Use calculus, NOT calculators!
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- Опубліковано 11 лип 2024
- Use calculus, NOT calculators! We will use a tangent line approximation and differentials to approximate sqrt(8.7). This is called local linear approximation which is an application of derivatives in Calculus 1. Subscribe for more calculus tutorials 👉 bit.ly/just_calc
0:00 use tangent line approximation for sqrt(8.7)
4:49 use differentials to approximate sqrt(8.7)
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Too slow? Try the fast version: ua-cam.com/video/Wx7lNCNN1y8/v-deo.html
Nice
Day 32 wishing you that you can pass calculus!!!
Thanks a lot
How are you billant at maths???
I would like to know!!!!
(Serious question)
Thank u sir
U tought what 12 years of school didn't
@@Mdead26 I don’t understand you
=(
Just here to mention that the square root function is a function whose slope gets lower lower and (albeit always bigger than 0), so this means, this method works incredibly well for larger x
you're also less likely to be near a square root at higher values though
@@Graknorke even the worst case gets better and better
Aaaa seni biyerde daha görmüştümm
@@trangium
fair enough I didn't actually work it out, just pointing out it's a bit more complex than just being more accurate because the second derivative gets smaller at higher x
Yep also 2nd derivative is always negative (when x>0) as well so every time you use linearization for sqrtx it’s gonna be an overestimation
The way he switches his markers is so smooth
yea man his writing while holding 2 markers is even better than mine writing normally
Focus on his points not how he writes. Or can't you understand the content
i didn’t even spot that he was changing markers
Should name the channel after the pens. 😄
@@fatsquirrel75 Naa there's already another youtuber that did that already
It's worth mentioning here that the two methods are completely equivalent
looking at the answers, it seems like we could say so, but still I don't see why these methods are equivalent
@@learpcss9569 Because the method of differentials isn't really rigorous, and to make it rigorous is really complicated, if you look closely, all we're claiming in the method is, in a sufficiently small neighborhood, delta(y)/delta(x)≈ dy/dx at that point. And that's the same thing as saying the curve is close to it's tangent near that point
@@learpcss9569 it's come form linear equation itself
Y2-Y1 = m(X2-X1)
Y2 = Y1 + m(X2-X1)
We know that Y1 = f(x), m = f`(x), and X2-X1 = dx
Hence
Y2 = f(x) + f`(x)dx
Yes. And the actual value of the square root will always be smaller than the one they give.
@@volodymyrgandzhuk361 True, but it depends on the slope. When using linear approximation, whether the approximate value is over or under the true value depends on the slope of the function. In this case, because the second derivative of Sqrt(x) is always positive, that means it will always be below the tangent line, which is equivalent to saying the estimate will always be bigger. Given a function with a second derivative that is always positive, like e^x, the tangent line will always be below the function, and the result will always be lower.
This differential method was very interesting to see. I’ve never seen a differential used like this before, thanks for showing this to us!!!
I'm pretty sure you already knew this stuff, who are you pranking?
ok "albert einstein" sure you haven't (ꐦ○_○）
I used Algebra, which is much much easier
3^2 = 9 > 8.7
(3 - x)^2 = 8.7
3^2 - 2 * 3 * x + x^2 = 8.7
9 - 6 x + x^2 = 8.7
6 x - x^2 = 9 - 8.7
6 x - x^2 = 0.3
Since x is very small, x^2
Try using this formula a^1/2 = (a+b)/(2*b^1/2). Where a is the number you want to square and b is the closest perfect square
@@Ok-fu5yi why does this work?
@@alial-khalili9232 it assumes that the square root of ab is the same as the average of a and b ((a+b)/2). This is not true normally but becomes more accurate as a and b are approaching each other. That is why we need b to be the closest perfect square
I wouldn't really call that using algebra. What you just did was basically calculus. The act of ignoring the x^2 term because it's "very small" is the basis of limits. And your whole method is literally just the differential method from the video just rearranged.
The algebraic method, would be to do what you did, but then to NOT ignore the x^2 and instead solve the quadratic equation you came to, but that would have still left you with a radical in the solution.
@@phiefer3 It's calculus, but it's calculus for babies. Lol.
If you work out the differential method for a general function y=f(x), you eventually get the exact same result, namely that f(x) ≈ f(x*)+f'(x*)(x-x*)
What do the asterisks mean?
