Binomial expansion of (x+y)^4 with x = 2 and y = -0.002 and picking the zero-th and first order: (x+y)^4 -> x^4 + 4x^3 y = 16 - 4*2*2*2*0.002 = 16 - 0.032 = 15,936 Which is exactly the same result as yours. In fact, that's exactly the same thing because you consider the derivative of (x+y)^4 with respect to y to be a constant, so you pick up to the first order. One could argue that you can use Taylor expansion of (2+x)^4 in x=0. You get (2+x)^4 = 16 + 32 x Plugin in x=-0.002 and you get 15,936 Oh, what a surprise, we also find the same result. How odd! In fact, all these methods are equivalent.
I can see this working with small exponents, but at larger exponents it's a bit harder to calculate the binomial coefficients or remember the Pascal Triangle at higher degrees. In terms of mental math, I'd prefer using calculus. Unless there is a trick that I might not know. If so please tell. :D
First thing came into my mind was using Binomial Theorem; (2-dx)^4= 2^4 - 4. 2^3 .(dx) ........ so on. Rest of the terms include dx to the power greater than 1 so we can ignore them for any practical purposes since they will be negligibly small. So 16 -0,064 = 15,936 Thank you for this problem, was interesting to see. #YAY
sir you are so great, the best thing I like about you is you always teach us happily which makes us understand maths easily, keep going sir never let us down, thankyou. H
A lot of people are making too much of how this particular case lends itself to a variety of approaches. But the derivative approach can be used for just about any situation where the function is differentiable. For example, back in physics class, we used to calculate very small time dilation effects (which involved square roots of differences of squares) by differentiating the time dilation function and using that to calculate the delta. Also, this ties into the Taylor Series, which can be used to approximate complicated functions with polynomials: en.wikipedia.org/wiki/Taylor_series
I took a more direct approach using a bit of precalculus and knowing the binomial expansions of (a+b)^n: (1.998)^4 = (2-0.002)^4 a = 2 b = -0.002 = -2E-3 (scientific notation makes this process a bit easier) (a+b)^4 = 1a^4b^0 + 4a^3b^1 + 6a^2b^2 + 4a^1b^3 + 1a^0b^4 Powers of a from 0 to 4: 1, 2, 4, 8, 16 Powers of b from 0 to 4: 1, -2E-3, 4E-6, -8E-9, 16E-12 From there, plug in for a and b: 1x16x1 + 4x8x-2E-3 + 6x4x4E-6 + 4x2x-8E-9 + 1x1x16E-12 16 + 32x-2E-3 + 24x4E-6 + 8x-8E-9 + 16E-12 16 + -64E-3 + 96E-6 + -64E-9 + 16E-12 From there I just expanded the scientific notation into full decimal representations and added the positives together, then the negatives, and then I subtracted: 16.000 000 000 000 00.000 096 000 000 00.000 000 000 016 + --------------------------------------- 16.000 096 000 016 0.064 000 000 000 0.000 000 064 000 + ----------------------------------- 0.064 000 064 000 16.000 096 000 016 00.064 000 064 000 - ------------------------------------ 15.936 095 936 016 Exact value without having to manually multiply 1.998 by itself four times over and having to waste time with long-form multiplication. c: Although your method is a lot more eloquent, a whole lot faster, and if you're just doing quick back of the envelope math for a crude engineering calculation just to get a quick idea of what's going on or because your tooling just isn't that precise anyways... it's perfectly A-O-K to use.
It can even be made simpler: To square x-a you get approximately x^2-2ax once that a is very small and so a^2 will be smaller. In our case, we get 4-2 . 2 . 0.002 which is 4 - 0.008. Repeating (because we want the fourth power), we get 16 - 2 . 4 . 0.008 = 16 - 0.064.
