Need some help with understanding linear approximations

In summary: Linear Approximation: f(x+Δx)≈f(x)+f '(x)Δx or Δy≈f '(x)Δx.In summary, linear approximation can be used when calculating a function at a certain point, but it can be tricky to use because you need to understand the basic concept of derivatives. This problem may seem trivial to you, but to someone who is unfamiliar with the concept, it can be quite difficult.
  • #1
KFSKSS
14
0
Hello. My problem is that I began with Linear Approximation and I'm terribly stuck. I have problems understanding its very concept and with calculations. (It may sound stupid but I'm autodidact and I'm studying mathematics in english (not my mother tongue) and sometimes it gets hard).

It would really help this: In the book I'm using in the exercises there's one that says "Using linear approximation, find an approximate value for the following quantities" one of those quantities is (2.94)^4. The solution must be 74.52. The problem is that I don't know how to use the equations. I know what they mean though. It's annoying.
If you could just solve this so I can see how you do it and learn it I would really appreciate it.
Linear Approximation: f(x+Δx)≈f(x)+f '(x)Δx or Δy≈f '(x)Δx.
Thanks in advance.
 
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  • #2
KFSKSS said:
Hello. My problem is that I began with Linear Approximation and I'm terribly stuck. I have problems understanding its very concept and with calculations. (It may sound stupid but I'm autodidact and I'm studying mathematics in english (not my mother tongue) and sometimes it gets hard).

It would really help this: In the book I'm using in the exercises there's one that says "Using linear approximation, find an approximate value for the following quantities" one of those quantities is (2.94)^4. The solution must be 74.52. The problem is that I don't know how to use the equations. I know what they mean though. It's annoying.
If you could just solve this so I can see how you do it and learn it I would really appreciate it.
Linear Approximation: f(x+Δx)≈f(x)+f '(x)Δx or Δy≈f '(x)Δx.
Thanks in advance.

Better yet, why don't you try to solve it first? Put ##f(x)=x^4##, ##x=3## and ##\Delta x=(-0.06)##. It's really not that hard.
 
  • #3
If you choose x = 3, which is close to 2.94, then what is
Δx? What is f' ?
 
  • #4
Can you think of a whole number that is close to 2.94?

Ah, I arrived late.
 
  • #5
Dick, I tried many things and sometimes I get a number under the result or over it.
Scottdave, f ' is the derivative of the function f. f '=df/dy.
I have no idea how it can give 74.52.
Thanks for answering anyway.
 
  • #6
You wrote this --
KFSKSS said:
Linear Approximation: f(x+Δx)≈f(x)+f '(x)Δx ...
Use Dick's hint from post #2. I get ##f(2.94) \approx 74.52##.

##f(x) = x^4##, so f'(x) = ?
 
  • #7
Show us what you got when you tried Dick's suggestion so we can see what's wrong.
 
  • #8
What I have been doing until now was this: (2.94)^4=(x+Δx)^4=x^4+4x^3Δx+6x^2Δx^2+4xΔx^3+Δx^4; d/dx(x^4+4x^3Δx+6x^2Δx^2+4xΔx^3+Δx^4)=4x^3+12x^2Δx+12xΔx^2+4Δx^3; now if x=2 and Δx=0.94=4(2)^3+12(2)^2(0.94)+12(2)(0.94)^2+4(0.94)^3=101.648736.
Honestly I think the book has an awful explanation of this subject. Mark44 how did you do it? Thank you.
 
  • #9
Dick how did you do it?
 
  • #10
The convenient Δx is not 0.94. I and others mentioned take a whole number nearest 2.96 .
 
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  • #11
KFSKSS said:
Dick how did you do it?

As epenguin suggests, make ##\Delta x## negative and ##x=3## not ##2##.
 
  • #12
Do you know how to calculate df/dx when f = x^4 ?
 
