n)*x^(1/n - 1) Evaluate at your reference point, x0, and call this slope k: k = 1/n*x0^(1/n - 1) Build your linear approximation L(x) from k and the reference point: L(x) = k*(x - x0) + x0^(1/n) For n=2 for square roots: k = 1/2*x0^(1/2 - 1) = 1/(2*sqrt(x0)) L(x) = 1/(2*sqrt(x0)) * (x - x0) + sqrt(x0) For n=3 for cube roots: k = 1/3*x0^(1/3 - 1) = 1/(3*cbrt(x0)^2) L(x) = 1/(3*cbrt(x0)^2) * (x - x0) + cbrt(x0)
@philippecanepa4509
14 күн бұрын
It’s tricky but you don’t really give the mathematical explanation. Maybe the Taylor expansion !
@carultch
11 күн бұрын
This is the first term of the Taylor series, which is called linear approximation. You construct a tangent line at the reference point, and use it to approximate a square root in the same neighborhood.
@wraith6776
13 күн бұрын
Since it's an approximate answer 8 is acceptable. There isn't anything saying where to round to.
@PrithwirajSen-nj6qq
12 күн бұрын
I beg to mention this process has limitation. As for example According to ur process √36=5 + 11/5*2=6.1 It is higher than actual so root of 36
@timeonly1401
2 күн бұрын
Well… In this approximation technique, you’re supposed to start with the CLOSEST perfect square to the number you’re trying to find the square root of. The closest perfect square to 36 is… 36 (= 6^2) with a difference of 0. So, this technique give approximation: 6 + 0/(2*6) = 6 + 0 = 6, which is as close to the square root of 36 as you could get!! 😂
@PrithwirajSen-nj6qq
2 күн бұрын
@@timeonly1401 I agree. Thanks.
@Hallaj_Dream_iisc
19 күн бұрын
√35=√(25+10) =√25+10/(2×5) . =5+1 . =6 But 6²=36🤔
@tvesaatamannamohanty5948
18 күн бұрын
This is an approximation. I am pretty sure since root35 is 5.9, it vecomes 6
@superacademy247
18 күн бұрын
You have violated the rule of nearest perfect square. the nearest perfect square to 35 is 36
@samueldeandrade8535
16 күн бұрын
@@superacademy247 he probably knows, fool.
@rainerzufall42
7 күн бұрын
First: Use the nearest sqare, which is 36. => sqrt(35) ~= 6 - 1/12 = 5.91666... Second: For b = 5 and d = 10, use my method above with d/2b = 1. d²/8b³ = 1^3 / 10 = 0.1. Subtract this from 6 and get 5.9 (even with b = 5). Exact: 5.916079...
@wilmenmedina9615
19 күн бұрын
Shame on you for those who say “still need calculator for fraction” 😮
@gregc.mariano9226
15 күн бұрын
Shame on you too. Others' minds are not the same as yours (coconut). Pls don't say those words as it's an insult to others.
@wilmenmedina9615
15 күн бұрын
@@gregc.mariano9226 sorry for hurting your feelings but I keep what I said. It is not about mind or superminds, it about paper + pencil and any method you learned in school.. if you don’t know fractions why are you trying to solve square roots without calculator 😅
@muhammadulwan24
15 күн бұрын
@@gregc.mariano9226 Yes i kind of agree with what you saying, but if you use calculator to calculate a decimal of a fraction then shouldnt you just use it to calculate sqrt as well? No offence but I just questioned this
@LuisCarlosManrique314
13 күн бұрын
Hahaha
@ratpack3247
10 күн бұрын
Lol
@tandemcompound2
17 күн бұрын
what happened to Route 66?
@mrhtutoring
17 күн бұрын
That's a good one!
@Julio05
19 күн бұрын
This method was discovered by Emmy Noether when she was at school.
@t.b.4923
18 күн бұрын
nope, issac newton discovered the generalized binomial theorem a few centuries earlier, this is just the first term
@samueldeandrade8535
16 күн бұрын
@@t.b.4923 this method exists since ancient times, fool.
@ToTheWolves
8 күн бұрын
My lab math: 0/1 points
@HvanSoolen
17 күн бұрын
what about if the number is less than one, say sq root of 3/10
@mrhtutoring
17 күн бұрын
You'd take square root of 3 and 10 separately.
@whoff59
16 күн бұрын
Or: 3/10 = 0.3 = 30/100 So: take square root of 30 and then divide by 10 (sqrt of 100)
@nikhiljagane5713
19 күн бұрын
In the last example many people will stuck in that fraction 3/20..is there any simple method to convert this to decimal?
@rajdhonsinghngangbam1848
19 күн бұрын
You can change 3/20 into (300/20)*(1/100) which can simplified into (30/2)*(1/100) and that will be 15/100 which will be 0.15
@carultch
19 күн бұрын
Recongize that if it were 3/10, that it would be 0.3. So cut 0.3 in half, to get 0.15.
@timeonly1401
2 күн бұрын
When you see fractions with denominators that easily divide into a power of ten, get an equivalent fraction by multiplying top & bottom by the number that will get that power of 10 in the bottom. In fraction 3/20, 20 goes into 100 five times. So, multiplying top & bottom each by 5 gives equivalent fraction 15/100. Multiplying or dividing by powers of 10 is easy: move decimal as many places as that power of ten has zeroes, in the direction that makes sense. Here 15 is divided by 100, which has two zeroes. So, we move the decimal of 15. two places TO THE LEFT (b/c dividing by 100 makes a number smaller, and moving decimal to the left makes a number smaller…). Without all the talking: 3/20 = = (3*5)/(20*5) = 15/100 = 0.15 Done!
@cyruschang1904
14 күн бұрын
8 x 8 = 64 8.1 x 8.1 = 64.8 + 0.81 = 65.61 8 < ✓65 < 8.1 8.05 x 8.05 = 64.4 + 0.4025 = 64.8025 8.05 < ✓65 < 8.1 continue until you have obtained the number of digits to your satisfaction
@mr.unusual8509
19 күн бұрын
*uses calculator to calculate the fractions
@mrhtutoring
19 күн бұрын
No need for calculator for that either.
@mr.unusual8509
19 күн бұрын
@@mrhtutoring right 😅
@SAMIRKHAN-jk2rq
19 күн бұрын
Still need calculator for the Fraction But still it is a crazy method🎉
@ratamacue0320
12 күн бұрын
That's approximating, not "finding" the roots.
@gregc.mariano9226
15 күн бұрын
For the square root of a non-perfect square number you still need a calculator like the 1/16. Others might have a hard time doing this without a calculator. If you have memorized the all the decimal number equivalent of fractions then so be it, otherwise, you have to do the manual division.
@pnachtwey
18 күн бұрын
Think of a real square. X is the initial guess. Y is what must be added. So the area is (x+y)^2=x^2+2*x*y+y^2. What is added is the 2*x*y so after one iteration the estimate is short by y^2. Another iteration will be closer yet but the new estimation is always short by the current value of y^2.
Пікірлер: 100