It easier to use the linear approximation using first two terms in taylor series L=f(a) +d/dx f(a) (x-a)..... 1 where a is constant, and L stand for linear approximation of the original function f(2)=2^4=16......2 d/dx f(2) =4(2)^3=32.......3 Substitute 2 and 3 in 1 L=16+32(x-2) ....4 SUBSTITUTE x=1.998 in 4 implies L=15.936
I was thinking about this a little differently... In general, we have: (a + b)^n = C0*a^n + C1*a^(n-1)*b + ... + Cn*b^n, where C0, C1, ..., Cn are the binomial expansion coefficients. If 0 < |b|
Binomial Theorem can also solve it. (2-0.002)^4=16-4(8)(0.002)+6(4)(0.000004)-4(2)(0.000000008)+(0.000000000016) =16-0.064+0.000096-0.000000064+0.000000000016 =15.936095936016
@@jadebriones1633 the only multiplication involved (aside from one factor of 3) are powers of 2 and 10, and the addition/subtraction all line up nicely so it's quite quick
Here in Poland we can't have calculators on exams at university so I've learned this on the beginning. And this is simple example. I remember people who was so angry with professor :) very useful approach not only for power. Try to calculate 4th root of 1558.57 without this method.
(2-0.002)^4=(2-0.002)^2^2 ~(4-0.008)^2 (note: 0.002^2 is too small, we regard it as 0) ~(16-0.064) (note: 0.008^2 is too small, we regard it as 0) =15.936
Clever tool that I can use in computing harder examples. Those in the comments section missed out the point. He knows he can use algebra in breaking apart the given but what he wants to teach you is to approximate a function using differentiation.
I converted 1.998 to 999/500 then squared it in my head to 998,001 / 250,000 which is about 3.992. Notice how the delta of 0.002 was made into a new delta of 0.008 so there is a cube factor in there (2*2*2 = 8). So by squaring the intermediate result again, the new delta should be 0.064 (8*2*2*2).
I just wonder why didn't my teacher tell this very basic question when we were being taught calculus (it's been more than even a month since we started doing differentiation). Thanks to u I am able to understand this concept more!
This is how I've always done mathematics since I began learning numbers. Never showing my work other than a couple of numbers I needed to remember along the way was always a problem in school.
My first thought was Taylor series, which would essentially give you the same result (first order approximation) . When you wrote down 2-0.002 I thought you might use the Binomial theorem. But your way of explaining it is nice for students that don't know about Taylor series yet.
I noticed that delta x / x is 1 per mille (1 part per thousand) I know that the proportion is multiplied by the power, so looking for delta y approx 4 per mille. 4 x 16 is 64 so have to subtract 64/1000 from 16. 15.936. Same as your answer but using an extra short cut
0:14 - took me 5 minutes. Starting with 1998 = 2*(1000-1) 1998^4 = 2^4 * (1000 - 1)^4 [a:= 1000, b :=1] = 16* (a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4 = 16 * (1,000,000,000,000 - 4,000,000,000 + 6,000,000 - 4,000 + 1) = 16,000,000,000,000 - 64,000,000,000 + 96,000,000 - 64,000 + 16 = 16,000,096,000,016 - 64,000 - 64,000,000,000 = 16,000,096,000,016 - 1,000,000 + 936,000 - 1,000,000,000,000 + 936,000,000,000 = 15,936,095,936,016 EDIT: OH WAIT, lol, confused your notation for meaning 1998, since I’m not that used to using "." for decimal points (although, I know, I adopted this very comment to US notation [but remember, I did everything on paper first]). Anyways, then the result would be 15.936095936016. And actually calculation was a lot easier with the separators every 3 digits.
That's correct, but the goal of the exercise was to get an approximation. Your result should have been truncated to the first order in the binomial expansion.
Well if you plan to comment this under every comment here, that presents or states they did a full computation, too, then go ahead and waste your time.
use formula y-f(a)=f'(a)(x-a); transform to y=f'(a)(x-a)+f(a) where the y is the final value; a=2 x=1.998 f(a)=a^4 f'(a)=4a^3 the y value will be the same as the video.
