Linear Approximation Part II

[Hckyplayer8]

Estimate the Sin 31

f(x) = sin x
f'(x) = cos x

f(sin 30) + f'(sin 30) * (sin 31 - sin 30) = sin 30 + cos 30 * (sin 31 - sin 30) = 1/2 + cos square root 3 over 2 * (sin 31 - sin 30)

Does everything look in order thus far?

 

[MarkFL]

I would begin by stating:

[MATH]\frac{\Delta f}{\Delta x}\approx\d{f}{x}[/MATH]
This implies:

[MATH]f(x+\Delta x)\approx\d{f}{x}\Delta x+f(x)[/MATH]
For this problem, we have:

[MATH]f(x)=\sin(x)\implies \d{f}{x}=\cos(x)[/MATH]
[MATH]x=\frac{\pi}{6}[/MATH]
[MATH]\Delta x=\frac{\pi}{180}[/MATH]
Hence:

[MATH]\sin\left(\frac{\pi}{6}+\frac{\pi}{180}\right)\approx\cos\left(\frac{\pi}{6}\right)\cdot\frac{\pi}{180}+\sin\left(\frac{\pi}{6}\right)\approx0.5151149947[/MATH]
For comparison, W|A gives an approximation of:

[MATH]\sin\left(31^{\circ}\right)\approx0.5150380749100542[/MATH]

 

[HallsofIvy]

It's not a matter of "thus far". What you have written makes no sense. You want to "approximate sin(31)" but you have sin(31) in the formula!

A linear approximation for f(x) about $\displaystyle x_0$ is $\displaystyle f(x)= f(x_0)+ f'(x_0)(x- x_0)$

NOT $\displaystyle f(x)= f(x_0)+ f'(x)(f(x)- f(x_0))$

 

[Hckyplayer8]

It's not a matter of "thus far". What you have written makes no sense. You want to "approximate sin(31)" but you have sin(31) in the formula!

A linear approximation for f(x) about $\displaystyle x_0$ is $\displaystyle f(x)= f(x_0)+ f'(x_0)(x- x_0)$

NOT $\displaystyle f(x)= f(x_0)+ f'(x)(f(x)- f(x_0))$

Is my last thread we found an estimate for 26.8^2/3 which was

f(x) + f'(x) *(26.8-27)

which is what I was trying to replicate in this thread.

 

[ksdhart2]

Correcting for what I'm sure is a typo, your solution seems to boil down to:

$\displaystyle \sin(31^{\circ}) \approx \frac{1}{2} + \frac{\sqrt{3}}{2} \times \biggr( \sin(31^{\circ}) - \sin(30^{\circ}) \biggr)$

As Halls pointed out, this is problematic because $\sin(31^{\circ})$ appears on both sides of the equation. If we use the symbol $S$ to stand for this value, then we get:

$\displaystyle S = \frac{1}{2} + \frac{\sqrt{3}}{2} \times \biggr( S - \frac{1}{2} \biggr)$

A little of a algebraic manipulation shows us that this can only be true if $S = \frac{1}{2}$. In other words, your approximation says that $\sin(31^{\circ}) = \sin(30^{\circ}) = \frac{1}{2}$. Does that make any sense?

Is my last thread we found an estimate for 26.8^2/3 which was

f(x) + f'(x) *(26.8-27)

which is what I was trying to replicate in this thread.

The previous problem is entirely consistent with this one. Nothing changed behind the scenes, it's just a little bit of thought may be needed to understand why. In your previous problem, you were trying to estimate the value of $26.8^{2/3}$. That means that $f(x) = x^{2/3}$. This function is then evaluated at the points $x = 26.8$ and $x_0 = 27$. Therefore, when creating a linear approximation, you used the formula:

$\displaystyle f(x) \approx f(x_0) + f^{\prime}(x_0) \times (x - x_0)$

It logically follows then that you'd want to follow the exact same process. Plugging the present values of $\displaystyle f(x) = \sin(x)$, $\displaystyle x = \frac{31\pi}{180}$, and $\displaystyle x_0 = \frac{\pi}{6}$:

$\displaystyle \sin \left( \frac{31\pi}{180} \right) \approx \sin \left( \frac{\pi}{6} \right) + \cos \left( \frac{\pi}{6} \right) \times \left( \frac{31\pi}{180} - \frac{\pi}{6} \right)$

At the heart of the matter is the distinction between $\displaystyle x = \frac{31\pi}{180}$ and $\displaystyle f(x) = \sin \left( \frac{31\pi}{180} \right)$. Those two are definitely not the same expression.

 

[JeffM]

Estimate the Sin 31

f(x) = sin x
f'(x) = cos x

f(sin 30) + f'(sin 30) * (sin 31 - sin 30) = sin 30 + cos 30 * (sin 31 - sin 30) = 1/2 + cos square root 3 over 2 * (sin 31 - sin 30)

Does everything look in order thus far?

Let's start over.

Paraphrasing Mark slightly, the general thought is

[MATH]\Delta x \ne 0 \text { but } \Delta x \approx 0 \implies \dfrac{f(x + \Delta x) - f(x)}{\Delta x} \approx f'(x) \implies[/MATH]
[MATH]f(x + \Delta x) - f(x) \approx f'(x) \Delta x \implies f(x + \Delta x) \approx f(x) + f'(x) \Delta x.[/MATH]
Now Mark did not say this, but I shall: when dealing with trig functions in calculus, work in radians to avoid error.