  • #13
KFSKSS said:
What I have been doing until now was this: (2.94)^4=(x+Δx)^4=x^4+4x^3Δx+6x^2Δx^2+4xΔx^3+Δx^4; d/dx(x^4+4x^3Δx+6x^2Δx^2+4xΔx^3+Δx^4)=4x^3+12x^2Δx+12xΔx^2+4Δx^3; now if x=2 and Δx=0.94=4(2)^3+12(2)^2(0.94)+12(2)(0.94)^2+4(0.94)^3=101.648736.
Honestly I think the book has an awful explanation of this subject.
The idea isn't to expand ##(x + \Delta x)^4##, as you are doing above.
KFSKSS said:
Mark44 how did you do it? Thank you.
I did exactly what you wrote in post #1. This is the linear approximation.
KFSKSS said:
Linear Approximation: f(x+Δx)≈f(x)+f '(x)Δx ...
Since you want f(2.94), you want x to be close to 2.94, but easy to work with -- an integer. ##\Delta x## isn't necessarily going to be positive.
 
  • #14
Scottdave, f=x^4, df/dx= 4x^3.
 
  • #15
KFSKSS said:
What I have been doing until now was this: (2.94)^4=(x+Δx)^4=x^4+4x^3Δx+6x^2Δx^2+4xΔx^3+Δx^4; d/dx(x^4+4x^3Δx+6x^2Δx^2+4xΔx^3+Δx^4)=4x^3+12x^2Δx+12xΔx^2+4Δx^3; now if x=2 and Δx=0.94=4(2)^3+12(2)^2(0.94)+12(2)(0.94)^2+4(0.94)^3=101.648736.
Honestly I think the book has an awful explanation of this subject. Mark44 how did you do it? Thank you.
So the linear approximation uses the fact that for small Delta x, you assume that (Delta x)^2 and higher powers is approximately zero.
 
  • #16
Pick x = 3 and you can get a relatively small Delta x
 
  • #17
So x must be a number close to 2.94, which is three. In case 1.2, x=1 right? What about Δx? How do I choose its value?
 
  • #18
So if you are wanting to find the function at 1.2, you could choose x=1 and Δx = +0.2. Just note that as Δx increases, your approximation will be farther away from the true value. Looking back at 2.94, you can choose x=3, and Δx = -0.06, so for the term (Δx)2, you are on the order of 0.0036 multiplied by a number. When Δx = 0.2 you are on the order of 0.04

It may seem trivial to you. Maybe you think "Why can't I just type 2.94 and raise to power 4?" That seems pretty simple.

There are other functions like square root, and log, which linear approximations can be used. If you have a computer having to analyze millions of data points, and the linear approximation can be made "close enough" for the application, then there can be a real advantage to using these approximations.
 
  • #19
KFSKSS said:
So x must be a number close to 2.94, which is three.
Yes. x = 3. You're picking a number that's reasonably close to 2.94, and for which it's easy to calculate f(x) and f'(x). We're assuming that it's "hard" to calculate ##2.94^4##

KFSKSS said:
In case 1.2, x=1 right?
Yes, but let's stay with the current problem.

KFSKSS said:
What about Δx? How do I choose its value?
You want ##x + \Delta x## to be 2.94.
 
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  • #20
Where am I failing? 2.94; x=3, Δx=2.94-3=-0.06. But (3)^4+4(3)^3·(-0.06)=11.34??
 
  • #21
To get x I choose a whole number close to value on the function and subtract; 2.94, x=2.94-3=-0.06. Is this right?
 
  • #22
3^4 = 81.
3^3 = 27.
27*4 = 108.
108*(.06) = 6.48
81-6.48 =
 
  • #23
So for example, the linear approximation of (4.62)^3=x=5; Δx=4.62-5=-0.38; (5)^3+3(5)^2·(-0.38)=125+75(-0.38)=125-28.5=96.5. Is this correct?
 