What I would do is finding the tangent line r to the curve y=x⁴ at x=2. If r: y=ax+b (which I can determine using the power invested on me by HS calculus) Then I'd say f(1.998)=a(-0.002)+b. I never used the differential itself for approximations but I can always rely on good ol' HS calculus
I find differentials make more sense with a picture showing that you're basically just multiplying slope by deltaX to get the change in y due to the tangent (and also why the points must be close)
L(x) = f(2) + f’(2)(x-2) near x=2. I learned that a few weeks ago in my calculus class. L(1.998) = 16 + 32(-0.002) L(1.998) = 16 - 0.064 L(1.998) = 15.936
Tldr; we know 2^4=16, and if we model that as y=x^4 we can get the slope at x=2, multiply that slope by DeltaX where DeltaX = 1.998-2 to get a linear approximation of the difference in Y from x=2 to x=1.998, and that to 16, et voila
We can use (1+x)^n is approximately 1+nx for small numbers. It gave me the answer under half a minute 16(1-0.001)^4. = 16(1-4(0.001)) = 16(0.996) =15.936
Done it (using powers of 10 notations to simplify 0.002 and powers computing). Took less than 5 minutes to get the exact value : 15.936095936016 . Calculus isn't needed, only square identity is ( (A+B)^2 = A^2+2AB+B^2 ).
I think using the binomial theorem and just doing the first two orders of it would give you the exact same answer you had, but with it you can got even more exact, if you want to do it for some reason
There is a nice relation to statistics. This way showed in the video is used in "propagation of uncertainty" in physics and statistics. It's called "variance".
Another slightly different approach: We know the fourth order Taylor expansion will be a polynomial in x-2 that's equal to x^4, so to find that expansion we can do x^4 = (2 + (x-2))^4 = 2^4 + 4*2^3*(x-2) + 6*2^2*(x-2)^2 + 4*2*(x-2)^3 + (x-2)^4 = 16 + 32*(x-2) + 24*(x-2)^2 + 8*(x-2)^3 + (x-2)^4 And truncate to your desired accuracy
In my studies of chemistry we had to attend lectures in statistical thermodynamics as well. So we had to calculate a lot with the amount of permutations etc. Weirdest part for us was calculating with numbers no computer on this planet can calculate. Like: What is the result of N! with N=10EXP(23) We had to estimate as well using analytical maths.
Taking 2 in factor, we have (1-1/1000)^4, approximately 1 - 4*1/1000, which gives immediately the correction to 16 : -64/1000. Your interesting demo gives the shortcut to the Taylor dev. on first order...
Guys, does anybody remember about Taylor rows? (1+x)^a≈1+a*x so, (2-0.002)^4=2^4*(1-0.001)^4≈ ≈16(1-0.004)=16-0.064=15.936 it takes me 20 sec to calculate
My "simple" solution. (1.998)^4 is the same as (2 * 0.999)^4 which is the same as 16 * (.999^4). I remember that 0.999 * 0.999 = 0.998001 and I can see by visual inspection that 0.998001 ^ 2 is about 0.996 (following the pattern of the 3rd digit after the decimal). So the answer is 16 * 0.996 which is 15.936.
I used Local Linearity principle and found the equation of tangent line at 2 using derivatives and then calculated function value of tangent line at 1.998 to have a better approximation of this problem
At 4:03 , it's _wrong_ to suggest that Δy is (approximately) equal to the derivative dy . (Note that if instead of (1.998)⁴ we had to calculate (1.997)⁴ , then we could use the _same_ relation y = x⁴ and the _same_ "starting point" 2⁴ , so by definition the _actual derivative_ should also remain the same.) I think it's more proper to say Δy ≈ (dy/dx) * Δx where dy/dx (and not dy) is actually the derivative. Since y = x⁴, we have dy/dx = 4x³ , so we get Δy ≈ 4x³ * Δx (This way, we're also not dealing clumsily with the distinction between dx and Δx .) With x = 2 and Δx = -0.002 , this results in Δy ≈ 4*2³ * (-0.002) = -0.064 hence y = 16 + Δy ≈ 16 - 0.064 = 15.936 - - - - - By the way, the video's approach is all just an application of the linearization of a (smooth) function f(x) near a point x=c , f(c+Δx) ≈ f(c) + f'(c)*Δx where f'(c) is defined as the value of (df/dx) at x=c. (In this particular example, we have f(x) = x⁴ and c=2 , and so f(1.998) ≈ f(2) + f'(2)*(-0.002) .)
Binomial theorem. 1.998 = 2 * (1 - 10^-3) so 1.998^4 = 16 * (1-10^-3)^4 = 16*(1 - 4*10^-3 + 6*10^-6 - 4*10^-9+10^-12) = 16 - 0.064 + 0.000096 - 0.000000064 + 0.0000000000016 = 15.936095936016. Done in approximately 1 minute longer than the video, which I'll now have to go back and watch, from the point you set the challenge. I just hope I've got the number of decimal places right as everything was done using mental arithmetic and written here.
