Linear Approximation test

In summary: Then f(a)= a5= 25 and f'(a)= 5a4= 5(22.001)4= 5(2.0025)(2.0025)= 5(2.0025)5. Since you are interested in f(2.001) (approximately), you need to calculate f(2.001)= f(2)+ f'(2)(0.001) using those values.
  • #1
Rizzice
3
0
So I have this as the last thing I don't understand before tommorrow's test.. I have tried reading in the book and online, but it's just not clicking for me!

There are so many numbers, and it they seem to just plug them in from nowhere.

Like some example problems would be like sqrt(25.1) and 2.001^5.

Supposedly I use the equation L(x) = f(a) - f'(a)(x-a)? But I have no idea which numbers I plug in where.. is this easier than I'm making it to be?
 
Physics news on Phys.org
  • #2
It would be easier if you read it as:

L(a+e)= f(a) +f'(a)e

that is the value at 'a' plus a little bit (e) is approximately linearly related to the value at 'a' using the derivative at 'a'.

so the square root of 25.1 is approximately the square root at 25 plus the derivative at 25 (of the square root function) times 0.1

All you're doing is using the first two terms of the Taylor/MacLaurin series to estimate the whole sum. For small 'e' this is approximately good enough.
 
  • #3
Rizzice said:
So I have this as the last thing I don't understand before tommorrow's test.. I have tried reading in the book and online, but it's just not clicking for me!

There are so many numbers, and it they seem to just plug them in from nowhere.

Like some example problems would be like sqrt(25.1) and 2.001^5.

Supposedly I use the equation L(x) = f(a) - f'(a)(x-a)? But I have no idea which numbers I plug in where.. is this easier than I'm making it to be?
I think it might help you to read the problem correctly! You say "Like some example problems would be like sqrt(25.1) and 2.001^5."

I hope you understand that "sqrt(25.1)" and "2.001^5" are numbers, not problems! State (at least to yourself) clearly what you are supposed to do with those numbers!

I suspect that the problem really is something like this:
"Use the tangent line approximation to give a simple approximation to sqrt(25.1)" or "approximate sqrt(25.1) without using a calculator".

Then you should think to yourself, "Alright, I know that sqrt(25)= 5 so I don't need a calculator for that! I also know that the tangent line to f(x) at x= a is given y(x)= f(a)+ f'(a)(x-a). Since I am asked to find sqrt(25.1), I had better take x= 25.1! Since I need to be able to find f(a)= sqrt(a) (with a as close to the given point as possible) , I had better take a= 25.

Similarly with "Use the tangent line approximation to give a simple approximation to 2.0015. Since you are specifically asked to find f(2.001) (approximately), you should take x= 2.001. Since you need to be able to find f(a)= a5 easily, and have a close to 2.001, you should take a= 2.
 

1. What is a linear approximation test?

A linear approximation test is a statistical method used to estimate the relationship between two variables. It assumes that the relationship between the variables is linear, meaning that the data points can be approximated by a straight line. This test is commonly used in regression analysis to make predictions or to determine the strength of the relationship between two variables.

2. How is a linear approximation test performed?

To perform a linear approximation test, the data points are plotted on a scatter plot and a line of best fit is drawn through the points. The closer the data points are to the line, the stronger the relationship between the variables. The test also calculates the correlation coefficient, which indicates the strength and direction of the relationship between the variables. The closer the correlation coefficient is to 1 or -1, the stronger the relationship.

3. What is the purpose of a linear approximation test?

The purpose of a linear approximation test is to determine the strength of the relationship between two variables and to make predictions based on this relationship. It helps to identify patterns and trends in the data and can be used to make informed decisions in various fields such as economics, social sciences, and engineering.

4. What are the assumptions of a linear approximation test?

The main assumptions of a linear approximation test are that the relationship between the variables is linear, the data is normally distributed, and the variability of the data is constant. Additionally, the data should be independent and there should be no outliers or influential points that may skew the results of the test.

5. What are the limitations of a linear approximation test?

One limitation of a linear approximation test is that it can only determine the relationship between two variables and cannot account for any other factors that may influence the data. Additionally, the test assumes a linear relationship, so it may not be suitable for non-linear data. It also requires a sufficiently large sample size to ensure accurate results.

Similar threads

Replies
14
Views
1K
  • Calculus
Replies
12
Views
838
Replies
3
Views
1K
Replies
30
Views
2K
Replies
3
Views
2K
Replies
5
Views
2K
Replies
11
Views
2K
  • Calculus
Replies
3
Views
745
Replies
3
Views
1K
  • Calculus
Replies
6
Views
1K
Back
Top