Thanks for teaching me. That"s is interesting process. Enjoy your every lesson again and again. This is NOT the answer I learn, but the PROCESS I truly enjoy it. So amazing !
This works because derivative is defined so it gives the best linear approximation: f(x) = f(a) + f'(a)(x-a) + o(x-a). In this particular case, sqrt(n^2+x) = n + x/(2n) + o(x).
Yeah, that's a given. But this works great for figuring out the approximate value of smaller numbers, which will of course help you become faster at hand calculation (this is helpful especially for us Asians haha)
Buy the Trachtenberg book on mathematics. Devised by Polish mathematician whilst POW in Germany. Wrote theories on cigarette papers to stay sane. Escaped and eventually married rich Countess then dedicated his life teaching under privileged children easy ways of mathematics in academies in Switzerland.
Wow I started looking at your videos to keep my mind sharp. I am now forwarding them to my children. Amazing - you are a fantastic teacher thank you 🙏🏽
Im 60 and have BS in Elementary and Special Education.Now that Im retired I'm enjoying learning algebra. I soooo wish i could have had you as a teacher. You're so good at explaining each step. Thanks so much❤
I would like to offer a similar method. Let G = guess and E = error and we want to find the square root of C. C = (G+E)² = G²+2GE+E². With a small enough E value, E² will be close to zero. Our equation now becomes an approximation. C~G²+2GE and E~(C-G²)/(2G). Now that you have your approximate error, simply add it to G to get your final estimate. If you choose to get more accuracy, you can revise your guess. This method can also be extended to cube roots, etc by knowing the binomial expansion formula and eliminating the terms that have powers of E greater than 1.
@@navamgarg I graduated with a minor in math in 2004. I took over 3 years worth of university math. But the funny thing is, I learned this on my own AFTER my schooling because I kept my old books. There was an interesting section on approximating square roots in my Numerical Methods book that never got covered and this was it.
This is actually really useful because sometimes i forget my calculator at home and everyone in my class doesn't want me to use the one they own Thank you sir
We can use the tailor's series , f(x) = ✓x f'(x)= 1/✓x For each number we must find the perfect square a which is closer ( x > a) f(x) ~ f'(a) + 1/2 × 1/✓a × (x-a) f(138) ~ f(121) + 1/2 × 1/11 × 17 f(138) ~11,7 We get the first decimal of ✓x , for more precision check the tailor's series .
Nice, but f'(x) = 1/2 * 1/✓x. The first order Taylor approximation is f(x) ~ f(a) + f'(a) * (x-a). You do not need x>a, but when x is close to a the approximation is best.
I remember when I was in 4th grade and questioned a teacher how to solve roots that are irrational numbers and they had no clue, now RU-vid gives it to me.
I think you have to show this Taylor series expansion if you want to teach this approximate solution to a square root. Otherwise it is better to just teach the linear interpolation solution, which is intuitive just by looking at the number line. It is not as accurate, but it doesn't require any calculus to understand. And we're talkin about approximate solutions either way. In this example, 12 + (138-144)/(144-121) = 11.74. That's still a pretty good approximation.
Ralph-Newton iteration gives you this. Let √N = x N = x² N - x² = 0 f¹(x) = -2x x₁ = x₀ + f(x)/f¹(x) For this case, N = 138, x₀ = 11 or 12, which ever you want to start from. x₁ = 11 - [138 - 11²/(-2 × 11)] x₁ = 11 - [138 - 121/(-22)] x₁ = 11 - [17/(-22)] x₁ = 11 + 0.77 ≈ 11.77 Binomial expansion can also be used
for a rougher but quicker one, take the lower of the 2 close roots (121), take that as your whole number, multiply it by 2 and that’s your denominator, find the difference of your original and the perfect square and that’s your numerator 11 and 17/22 = around 11.71
He used the first two terms of the Taylor series. It's not normally used for stuff like this, but the series can be used to approximate any function you want.
Linear interpolation is easier to remember and just about as close. Since 144-121=23 and 138-121=17 so 17/23 gives us 0.739 which is the fractional part of 11.739
Somewhere between 11/12, I think, but armed with phone I don't estimate anymore, you can actually just straight up google that or even ask siri or alexa.
This is the first order Taylor expansion of sqrt(x+h). It gives better results smaller the h, unless you go a bit further and add a second order correction √(x+h) = √x + h/(2√x) - h²/(8h^(3/2)) + O(h³)
this is really handy! especially because i have a national math exam at the end of this year, and one of the subjects that will be included in the exam are roots, on top of that they dont allow calculators. thank you so much, mrhtutoring!
There's also one easier method to do this that almost yields the same results, albeit the margin of error is higher than this method. For the sqrt(138), find the closest smaller square root, which is 11. And then find the difference between the perfect square and the number; hence, giving us 17. Make a fraction and put the difference on the numerator while twice the square root is the denominator, giving us 22. This would yield to a mixed fraction of 11 and 17/22, or 11.77
I have learned it another way... The part of writing it down is more complicated, but there is hardly making square root calculating in it.. mostly devisions and adding or substracting...
Just do thia :- if you want to find the square of 95 :-do 5 square write one no and carry second then do 9 square and double up the value and add the carried no.
The next time I have a math problem I will go to this guy for help, because after watching his videos, I have learned that he makes a math problem, no problem and no problems help me to go to sleep at night.