Why not just use Pascal’s triangle at 1 4 6 4 1 and skip all the function blabber? You’ll get an EXACT answer from the expansion of a^4 + 4 a^3•b + 6•(a^2)(b^2) + 4 a• b^3 + b^4 . What’s an error estimate for the procedure used?
Mathematician: "Let's use calculus to solve this."
Engineer: "It's 16."
its 2 minus (.002 ^4)
2- .0000016
1.9999984
15.9999984
@@williamhewitt4748 Bruh if you don't get the joke don't say anything.
I think he uses Integral Calculus in grocery shopping.
He saves a cent okay, that’s a lot for us asains
Asians*
Engineer: it's about 16. Add 25% contingency factor. Let's make it 20.
As long as there's a safety factor you can't be wrong
No, it is LESS than 16!!!
Engineers make me cringe
GLaDOS some engineering students become physicists, and viceversa
Doug didn't understand the joke.
Next: solve a calculus problem using only arithmetic. 😁
Michel ten Voorde Actually its been done. Sir Isaac Newton used basic arithmetics applied to calculus to find out the motion of heavenly bodies
Trapezium rule for area
Approximately 16, you’re welcome
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Don’t take it seriously lol
ok.
Weird flex but ok
Grievance studies graduate: But I don't want x^4 to grow. That's racist.
Binomial expansion of (x+y)^4 with x = 2 and y = -0.002 and picking the zero-th and first order:
(x+y)^4 -> x^4 + 4x^3 y = 16 - 4*2*2*2*0.002 = 16 - 0.032 = 15,936
Which is exactly the same result as yours.
In fact, that's exactly the same thing because you consider the derivative of (x+y)^4 with respect to y to be a constant, so you pick up to the first order.
One could argue that you can use Taylor expansion of (2+x)^4 in x=0.
You get (2+x)^4 = 16 + 32 x
Plugin in x=-0.002 and you get 15,936
Oh, what a surprise, we also find the same result. How odd!
In fact, all these methods are equivalent.
This is because this is a first degree Taylor polynomial in disguise...
Yes, that's what I said.
PackSciences yep...agreed... as a side note I'm pretty sure this vid is for calc 1 students
Little typo there in the first paragraph; it's (of course) 16 - 0.064; you basically just multiplied by 0.001 instead of 0.002
I can see this working with small exponents, but at larger exponents it's a bit harder to calculate the binomial coefficients or remember the Pascal Triangle at higher degrees. In terms of mental math, I'd prefer using calculus. Unless there is a trick that I might not know. If so please tell. :D
You know someone is smart when they use Wolfram Alpha as a calculator
First thing came into my mind was using Binomial Theorem;
(2-dx)^4= 2^4 - 4. 2^3 .(dx) ........ so on.
Rest of the terms include dx to the power greater than 1 so we can ignore them for any practical purposes since they will be negligibly small.
So 16 -0,064 = 15,936
Thank you for this problem, was interesting to see.
#YAY
This is a more conplicated demonstration of tangent line approximations.
Quicker using binomial expansion.
X = 16(1- d)^4 d=10^-3
X=. 16(1-4d + 6d^2...)
X= 16 -64d + 96d^2
X= 16- 0.064 + { 100×10^-6 = 10^-4 }
X = 15.9361
sir you are so great, the best thing I like about you is you always teach us happily which makes us understand maths easily, keep going sir never let us down, thankyou. H
The relative deviation of your approximation from the real value is 0.0006%. For practical purposes, this is usually neglectable.
Ralf Bodemann its even nicer that the relative difference between 2 and the actual input is *bigger* than the result!
*negligible
Why is deviation between (0.99)^2 and (1.01)^2 too high
@@wontpower bruh you do people always correct someone else's spelling mistakes. BRO ENGLISH IS NOT MOST PEOPLE'S FIRST LANGUAGE.
Is it only me who heard *Doraemon* tune in the intro? Btw thanks for this amazing video
Tushar Patel
Thank you! And you were right about the intro tune
@@blackpenredpen my life was a lie,
Now I cannot unhear it!!!
