Linear Approximation: Approximating \sqrt{4.1}-\sqrt{3.9}

In summary, the student attempted to solve a linear approximation problem by approximating each square root separately, and then taking the sum of the squares. However, the approximation of \sqrt{4.1} is very large and may not be accurate.
  • #1
antinerd
41
0

Homework Statement



OK, I'm doing this linear approximation problem:

Approximate [tex]\sqrt{4.1} - \sqrt{3.9}[/tex]

Homework Equations



f(a + h) ~ f(a) + hf`(a)

The Attempt at a Solution



This is what I have done so far:

I approximated each square root separately...

4.1 = 4 + h
h = .1
f(x) = [tex]\sqrt{x}[/tex]
f`(x) = [tex]1/(2\sqrt{x}[/tex]

then I got:

[tex]\sqrt{x+h} ~ \sqrt{x} + h/(2\sqrt{x})[/tex]
so that the approximation of [tex]\sqrt{4.1}[/tex] is 2 + (.1/4)

I did the same thing for [tex]\sqrt{3.9}[/tex] and got

2 - (.1/4)

Then I just took them and subtracted

2 + (.1/4) - 2 + (.1/4)
and got .2/4 as the approximation. It seems like it's erroneous. Could someone help me out with setting up this problem if I did it wrong?
 
Physics news on Phys.org
  • #2
All you steps are right. Why do you think it is erroneous?
 
  • #3
neutrino said:
All you steps are right. Why do you think it is erroneous?

It seems to me that it is a large margin of error, I was just wondering why this is so.

Or maybe it's not too large, I'm not sure...
 
  • #4
antinerd said:
It seems to me that it is a large margin of error, I was just wondering why this is so.

Or maybe it's not too large, I'm not sure...

everything looks right to me.

check your calculator and calculate [tex]\sqrt{4.1} - \sqrt{3.9}[/tex]

it's very close to 0.2/4 = 0.05
 
  • #5
Thanks.
 
  • #6
antinerd said:
It seems to me that it is a large margin of error, I was just wondering why this is so.

Or maybe it's not too large, I'm not sure...

It's an approximation. Without a calculator, it's not an easy task for many to calculate the square root of an arbitrary number, unless it's a "nice" number like a perfect square. In those cases, you can use this method to approximate square roots to find a number reasonably close to the right one.
 
  • #7
So let's say I was doing something like e^(-0.1) - ln (1.1)

I can just go about doing it thus:

h = -0.1

so that:
-0.1 = 0 + h and 1.1 = 1 - h



for the first term:

f(x) = e^(x)
f`(x) = e^(x)

but then how do I go about it from there...

Could someone help me setup the problem?

I guess this would work, wouldn't it:

e^(0) - h/(e^(0))
1 - 0.1 = 0.9

Which seems reasonable...

And then for the ln(1.1) it would be
f(x) = ln(x)
f`(x) = 1/x

ln(1) - h/x
which would give me
0 - (-0.1/1)
= 0.1

That doesn't make any sense...

What'd I do wrong.
 
  • #8
What's the problem?
 
  • #9
Integral said:
What's the problem?

EDIT:

Should I use -0.1 for h or can i use 0.1...
 
Last edited:
  • #10
When I do:

e^0 - (h / e^0)
= 1 + (0.1/1)
= 1.1

It should be 0.9 ...

I have the correct approximation for -ln(1.1) = 0.1

but I made a mistake

I guess this would work, wouldn't it:

e^(0) - h/(e^(0))
1 - 0.1 = 0.9

Shouldn't it be 1 PLUS 0.1 to give me 1.1? Cuz h = NEGATIVE 0.1

What's wrong here?
 
  • #11
Holy crap man, nevermind. I've been up for like 3 days straight so I'm buggin' out.

Thanks for your insight, guys :)
 
  • #12
antinerd said:
So let's say I was doing something like e^(-0.1) - ln (1.1)

I can just go about doing it thus:

h = -0.1

so that:
-0.1 = 0 + h and 1.1 = 1 - h



for the first term:

f(x) = e^(x)
f`(x) = e^(x)

but then how do I go about it from there...

Could someone help me setup the problem?

I guess this would work, wouldn't it:

e^(0) - h/(e^(0))
1 - 0.1 = 0.9

Although the numbers come out right, I'm wondering why you are dividing by e^(0) instead of multiplying by e^(0)
 
  • #13
I would advise that you approximate each term separately so that you don't get confused with the signs... Also, define h consistently...

so I'd do this:

let h = [tex]x_{actual}-x_{approximate}[/tex]

So if I'm approximating e^(-0.1) by e^0...

h = -0.1 - 0 = -0.1

And then use this exact formula where a refers to the approximate x value:

f(a + h) ~ f(a) + hf`(a) (don't switch from h to -h or anything like that... be consistent)

So: f(0 + (-0.1)) ~ f(0) + (-0.1)f'(0)

ie: e^(0 + (-0.1)) ~ e^(0) + (-0.1)(e^(0)) = 0.9

So e^(-0.1) ~ 0.9

Now approximate ln(1.1) as a separate problem, defining h again...

It is important to be consistent... worry about signs at the end...
 

What is linear approximation?

Linear approximation is a method used in calculus to approximate the value of a function at a certain point by using the tangent line at that point. It is useful when the function is difficult or impossible to solve directly.

Why would we use linear approximation to approximate √4.1 - √3.9?

We would use linear approximation in this case because the square root function is not easily solvable by hand. By using the tangent line at a specific point, we can get a close approximation of the value without having to manually solve the function.

What is the formula for linear approximation?

The formula for linear approximation is y = f(a) + f'(a)(x-a), where a is the point at which we are approximating the function and f'(a) is the derivative of the function at that point.

How do we apply linear approximation to approximate √4.1 - √3.9?

We first find the derivative of the square root function, which is 1/(2√x). Then, we plug in the values of 4.1 and 3.9 into the formula y = f(a) + f'(a)(x-a). This will give us the equation of the tangent line at those points. Finally, we can plug in a value for x (such as 4) to get an approximation for √4.1 - √3.9.

Is linear approximation always accurate?

No, linear approximation is not always accurate. It is only an approximation and will not give us the exact value of the function. The accuracy of the approximation depends on how close the point we are approximating is to the actual value of the function.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
790
  • Calculus and Beyond Homework Help
Replies
14
Views
234
  • Calculus and Beyond Homework Help
Replies
21
Views
754
  • Calculus and Beyond Homework Help
Replies
4
Views
756
  • Calculus and Beyond Homework Help
Replies
1
Views
633
  • Calculus and Beyond Homework Help
Replies
5
Views
896
  • Calculus and Beyond Homework Help
Replies
20
Views
383
Replies
9
Views
660
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
Back
Top