@@nathanielnotbandy991 It is the point around which you consider the derivative, and make a linear approximation.
To go a bit more in depth, it also is basically taking the first term of the taylor series of the function.
@@nathanielnotbandy991 i used x* to denote an arbitrary point, just like x_0. But x* was a bit less messy in my opinion
Ahh what a lovely first order Taylor series expansion you got there
@@NZ-fo8tp jep LOL. Every approximation formula is basically the same
Differentials are strange objects.
In their original meaning they are infinitesimals, which mathematicians came to frown on. In modern mathematics they mean a lot of different things, including linear approximations. In first-year derivatives they pretend to be a ratio, but with limited algebraic properties. In integral notation they just kind of sit there as a reminder that we're dealing with limits, and otherwise just provide some mental shortcuts for manipulating differential equations.
And then it becomes the Jacobian, and so on...
Then comes vector calculus with gradient, divergences and curls, and now you have vectors made of differentials
@@sniperwolf50 Then tensors are like: we heard you like vectors so we put vectors in your vectors so you can map while you map...
@@stapler942 Finally, Clifford algebra comes along saying: I've heard you like dimensions, so I brought all of them
I have no clue what you are talking about lol
There's also some versions of nonstandard analysis where they are taken to be infinitesimals again.
Can we appreciate how smoothly he switches markers?
Blew my mind, love to see the many methods out there that are typically overlooked
Wait.
You could improve this method by using it recursively, right?
So, for example, to find the root of 8.7, instead of just using the derivative at 9, you could use the derivative at 9 to find the root of 8.9.
Then using use root 8.9, find root 8.8, and then finally reach 8.7.
That should, in theory, provide a closer approximation.
That is correct. Its the same as eulers method with a step size of -0.1 where you "track" the function with tangent lines.
Exactly. This can be used for numerical solutions of differential equations, given a starting point. If I remember correctly this is called "Euler's Method".
Yes!
man i love the maths community on youtube fr
You should add the second derivative devided by 2 factorial times (X1-X2)^2 , the third derivative devided by three factorial (X1-X2)^3 etc. etc. Where the derivative is filled in like X1=9. And X2 being 8,7. Keep doing it and you will add up with square root of 8,7. It's called the Taylor series. In this video only the first derivative of the linerisation of the function is used. Which is enough most of the times.
Thank you! You have a very refreshing style. Using the differential was very insightful.
I'm gonna put this in excel and iterate it quickly, that way I don't have to use a calculator
This is so awesome! Finished Calculus 1 and just love watching calculus videos now😅
very cool! I loved the video man, keep it up!
I really loved this concept.
I actually like to use quadratic approximation for one step closer.
Thanks for this intuitive explaination. Though I used numerical techniques numerous times, This is is a new perspective I learnt from you. Thanks again.
i just learnt this in my calc class last week but you made me understand it way more, tqsm
You just earn a new subscriber bro👍🏻
How I think about it: The derivative of sqrt(x) is 1/(2sqrt(x)) so the derivative at x = 9 is 1/6. So by definition of the derivative
sqrt(9 + h) = sqrt(9) + h/6 + o(h) ~ 3 + h/6
Now plug in h = -0.3 to get
sqrt(8.7) ~ 3 - 0.3/6 = 2.95
That's actually a great way to think about it
That's the underlying theory behind the calculation of derivatives, but it's really unnecessary
@@scj8863 Why is unnecessary, it's literally what approximation by differentials is saying
@@scj8863 In my mind f'(x) is, by definition, the number satisfying f(x+h) = f(x) + f'(x)h + o(h). So it's all the other methods that would need to prove themselves "necessary".
√8.7 = 3√(1-0.3/9) ~ 3(1-1/2×0.3/9)=2.95
Such simple explanation.... Thank you so much
I never thought to use local linearity (or even differentials) to approximate square roots before (outside of calculus class, that is). This is a very handy technique for those who have studied calculus but are, for whatever reason, solving arithmetic or algebra problems on a tight time constraint and without a calculator. Thank you so much for sharing!
It's great how you showcased both methods to the one problem in a single video. It would also be good to know whether the two methods are always interchangeable, or whether there are problems where only one of the two methods can be applied. Thanks!
You can take this approximation even further by using more and more terms of the taylor expansion of sqrt(x) around the point 9...