A lot of people are making too much of how this particular case lends itself to a variety of approaches. But the derivative approach can be used for just about any situation where the function is differentiable. For example, back in physics class, we used to calculate very small time dilation effects (which involved square roots of differences of squares) by differentiating the time dilation function and using that to calculate the delta.
Also, this ties into the Taylor Series, which can be used to approximate complicated functions with polynomials:
en.wikipedia.org/wiki/Taylor_series
Excellent video. Also, it's really awesome to read all these comments offering other solutions as well! Math is so fun!
This is really interesting how calculus can be used to solve such problems!
This is an excellent explanation for local linear approximation for anyone who has basic knowledge of derivatives
I took a more direct approach using a bit of precalculus and knowing the binomial expansions of (a+b)^n:
(1.998)^4 = (2-0.002)^4
a = 2
b = -0.002 = -2E-3 (scientific notation makes this process a bit easier)
(a+b)^4 = 1a^4b^0 + 4a^3b^1 + 6a^2b^2 + 4a^1b^3 + 1a^0b^4
Powers of a from 0 to 4: 1, 2, 4, 8, 16
Powers of b from 0 to 4: 1, -2E-3, 4E-6, -8E-9, 16E-12
From there, plug in for a and b:
1x16x1 + 4x8x-2E-3 + 6x4x4E-6 + 4x2x-8E-9 + 1x1x16E-12
16 + 32x-2E-3 + 24x4E-6 + 8x-8E-9 + 16E-12
16 + -64E-3 + 96E-6 + -64E-9 + 16E-12
From there I just expanded the scientific notation into full decimal representations and added the positives together, then the negatives, and then I subtracted:
16.000 000 000 000
00.000 096 000 000
00.000 000 000 016 +
---------------------------------------
16.000 096 000 016
0.064 000 000 000
0.000 000 064 000 +
-----------------------------------
0.064 000 064 000
16.000 096 000 016
00.064 000 064 000 -
------------------------------------
15.936 095 936 016
Exact value without having to manually multiply 1.998 by itself four times over and having to waste time with long-form multiplication. c:
Although your method is a lot more eloquent, a whole lot faster, and if you're just doing quick back of the envelope math for a crude engineering calculation just to get a quick idea of what's going on or because your tooling just isn't that precise anyways... it's perfectly A-O-K to use.
@Calyo Delphi: I did only one addition using nine's-complement.
Calyo Delphi Can you do a video of this ? I'm sure it would be more explanatory.
Engineer would simply say 16lol
no u
@@HandledToaster2 succ
It can even be made simpler: To square x-a you get approximately x^2-2ax once that a is very small and so a^2 will be smaller.
In our case, we get 4-2 . 2 . 0.002 which is 4 - 0.008.
Repeating (because we want the fourth power), we get 16 - 2 . 4 . 0.008 = 16 - 0.064.
It easier to use the linear approximation using first two terms in taylor series
L=f(a) +d/dx f(a) (x-a)..... 1
where a is constant, and L stand for linear approximation of the original function
f(2)=2^4=16......2
d/dx f(2) =4(2)^3=32.......3
Substitute 2 and 3 in 1
L=16+32(x-2) ....4
SUBSTITUTE x=1.998 in 4 implies
L=15.936
This is very interesting! This approximation method is very precise and you can always experiment with other values as well.
I was thinking about this a little differently...
In general, we have: (a + b)^n = C0*a^n + C1*a^(n-1)*b + ... + Cn*b^n, where C0, C1, ..., Cn are the binomial expansion coefficients.
If 0 < |b|
Awesome Video, Really Informative and Useful.Thank you so much.
Binomial Theorem can also solve it.
(2-0.002)^4=16-4(8)(0.002)+6(4)(0.000004)-4(2)(0.000000008)+(0.000000000016)
=16-0.064+0.000096-0.000000064+0.000000000016
=15.936095936016
GG to everyone who used this!
Yea, but it's more of a general method to use the derivative.
tipoima I know I know, but this is a special approach for the power functions, and yes, the derivatives are a bit easier for approximation
M. Shebl just multiplying 1.998 by itself four times is probably as fast as this method though...
@@jadebriones1633 the only multiplication involved (aside from one factor of 3) are powers of 2 and 10, and the addition/subtraction all line up nicely so it's quite quick
Cal 2 pays off! No kidding. Great video, it's great to see application of calculus, love how alive you look and how the brain is being used!
