Linear Approximation

[Hckyplayer8]

Visually, I understand what is going on with this topic. Since a tangent line only touches a curve at one spot, as one moves further and further away from that specific point, the approximation error gets large and large as more "space" develops between the curve and the tangent line to the aforementioned point. But for secondary points relatively close to the original point, the fact the tangent line is fairly close to that of the curve means we can approximate some tougher functions to some success.

The linear approximation equation is L(x) = f(a) + f'(a) (x-a).

Estimate 26.8^2/3

Handling the fractional power results in (26.8^1/3)² = (3 square root 26.8)²

But now what? If the number was rounded to 27, the Algebra would bring me to the cube root of 27 is 3. Do I use the fact I know that and apply 27 as my secondary x coordinate?

 

[Dr.Peterson]

It's a good observation that a perfect cube would be easier to work with, and that 26.8 is close to 27.

So one way to do this would be to treat it as (27 - 0.2)^2/3. That is, 0.2 is the small change in x away from 27.

So let f(x) = x^2/3, and do a linear approximation with a = 27 and x = 26.8.

By the way, the cube root isn't "3 square root"; that would be 3 times the square root. But I do understand what you meant: [MATH]\sqrt[3]{26.8}^2[/MATH].

 

[JeffM]

Building on what Dr. Peterson had to say

[MATH]f(x) = x^{2/3} = \sqrt[3]{x^2} \implies f'(x) = \dfrac{2}{3} * x^{-1/3} = \dfrac{2}{3\sqrt[3]{x}}.[/MATH]
Now we get for an estimate of f(26.8)

[MATH]f(27) + f'(27) * (26.8 - 27) = \sqrt[3]{(3^3)^2} + \dfrac{2 * (-\ 0.2)}{3\sqrt[3]{27}} = \sqrt[3]{(3^2)^3} - \dfrac{0.4}{3 * 3} = [/MATH]
[MATH]3^2 - \dfrac{.4}{9} \approx 9 - 0.0444 = 8.9556.[/MATH]
So everything is very easy arithmetically. How good is the approximation?

[MATH]\sqrt[3]{26.8^2} = \sqrt[3]{718.24} \approx 8.9555.[/MATH]
Very good approximation for most purposes.

 

[Hckyplayer8]

It's a good observation that a perfect cube would be easier to work with, and that 26.8 is close to 27.

So one way to do this would be to treat it as (27 - 0.2)^2/3. That is, 0.2 is the small change in x away from 27.

So let f(x) = x^2/3, and do a linear approximation with a = 27 and x = 26.8.

By the way, the cube root isn't "3 square root"; that would be 3 times the square root. But I do understand what you meant: [MATH]\sqrt[3]{26.8}^2[/MATH].

Ah yeah. Should have said radical. Thanks for catching that and for the help!

 

[Hckyplayer8]

Building on what Dr. Peterson had to say

[MATH]f(x) = x^{2/3} = \sqrt[3]{x^2} \implies f'(x) = \dfrac{2}{3} * x^{-1/3} = \dfrac{2}{3\sqrt[3]{x}}.[/MATH]
Now we get for an estimate of f(26.8)

[MATH]f(27) + f'(27) * (26.8 - 27) = \sqrt[3]{(3^3)^2} + \dfrac{2 * (-\ 0.2)}{3\sqrt[3]{27}} = \sqrt[3]{(3^2)^3} - \dfrac{0.4}{3 * 3} = [/MATH]
[MATH]3^2 - \dfrac{.4}{9} \approx 9 - 0.0444 = 8.9556.[/MATH]
So everything is very easy arithmetically. How good is the approximation?

[MATH]\sqrt[3]{26.8^2} = \sqrt[3]{718.24} \approx 8.9555.[/MATH]
Very good approximation for most purposes.

Thank you for the help. I have two questions.

1. When converting a fractional power to a radical, is attaching the power to the variable inside of the radical saying the same thing if it is attached outside of the radical?

2. When you inserted 27 into f(x), is there a specific reason you went with the 3 to the 3rd power notation instead of just leaving it 27?

 

[JeffM]

[MATH]\left (\sqrt[3]{a} \right)^2 \equiv \sqrt[3]{a} * \sqrt[3]{a} \equiv \sqrt[3]{a * a} \equiv \sqrt[3]{a^2}.[/MATH]
Clear now?

I just translated 27 into 3^3 to make clear what we were doing with the cube roots. It makes no fundamental difference, but it reflects your perception that 27 is a perfect cube.

 

[Hckyplayer8]

[MATH]\left (\sqrt[3]{a} \right)^2 \equiv \sqrt[3]{a} * \sqrt[3]{a} \equiv \sqrt[3]{a * a} \equiv \sqrt[3]{a^2}.[/MATH]
Clear now?

I just translated 27 into 3^3 to make clear what we were doing with the cube roots. It makes no fundamental difference, but it reflects your perception that 27 is a perfect cube.

Yes. Thanks!