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

[antinerd]

Homework Statement

OK, I'm doing this linear approximation problem:

Approximate $$\sqrt{4.1} - \sqrt{3.9}$$

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) = $$\sqrt{x}$$
f`(x) = $$1/(2\sqrt{x}$$

then I got:

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

I did the same thing for $$\sqrt{3.9}$$ 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?

 

[neutrino]

All you steps are right. Why do you think it is erroneous?

 

[antinerd]

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...

 

[learningphysics]

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 $$\sqrt{4.1} - \sqrt{3.9}$$

it's very close to 0.2/4 = 0.05

 

[antinerd]

Thanks.

 

[neutrino]

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.

 

[antinerd]

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.

 

[Integral]

What's the problem?

 

[antinerd]

What's the problem?

EDIT:

Should I use -0.1 for h or can i use 0.1...

 

[antinerd]

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?

 

[antinerd]

Holy crap man, nevermind. I've been up for like 3 days straight so I'm buggin' out.

Thanks for your insight, guys :)

 

[learningphysics]

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)

 

[learningphysics]

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 = $$x_{actual}-x_{approximate}$$

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...