Here in Poland we can't have calculators on exams at university so I've learned this on the beginning. And this is simple example. I remember people who was so angry with professor :) very useful approach not only for power. Try to calculate 4th root of 1558.57 without this method.
Really enjoyed this one. Thanks for refreshing my knowledge of differentials!
(2-0.002)^4=(2-0.002)^2^2
~(4-0.008)^2 (note: 0.002^2 is too small, we regard it as 0)
~(16-0.064) (note: 0.008^2 is too small, we regard it as 0)
=15.936
Clever tool that I can use in computing harder examples. Those in the comments section missed out the point. He knows he can use algebra in breaking apart the given but what he wants to teach you is to approximate a function using differentiation.
I converted 1.998 to 999/500 then squared it in my head to 998,001 / 250,000 which is about 3.992. Notice how the delta of 0.002 was made into a new delta of 0.008 so there is a cube factor in there (2*2*2 = 8). So by squaring the intermediate result again, the new delta should be 0.064 (8*2*2*2).
I figured out what you were doing halfway in and just kind of reveled in the genius
this happens every video
Excellent refresher, sir. Thank you!
I just wonder why didn't my teacher tell this very basic question when we were being taught calculus (it's been more than even a month since we started doing differentiation). Thanks to u I am able to understand this concept more!
brilliant video
Your videos are addictive. I enjoy these ingenious connections and tricks you come up with. Keep it up:) Much love from Kent State Uni
If you wanna use approximation by calculus, it's better to use the form (1+x)^n ~ 1+ nx, where x
Thank you for excellent explanation
I loved it, beautiful
I like it. It's a very good way to use calculus
perhaps a shorter rouute is to transforn dy=4x^3 dx into
dy/y = 4 dx/x
dy= y( 4 * -0.001)
dy =16(-0.004)
Y = 16- 0.064 =15.936
Another excellent explanation....
Awesome. Now I understand the usage on differential calculus.
This is how I've always done mathematics since I began learning numbers. Never showing my work other than a couple of numbers I needed to remember along the way was always a problem in school.
That what we call him The mean value theorem , thanks for teach us .❤️
Loved the video!
Much better explanation than my prof gave me. Thanks!
Good teaching. Thank you, Sir.
Really nice one man !! Love from India
My first thought was Taylor series, which would essentially give you the same result (first order approximation) . When you wrote down 2-0.002 I thought you might use the Binomial theorem.
But your way of explaining it is nice for students that don't know about Taylor series yet.
y= 1.998^4 = (2 - 2/1000)^4
Factor out the 2:
y= 2^4(1 - 1/1000)^4
Let x= 1/1000
Using the aproximation: (1±x)^n = 1 ± nx, for |x|
Based upon your previous video, I thought you might do a video like this one! :)
duggydo yup!
Love it. Thank you
Very applaudable, sir :) thanks!
I noticed that delta x / x is 1 per mille (1 part per thousand)
I know that the proportion is multiplied by the power, so looking for delta y approx 4 per mille.
4 x 16 is 64 so have to subtract 64/1000 from 16.
15.936. Same as your answer but using an extra short cut
0:14 - took me 5 minutes.
Starting with 1998 = 2*(1000-1)
1998^4 = 2^4 * (1000 - 1)^4 [a:= 1000, b :=1]
= 16* (a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4
= 16 * (1,000,000,000,000 - 4,000,000,000 + 6,000,000 - 4,000 + 1)
= 16,000,000,000,000 - 64,000,000,000 + 96,000,000 - 64,000 + 16
= 16,000,096,000,016 - 64,000 - 64,000,000,000
= 16,000,096,000,016 - 1,000,000 + 936,000 - 1,000,000,000,000 + 936,000,000,000
= 15,936,095,936,016
EDIT: OH WAIT, lol, confused your notation for meaning 1998, since I’m not that used to using "." for decimal points (although, I know, I adopted this very comment to US notation [but remember, I did everything on paper first]).
Anyways, then the result would be 15.936095936016.