  • #24
KFSKSS said:
So for example, the linear approximation of (4.62)^3=x=5; Δx=4.62-5=-0.38; (5)^3+3(5)^2·(-0.38)=125+75(-0.38)=125-28.5=96.5. Is this correct?
Yes, this is correct.
Also, the smaller ##\Delta x## is in comparison to x, the better the approximation will be. So, for example, the approximation for (5.1)3 will be better than the one for (4.8)3. "Better" means "smaller error". For your approximation of 4.623, the true value (by calculator) is 98.611128, so the error was larger than 2, a bit more than a 2% error.
 
  • #25
Mark44 said:
The idea isn't to expand ##(x + \Delta x)^4##, as you are doing above.

I did exactly what you wrote in post #1. This is the linear approximation.

Since you want f(2.94), you want x to be close to 2.94, but easy to work with -- an integer. ##\Delta x## isn't necessarily going to be positive.

You can do this expansion according to the binomial formula to approximate a power of x (and hence also to approximate any polynomial) and the result is exactly the same as using differentiation. (.Just not so suitable for more general functions)

You say your book is bad, but if you needed to be told that the whole number nearest to 2.96 is 3 maybe it's not all the book's fault?... We all have our blind moments.

The idea is very simple: suppose you are traveling at 36 Km/h, that is10 m/s (f'(t)) and you are at some time 20 km (x) from home, Then after another 10 s (Δt) will be an extra 10×10 m (Δx = (dx/dt)×Δt) from home, in total 20,000 + 100 = 20100 m from home (x + Δx) = ... .
This calculation will give an approximately good result if the speed is not varying too much. If the speed is varying at that point really brusqely , it will need a shorter interval for the calculation to be any good.

In other words you calculate the function at a point near a point where it is known, as if it increased beyond the known point at the same rate that is increasing at the known point.
 
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  • #26
I like this book I'm learning relatively fast with it. Only sometimes the things in it are "not written the way I would understand". Anyway thank you all. And thanks for your patience.
 
  • #27
epenguin said:
You can do this expansion according to the binomial formula to approximate a power of x (and hence also to approximate any polynomial) and the result is exactly the same as using differentiation.
No, the result will be different.

The binomial expansion gives
##(x + \Delta x)^4 = x^4 + 4x^3\Delta x + 6x^2(\Delta x)^2 + 4x(\Delta x)^3 + (\Delta x)^4##
The linear approximation involves only the first two terms of the right side. IOW, only the terms up to (and including) the linear (i.e., first-degree) term in ##\Delta x##).
In any case, owing to the title of the thread, the intended approximation here is this one: ##(x + \Delta x)^4 = x^4 + 4x^3\Delta x##.
 
  • #28
I was taking as understood you only take the process as far as the second term.
 
  • #29
epenguin said:
I was taking as understood you only take the process as far as the second term.
That wasn't clear to me in what you wrote.
 
  • #30
KFSKSS said:
Where am I failing? 2.94; x=3, Δx=2.94-3=-0.06. But (3)^4+4(3)^3·(-0.06)=11.34??
Apparently you've got it now, but I would suggest part of your problem is inability to do algebra/arithmetic.
 
  • #31
LCKurtz said:
Apparently you've got it now, but I would suggest part of your problem is inability to do algebra/arithmetic.
As epenguin said "we all have our blind moments".
 

1. What is a linear approximation?

A linear approximation is an estimation of a non-linear function using a linear function. It is used to simplify complex functions and make them easier to analyze and understand.

2. How is a linear approximation calculated?

A linear approximation is calculated using the tangent line to a point on the curve of the function. The slope of the tangent line is then used to create a linear function that closely approximates the original function at that point.

3. What is the purpose of using linear approximations?

The purpose of using linear approximations is to make complex functions more manageable and easier to analyze. It allows for a better understanding of the behavior of the function near a specific point.

4. Can linear approximations be used for any type of function?

No, linear approximations can only be used for functions that are differentiable at the point of approximation. This means that the function must have a well-defined tangent line at that point.

5. What are some real-world applications of linear approximations?

Linear approximations are commonly used in physics, engineering, and economics to approximate complex functions and make predictions. They are also used in computer graphics and animation to create smooth curves and surfaces.

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