Or by using itteration on the answer you got (newton's method exactly).
New channel? I'm not hesitating to subscribe.
Nice video just calculus!
Loved it!
We can also approach this with the (a+b)² method, write the sqrt value as a sum/difference (depending on the closest square) of the decimal component (set as unknown value 'b') and a whole number component and square it, remove the b² because it is negligible (square of a decimal component ≈0). Solve for b and then substract/add with the whole number and you get the same value
Thank you, I'm currently trying to learn calculus by myself and these videos help increase my understanding about the topic
Good luck
Watch professor leonard's calculus lectures. he has all his lectures for calc 1, 2 and 3 on youtube.
In your country... at what age do they start teaching calculus?
I'm indian and we start learning at 16 years of age...
(My age was zero the day I was born... mentioning because in some countries the age is 1 when child is born)
I've done bunches of these, and yes, it is very, very kewl!
watching this brings me right back to high school maths class; heart racing, clammy hands, rising sense of panic as I completely loose track of what is going on. I don’t know why UA-cam suggested this video, it took me over 20 years to make my peace with mathematics but now I’m an Architect. If this video make you panic don’t worry, it is possible to get on in the world without fully understanding calculus.
Nice video.
Thank you 😊
Oh my goodness I love this content 💖💖💖
diff variant is way more practicable - love it
Great video!
Just to add that this is a straightforward extension with limits of the secant line method where m = (3 - 2)/(9 - 4). This method gives you the decent approximation sqrt(8.7) = 2.94
Thought you were holding your coffee for the first 5 mins 😂
I understood everything, good explanation
Yet another equivalent method is to use the first order Taylor series at 9:
f(x) ≈ f(a) + f'(a)(x-a)
sqrt(8.7) ≈ sqrt(9) + (8.7-9)/(2*sqrt(9)) = 3 - 0.3/6 = 3 - 0.05 = 2.95
It’s equivalent to the two methods shown in the video, but more direct.
This is so cool!
Double marker use is so cool
This is freaking awesome
Well done!
Man i wish you were my math teacher, really wanted a teacher who’d tell me why im learning something and to actually use what im learning
You can do something similar to do 6/7/8 and 9 during multiplication.
Normally they're hard numbers to deal with by by splitting it into 5+1, 5+2, 10-1 and 10-2 you can solve them much quicker.
It may be first order, it’s still really cool how close the approximation is. But I’m biased as an engineering student (approximation go brrrrr)
That's greatly explained
Good job sir! Very nice tutorial for first or second year college students.
Thankyou!!
What about using the second iteration to get an even more accurate approximation?
The second iteration uses sqrt(8.7025) = 2.95
You're essentially using the Newton-Raphson method.
Or the first order Taylor Series.
Both of these methods give the same approximation, which can be expressed as √x≈(x+a²)/(2a), where x is the number whose square root we want to find and a² is the nearest perfect square to x (so a is an integer), x>0, a>0. Btw, the value that is obtained will always be greater than the actual value of the square root.
Nice and beautiful idea
Thanks!
I wasn't expecting to understand, but I understood. Dang.
Numerical differentiation, similar to interpolation. Assuming slope in the vicinity of point approximately same.
I'm a spanish native speaker and your accent helps me to understand better English i mean it's more difficult to understand... so i keep learning
4:58 thats the piece im playing on my piano rn XD (entertainer, scott joplin
)
In the time it took to do the long division of 8.7 by 6, you could have computed the first three digits by the digit by digit method.
There's another approximation (but very crude) for very large numbers that's used in computer science as a "first guess" before using Newton's method. Say we have an N digit number (call it A), then your first approximation for sqrt is the first N/2 digits (call it a) and then you average a and A/a
love this
The differential method is taught in AP Calculus AB
I like the differential approach
That is a fun way to do it .
You ever hear about Ed Marlo ?
Yeah the first one was also a cool thought!!
Very nice video
I love this guy's accent. And he's teaching amazingly well!
Thankss sir ❤️
It would be nice to specify error range, otherwise I can say sqrt(8.7) ~= 3.
You can use binomial expansion as well.
I love watching stuff that I can understand at the beginning, and that becomes super hard after 30 seconds LOL
Wow, this amazing
Wow so good!
In Japan, people(especially, high school students in entrance exam) often use "Kaiheihou" to calculate squared numbers.