And actually calculation was a lot easier with the separators every 3 digits.
That's correct, but the goal of the exercise was to get an approximation.
Your result should have been truncated to the first order in the binomial expansion.
I know. I was answering question right at the beginning, when he asked how long we’d need for a complete computation.
Well there is no point to do the whole computation, you just wasted your time
Well if you plan to comment this under every comment here, that presents or states they did a full computation, too, then go ahead and waste your time.
shots fired!
use formula y-f(a)=f'(a)(x-a); transform to y=f'(a)(x-a)+f(a)
where the
y is the final value;
a=2
x=1.998
f(a)=a^4
f'(a)=4a^3
the y value will be the same as the video.
love you so much sir...you are my inspiration... love you sir... Good bless.
The same procedure as considering a central value(2) and the error(0.002). The bottom limit gives your solution.
This is beautiful.
What I would do is finding the tangent line r to the curve y=x⁴ at x=2. If
r: y=ax+b (which I can determine using the power invested on me by HS calculus)
Then I'd say f(1.998)=a(-0.002)+b. I never used the differential itself for approximations but I can always rely on good ol' HS calculus
Konhat Lee Sakurai 4x^3 is the start of your tangent equation. Good intuition and I believe this is part of most HS Calculus curriculums
This is equivalent to computing the linear approximation of x^4 around x=2, which is y = 32x - 48.
I find differentials make more sense with a picture showing that you're basically just multiplying slope by deltaX to get the change in y due to the tangent (and also why the points must be close)
That's so cool!
L(x) = f(2) + f’(2)(x-2) near x=2. I learned that a few weeks ago in my calculus class.
L(1.998) = 16 + 32(-0.002)
L(1.998) = 16 - 0.064
L(1.998) = 15.936
Amazingly clever.
I read this in my high school calculus class
You can use the formilation of taylor and you well get an exact value becouse x^4 is defferentiable 4 times
Tldr; we know 2^4=16, and if we model that as y=x^4 we can get the slope at x=2, multiply that slope by DeltaX where DeltaX = 1.998-2 to get a linear approximation of the difference in Y from x=2 to x=1.998, and that to 16, et voila
We can use (1+x)^n is approximately 1+nx for small numbers. It gave me the answer under half a minute
16(1-0.001)^4. = 16(1-4(0.001))
= 16(0.996)
=15.936
Wah re chutiya(good luck for your maths exams you moron)
easier approach:
((2-0.002)^2)^2
done
u can use ^4 power right from start and binomial formula ;)
Done it (using powers of 10 notations to simplify 0.002 and powers computing). Took less than 5 minutes to get the exact value : 15.936095936016 . Calculus isn't needed, only square identity is ( (A+B)^2 = A^2+2AB+B^2 ).
Blu Deat who the fuck actually knows that
Blu Deat
(a+b)^4=
a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4
a=2
b=-0.002
16-0.064+0.000096-0.000000064-0.000000000016 =
15.936095936
*15.936095936016
Well explained 💪💪💪
I think using the binomial theorem and just doing the first two orders of it would give you the exact same answer you had, but with it you can got even more exact, if you want to do it for some reason
Took 4 mins 43 sec..... just by doing (1.998^2)^2.
Took under 2 mins by expanding (a - b)^4 and substituting a=2 and b=0.002.
Both answers were exact.
Interesting application
There is a nice relation to statistics.
This way showed in the video is used in "propagation of uncertainty" in physics and statistics.
It's called "variance".
Another slightly different approach: We know the fourth order Taylor expansion will be a polynomial in x-2 that's equal to x^4, so to find that expansion we can do
x^4 = (2 + (x-2))^4 = 2^4 + 4*2^3*(x-2) + 6*2^2*(x-2)^2 + 4*2*(x-2)^3 + (x-2)^4
= 16 + 32*(x-2) + 24*(x-2)^2 + 8*(x-2)^3 + (x-2)^4
And truncate to your desired accuracy
It’s beautiful how the universe works.
You get pretty much excited when doing math, and that's great!
Hey sir Vishal from India thank you for this lesson it help me lot in my mathematic now I can easily solve question like these
: )
Beautiful
In my studies of chemistry we had to attend lectures in statistical thermodynamics as well. So we had to calculate a lot with the amount of permutations etc. Weirdest part for us was calculating with numbers no computer on this planet can calculate.
