@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.
@incendiohawk17254 ай 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
@smalin4 ай бұрын
I'd like to see a geometric explanation of why this approximation works.
@bjorncedervall52914 ай 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...).
@anotherelvis4 ай 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))
@bjorncedervall52914 ай 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.
@Loots14 ай бұрын
@@bjorncedervall5291 cool!
@ockham1963Ай бұрын
Brilliant teaching. Succinct and very clear.
@mrhtutoringАй бұрын
Glad you think so!
@hywell_4 ай бұрын
Just learned linear approximation recently so it finally makes a bit more sense to me now why this method works.
@homosapien60314 ай бұрын
Just curious. Is that linear algebra or…?
@homosapien60314 ай бұрын
Asking cuz I’ve taken up to Calc 2 so far and haven’t really gotten there ywt
@hywell_4 ай 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.
@homosapien60314 ай 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.49234 ай 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.
@AizenSosukesama4 ай бұрын
Thats very helpful,thank you!!
@MeMadeIt4 ай 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. 😄
@dogslife483116 күн бұрын
Mr H's teaching method is very polished Good mixture of technology
@MohammedShayaanX4 ай бұрын
Thank you sir
@-.-44 ай 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.
@Phylaetra4 ай 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.
@paulstudier57064 ай бұрын
Use a slide rule to get the square root to 2 or 3 decimal places, then apply this method. This saves an iteration.
@Phylaetra4 ай бұрын
@@paulstudier5706 Assuming you have a slide rule (which _I_ do, but not everyone does...)
@samueldeandrade85354 ай бұрын
Hahahahaha. You saying "you are not actually finding the square root - you are approximating the square root" is so funny. Hahahahahahahahahaha.
@brnmcc014 ай 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
@StereoSpace4 ай бұрын
Awesome teaching, as always with Mr H.
@nafnist4 ай бұрын
This is not finding the squareroot, but doing an approximation.
@sarojthakur-zy9fkАй бұрын
Thank you Mr.H its really helpful...
@dannmann174 ай бұрын
You are an awesome teacher👍🏻🇺🇸
@robhill57324 ай бұрын
He is using the derivative of X^0.5 to find the fractional part added to the perfect square. Easy!
@nandopanda_ts4 ай бұрын
very nice
@kingbeauregard4 ай 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.
@bearohan4 ай бұрын
chad math teacher❤
@rogerphelps99394 ай бұрын
This is just Newton's method for successive approximaion of a square root.
@CommDao4 ай бұрын
I'm not smart, but this made me feel like I have the potential to be smart. 🙏
@bowlineobama3 ай 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.
@NavigatEric3 ай бұрын
Excellent, thank you.
@charlespartrick5284 ай бұрын
Fantastic - another great method I can teach my students.
@whatzause4 ай 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.
@gregwolter15364 ай бұрын
Thank you, sir.
@pnachtwey4 ай 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.
@shabbirghulam54544 ай бұрын
Thank you so much Sir ❤
@njd23424 ай бұрын
Square numbers until you get close.
@wilmenmedina96154 ай бұрын
Shame on you for those who say “still need calculator for fraction” 😮
@gregc.mariano92264 ай 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.
@wilmenmedina96154 ай 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 😅
@muhammadulwan244 ай 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
@LuisCarlosManrique3144 ай бұрын
Hahaha
@ratpack32474 ай бұрын
Lol
@Assassin41744 ай бұрын
Thank you Sir.
@SpenzerBrightestАй бұрын
Thank you sir you have saved my life
@rivenoak4 ай бұрын
woah, that was not discussed in my math lessons. but definitely should be in curriculum !
@kaijabyenkya4 ай бұрын
This is really cool!
@roland3et3 ай 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'.
@philippecanepa45094 ай бұрын
It’s tricky but you don’t really give the mathematical explanation. Maybe the Taylor expansion !
@carultch4 ай 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.
@claireli884 ай 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.
@okaro65954 ай 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)
@Hallaj_Dream_iisc4 ай бұрын
√35=√(25+10) =√25+10/(2×5) . =5+1 . =6 But 6²=36🤔
@tvesaatamannamohanty59484 ай бұрын
This is an approximation. I am pretty sure since root35 is 5.9, it vecomes 6
@superacademy2474 ай бұрын
You have violated the rule of nearest perfect square. the nearest perfect square to 35 is 36
@samueldeandrade85354 ай бұрын
@@superacademy247 he probably knows, fool.
@rainerzufall423 ай 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...
@naderhumood11994 ай бұрын
Great apprauch thanks v much
@JimRohn-u8c4 ай бұрын
So, why does this actually work? I understand how but not why…
@crix_h3eadshotgg9924 ай 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!
@kevinchen93894 ай 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.
@carultch13 күн бұрын
@@crix_h3eadshotgg992 Essentially what a Taylor series is, is a derivative matching polynomial. You create polynomial terms to match to a function based on each level of derivative. The zeroth term, is the value of the function. The first term, is the first derivative. From the zeroth and first term, we construct a linear equation that locally approximates the function. The second term, is the second derivative. From these term, we construct a parabola that locally approximates the function. The third term, is the third derivative. From theses terms, we construct a cubic that locally approximates the function. And the pattern continues, matching value, slope, curvature, and further degrees of differentiation.