Sorry, not "to calculate squared numbers" but "to calculate the square root of numbers".
What about Taylor series to calculate this using derivatives?
Cool like it💚
Great!!
I think if we just use the lever rule, it would be much easier 👌🏻
That's pretty cool!
My approach was 8.7 x 10 i.e. 870 which comes almost halfway between 841 (29^2) and 900 (30^2)
Therefore its sqrt would be halfway between 29 & 30 i.e. 29.5; dividing by 10 we'd get 2.95 [sqrt 8.7]
It's kinda hard for me to understand at first but now I get it, still, great methods, I guess it's useful to explain such mathematics question on some of exercise our mad math teacher one gave us lol
in elementary school i learn to approximate square root using linear interpolation, even without knowing what it is called. in this case 8.7 is between 4 and 9 so the square root has to be more than 2, for the decimal it is (8.7-4)/(9-4) which is equal to 0.94. combine both and you will get 2.94.
the problem with this method is that the estimate will always be underestimated because instead of using the tangent line at x=9 this method use the line that cross the curve at (4,2) and (9,3), but it certainly easier cuz even an elementary student can understand them.
Verry excellent ❤
I’m not even taking Math right now but I was intrigued to watch lol
sir mad respect
I love it ❤
Just a question, why did you changed the general axis notations?
Thank you this maths is nice
I just wondering what if I have to find out an approximate root value with calculus where the nearest square number is far away. For instance if we have to approximate root over 598.6 where the nearest square number are not that close lowest square number 576 and highest square number is 625.
Now my question is, will it yield as accurate results as it did for root over 8.7??
L(x)=f(a)+f'(a)(x-a) it helps me to think about a as the number you're using to approximate and x as the actual value that you're finding
One can use Pade approximation for square root. rewrite sqrt(8.7)=3*sqrt(1-1/30)
1st term is the same as in Taylor series sqrt(1-x)~a1=1-0.5*x
next use recurrence a_{n}=1-x/(1+a_{n-1})
so for x=1/30 a1=59/60 ; a2=117/119
let's try 2nd term sqrt(8.7)~3*117/119~2.94958 correct up to 5th sign
this is just beautiful
Throwing in the linearization formula would be helpful!
What happened to the 3 in the tangent line step?
The differential method is pure beauty
What if the given value was sqrt(6) ? It's not so close to 3, almost in the middle. How would you determine reference point to determine tangent line?
Is the first method not basically taylor aproximation of the first order?
Wow, this is pretty cool
what if you take the two square roots before and after the number you are approximating, find the equation of the straight line of each tangent, and consider the two points of tangency and the point of their intersection, and then figure out the equation of a straight line that is equidistant from these three point, and use that new straight line for a much better approximation :D
There's an other way to approximate square roots, more accurate for larger numbers:
4 is the square of 2, 9 is the square of 3.
9 - 4 = 5, 3 - 2 = 1
8.7 - 4 = 4.7
Let d be the difference between √8.7 and √4.
4.7/5 is approximately equal to d/1.
After calculation, we get d to be 0.94.
2 + 0.94 = 2.94
Therefore, √8.7 ~ 2.94.
Basically, the two square numbers, between which is the number (let this be x) whose square root is to be found, are taken, with their square roots, which are integers.
(x - Smaller square number/The difference between the two square numbers) is approximately equal to (√x - Smaller square root/The difference between the square roots).
Eg: Let x be 3. 1 and 4 are the required square numbers, and 1 and 2 are their square roots.
According to the formula,
(3 - 1/4 - 1) ~ (√3 - 1/2 - 1)
-> (2/3) ~ (√3 - 1/1)
-> √3 - 1 ~ 0.666...
-> √3 ~ 1.666... (~1.732)
As I said, this method is more accurate for larger numbers.
First one is pretty similar to newtons method for finding roots of a polynomial
thank yooooou
tbh I think that the simple (a-b)^2 = a^2-2ab+b^2 approach is a bit easier, a^2 being 9, so a is 3, and you have b(a-2b) = 0.3, a being 3 means that so 6b-b^2 = 0.3, and you can say that b^2 is very small, so 6b(slightly b= 0.5- change. this isn't a calculus approach but more of an algebric one, another way is 2.9^2 is 841, and then by (a+b)^2 you can to a bit of trial and error and come around to approximately 2.94 (2.95 ^ is 8.7025 I think).