Like: What is the result of N! with N=10EXP(23) We had to estimate as well using analytical maths.
Great and useful video.
Meaning of YAY: y=a^y Solve for y
a...
Taking 2 in factor, we have (1-1/1000)^4, approximately 1 - 4*1/1000, which gives immediately the correction to 16 : -64/1000. Your interesting demo gives the shortcut to the Taylor dev. on first order...
Guys, does anybody remember about Taylor rows?
(1+x)^a≈1+a*x
so,
(2-0.002)^4=2^4*(1-0.001)^4≈
≈16(1-0.004)=16-0.064=15.936
it takes me 20 sec to calculate
Bro how can you equate (1-0.001)⁴ to (1-0.004) this is not possible simply🤔
@@samyakjain9295 that is the two first terms of Tailor's row of function (1+x)^a in zero
That was my answer thanks ! :)
That's the calculation he made, actually.
thank you awesome
We can write it as (2.000-0.002)⁴ or (2.000-0.002)²×(2.000-0.002)²
My "simple" solution. (1.998)^4 is the same as (2 * 0.999)^4 which is the same as 16 * (.999^4). I remember that 0.999 * 0.999 = 0.998001 and I can see by visual inspection that 0.998001 ^ 2 is about 0.996 (following the pattern of the 3rd digit after the decimal). So the answer is 16 * 0.996 which is 15.936.
Thank you so much
I learnt it today. Great. I learnt why I forgot calculus. Must be for good.
(2000-2)^4 use Pascal's triangle to easily do this, and then whatever you get move the decimal 12 places to the left
I used Local Linearity principle and found the equation of tangent line at 2 using derivatives and then calculated function value of tangent line at 1.998 to have a better approximation of this problem
That reminds of high school, such a good nostalgia
At 4:03 , it's _wrong_ to suggest that Δy is (approximately) equal to the derivative dy . (Note that if instead of (1.998)⁴ we had to calculate (1.997)⁴ , then we could use the _same_ relation y = x⁴ and the _same_ "starting point" 2⁴ , so by definition the _actual derivative_ should also remain the same.) I think it's more proper to say
Δy ≈ (dy/dx) * Δx
where dy/dx (and not dy) is actually the derivative.
Since y = x⁴, we have dy/dx = 4x³ , so we get
Δy ≈ 4x³ * Δx
(This way, we're also not dealing clumsily with the distinction between dx and Δx .)
With x = 2 and Δx = -0.002 , this results in
Δy ≈ 4*2³ * (-0.002) = -0.064
hence y = 16 + Δy ≈ 16 - 0.064 = 15.936
- - - - -
By the way, the video's approach is all just an application of the linearization of a (smooth) function f(x) near a point x=c ,
f(c+Δx) ≈ f(c) + f'(c)*Δx
where f'(c) is defined as the value of (df/dx) at x=c.
(In this particular example, we have f(x) = x⁴ and c=2 , and so f(1.998) ≈ f(2) + f'(2)*(-0.002) .)
Binomial theorem. 1.998 = 2 * (1 - 10^-3) so 1.998^4 = 16 * (1-10^-3)^4 = 16*(1 - 4*10^-3 + 6*10^-6 - 4*10^-9+10^-12) = 16 - 0.064 + 0.000096 - 0.000000064 + 0.0000000000016 = 15.936095936016.
Done in approximately 1 minute longer than the video, which I'll now have to go back and watch, from the point you set the challenge. I just hope I've got the number of decimal places right as everything was done using mental arithmetic and written here.
Yo, I've learn this thing b4 but don't know a thing, thanks for clearing this up after 1 year :)
It's amazing!!!!... And have a request can you make a video on different graphs.
Very cool thanks
It's rather elegant to take this as an accumulation, where f(b)=f(a)+int from a to b of f'(x)dx, dx~∆x.
Why not just use Pascal’s triangle at 1 4 6 4 1 and skip all the function blabber? You’ll get an EXACT answer from the expansion of a^4 + 4 a^3•b + 6•(a^2)(b^2) + 4 a• b^3 + b^4 . What’s an error estimate for the procedure used?