@MalenTopio4 ай бұрын
Wow thank you
@PrithwirajSen-nj6qq4 ай 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
@timeonly14013 ай 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-nj6qq3 ай бұрын
@@timeonly1401 I agree. Thanks.
@Hurtsoul-h5e4 күн бұрын
Big brain moment
@sharmota27604 ай бұрын
Thank you
@wraith67764 ай бұрын
Since it's an approximate answer 8 is acceptable. There isn't anything saying where to round to.
@stone4bread3 ай бұрын
awesome
@gregc.mariano92264 ай 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.
@Aeyo4 ай 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?
@carultch4 ай 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)
@tvesaatamannamohanty59484 ай бұрын
Thank you so much 😭🙏
@cyruschang19044 ай 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
@ElihuAsmerom7 күн бұрын
Is this newton's method
@Thauan70204 ай 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?
@heneyyamim834118 күн бұрын
where does the formula come from.or logic behind this fomula.
@carultch13 күн бұрын
It comes from the slope of the tangent line. You construct a line that matches the square root function at the nearby point, in this case at x=64 and y = 8. You then determine its slope, which for the square root function is 1/(2*sqrt(x)). Evaluate the slope at x=64, and get 1/16. So now you construct the line L(x), with a slope of 1/16, and at the point (64, 8): L(x) = 1/16*(x - 64) + 8 Where does 1/(2*sqrt(x)) come from? It comes from the power rule, which in general is d/dx k*x^p = p*k*x^(p - 1). k is a constant coefficient, and p is a constant for the power. Drop the power by 1, and have the previous power join the coefficient k to become k*p. For a square root, p = 1/2. For nth roots in general, p = 1/n, since reciprocal exponents are roots. This is the zeroth and first terms of the Taylor series, for the more general case, that allows you to get more precision, where you match a polynomial to the derivatives of a function. This is one way of calculating transcendental functions, like exponentials, logs, and trig. There are more computationally-efficient methods, but this serves as a great introduction to prepare you for understanding what calculators really do to get those functions, since it's the easiest to understand.
@ToTheWolves4 ай бұрын
My lab math: 0/1 points
@mahmoudibra58224 ай бұрын
Subtle Way...thanks
@cejII4 ай бұрын
Why is the constant always 2 in the denominator? And would this metbod work with cube root?
@samueldeandrade85354 ай бұрын
Because (√x)' = 1/(2√x)
@carultch4 ай 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)
@Julio054 ай бұрын
This method was discovered by Emmy Noether when she was at school.
@t.b.49234 ай бұрын
nope, issac newton discovered the generalized binomial theorem a few centuries earlier, this is just the first term
@samueldeandrade85354 ай бұрын
@@t.b.4923 this method exists since ancient times, fool.
@HvanSoolen4 ай бұрын
what about if the number is less than one, say sq root of 3/10
@mrhtutoring4 ай бұрын
You'd take square root of 3 and 10 separately.
@whoff594 ай бұрын
Or: 3/10 = 0.3 = 30/100 So: take square root of 30 and then divide by 10 (sqrt of 100)
@carultch13 күн бұрын
Find a nearby perfect square fractional number, and use it to follow this procedure. The nearby fractional square, is 25/81. We can find the square root of 25/81 = 5/9. And the difference: 3/10 - 25/81 = -7/810 The slope of the tangent line: 1/(2*(5/9)) = 9/10 So this means: sqrt(0.3) is approx 5/9 - 9/10*7/810 = 4437/8100, which is 0.547 with 7 repeating. actual square root of 0.3 is 0.547722558
@tandemcompound24 ай бұрын
what happened to Route 66?
@mrhtutoring4 ай бұрын
That's a good one!
@Paul-im7pd2 ай бұрын
Where does this come from? The theory please.
@carultch13 күн бұрын
It comes from the slope of the tangent line. You construct a line that matches the square root function at the nearby point, in this case at x=64 and y = 8. You then determine its slope, which for the square root function is 1/(2*sqrt(x)). Evaluate the slope at x=64, and get 1/16. So now you construct the line L(x), with a slope of 1/16, and at the point (64, 8): L(x) = 1/16*(x - 64) + 8 Where does 1/(2*sqrt(x)) come from? It comes from the power rule, which in general is d/dx k*x^p = p*k*x^(p - 1). Drop the exponent by 1, and have the previous exponent join the coefficient k to become k*p. For a square root, p = 1/2. For nth roots in general, p = 1/n, since reciprocal exponents are roots. This is the zeroth and first terms of the Taylor series, for the more general case, that allows you to get more precision, where you match a polynomial to the derivatives of a function.
@ratamacue03204 ай бұрын
That's approximating, not "finding" the roots.
@mr.unusual85094 ай бұрын
*uses calculator to calculate the fractions
@mrhtutoring4 ай бұрын
No need for calculator for that either.
@mr.unusual85094 ай бұрын
@@mrhtutoring right 😅
@nikhiljagane57134 ай бұрын
In the last example many people will stuck in that fraction 3/20..is there any simple method to convert this to decimal?
@rajdhonsinghngangbam18484 ай 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
@carultch4 ай бұрын
Recongize that if it were 3/10, that it would be 0.3. So cut 0.3 in half, to get 0.15.
@timeonly14013 ай 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!
@SAMIRKHAN-jk2rq4 ай бұрын
Still need calculator for the Fraction But still it is a crazy method🎉