Approximating (1.998)^4 by using differential

Approximating (1.998)^4 by using differential,

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blackpenredpen | 曹老師

[Phi]

Mathematician: "Let's use calculus to solve this."
Engineer: "It's 16."

[aryansant]

I think he uses Integral Calculus in grocery shopping.

[merubindono]

Engineer: it's about 16. Add 25% contingency factor. Let's make it 20.

[PackSciences]

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.

[michel_dutch]

Next: solve a calculus problem using only arithmetic. 😁

[gubby]

Approximately 16, you’re welcome

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Don’t take it seriously lol

[josephjackson]

You know someone is smart when they use Wolfram Alpha as a calculator

[tanelgulerman]

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

[Exachad]

This is a more conplicated demonstration of tangent line approximations.

[harshvardhangupta]

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

[tushar.mp]

Is it only me who heard *Doraemon* tune in the intro? Btw thanks for this amazing video <3

[elchingon]

This is an excellent explanation for local linear approximation for anyone who has basic knowledge of derivatives

[calyodelphi]

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.

[kingbeauregard]

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:

[gianlucamolinari]

This is really interesting how calculus can be used to solve such problems!

[ralfbodemann]

The relative deviation of your approximation from the real value is 0.0006%. For practical purposes, this is usually neglectable.

[collegemathematics]

It easier to use the linear approximation using first two terms in taylor series
L=f(a) +d/dx f(a) 1
where a is constant, and L stand for linear approximation of the original function

d/dx f(2)
Substitute 2 and 3 in 1
L=16+32(x-2) ....4
SUBSTITUTE x=1.998 in 4 implies
L=15.936

[deutschlandmeinvaterland]

Excellent video. Also, it's really awesome to read all these comments offering other solutions as well! Math is so fun!

[Deibler]

This is very interesting! This approximation method is very precise and you can always experiment with other values as well.

[anshumanagrawal]

If you wanna use approximation by calculus, it's better to use the form (1+x)^n ~ 1+ nx, where x<<1

(1.998)^4 = (2×0.999)^4 = 2^4 × (0.999)^4
= 16 (1- 0.001)^4
~ 16 × (1- 0.001×4) = 16 (1- 0.004)
= 16×0.996 or 16 - 0.064
=15.936