Now let's get to Halls' post, but again my paraphrase.

Following on from Mark, we get a general method of approximation:

[MATH]f(q) \text { and } f'(q) \text { are known exactly, } p \ne q, \text { but } p \approx q \implies f(p) \approx f(q) + f'(q) * (p - q).[/MATH]
Follow that? On one side of the approximation, we have what we want to approximate, and on the other side we only have things that we can express exactly.

So when we get to your specific problem, we arrive at

[MATH]sin \left ( \dfrac{\pi}{6} + \dfrac{\pi}{180} \right ) \approx sin \left ( \dfrac{\pi}{6} \right ) + cos \left ( \dfrac{\pi}{6} \right ) * \left \{ \left ( \dfrac{\pi}{6} + \dfrac{\pi}{180} \right ) - \dfrac{\pi}{180} \right \}.[/MATH]
That is NOT what you had. Ignoring the failure to use radians, you were not multiplying the cosine by delta x but rather by f(x + delta x) - f(x), which is totally wrong and not what follows from what Mark said. Furthermore, as Halls pointed out very cogently, it also makes no sense because you are trying to find an approximate value for f(x + delta x) in the first place so you cannot use it to find it.

Do you see now what you did wrong?

 

[Hckyplayer8]

Correcting for what I'm sure is a typo, your solution seems to boil down to:

$\displaystyle \sin(31^{\circ}) \approx \frac{1}{2} + \frac{\sqrt{3}}{2} \times \biggr( \sin(31^{\circ}) - \sin(30^{\circ}) \biggr)$

As Halls pointed out, this is problematic because $\sin(31^{\circ})$ appears on both sides of the equation. If we use the symbol $S$ to stand for this value, then we get:

$\displaystyle S = \frac{1}{2} + \frac{\sqrt{3}}{2} \times \biggr( S - \frac{1}{2} \biggr)$

A little of a algebraic manipulation shows us that this can only be true if $S = \frac{1}{2}$. In other words, your approximation says that $\sin(31^{\circ}) = \sin(30^{\circ}) = \frac{1}{2}$. Does that make any sense?



The previous problem is entirely consistent with this one. Nothing changed behind the scenes, it's just a little bit of thought may be needed to understand why. In your previous problem, you were trying to estimate the value of $26.8^{2/3}$. That means that $f(x) = x^{2/3}$. This function is then evaluated at the points $x = 26.8$ and $x_0 = 27$. Therefore, when creating a linear approximation, you used the formula:

$\displaystyle f(x) \approx f(x_0) + f^{\prime}(x_0) \times (x - x_0)$

It logically follows then that you'd want to follow the exact same process. Plugging the present values of $\displaystyle f(x) = \sin(x)$, $\displaystyle x = \frac{31\pi}{180}$, and $\displaystyle x_0 = \frac{\pi}{6}$:

$\displaystyle \sin \left( \frac{31\pi}{180} \right) \approx \sin \left( \frac{\pi}{6} \right) + \cos \left( \frac{\pi}{6} \right) \times \left( \frac{31\pi}{180} - \frac{\pi}{6} \right)$

At the heart of the matter is the distinction between $\displaystyle x = \frac{31\pi}{180}$ and $\displaystyle f(x) = \sin \left( \frac{31\pi}{180} \right)$. Those two are definitely not the same expression.

Let's start over.

Paraphrasing Mark slightly, the general thought is

[MATH]\Delta x \ne 0 \text { but } \Delta x \approx 0 \implies \dfrac{f(x + \Delta x) - f(x)}{\Delta x} \approx f'(x) \implies[/MATH]
[MATH]f(x + \Delta x) - f(x) \approx f'(x) \Delta x \implies f(x + \Delta x) \approx f(x) + f'(x) \Delta x.[/MATH]
Now Mark did not say this, but I shall: when dealing with trig functions in calculus, work in radians to avoid error.

Now let's get to Halls' post, but again my paraphrase.

Following on from Mark, we get a general method of approximation:

[MATH]f(q) \text { and } f'(q) \text { are known exactly, } p \ne q, \text { but } p \approx q \implies f(p) \approx f(q) + f'(q) * (p - q).[/MATH]
Follow that? On one side of the approximation, we have what we want to approximate, and on the other side we only have things that we can express exactly.

So when we get to your specific problem, we arrive at

[MATH]sin \left ( \dfrac{\pi}{6} + \dfrac{\pi}{180} \right ) \approx sin \left ( \dfrac{\pi}{6} \right ) + cos \left ( \dfrac{\pi}{6} \right ) * \left \{ \left ( \dfrac{\pi}{6} + \dfrac{\pi}{180} \right ) - \dfrac{\pi}{180} \right \}.[/MATH]
That is NOT what you had. Ignoring the failure to use radians, you were not multiplying the cosine by delta x but rather by f(x + delta x) - f(x), which is totally wrong and not what follows from what Mark said. Furthermore, as Halls pointed out very cogently, it also makes no sense because you are trying to find an approximate value for f(x + delta x) in the first place so you cannot use it to find it.

Do you see now what you did wrong?

Ah I see. Stupid mistake confusing x and f(x).

Thank you all for posting.