Newton's Method and Hoptial's Rule

[redbird555]

Hi all I have these problem on my test review and even after guessing one right I have no idea how one would arrive at that answer so your help is greatly appreciated.

Homework Statement

1.find the limit as x approaches 0 in the equation[(3/x^4)-(4/x^2)]

2.Use Newton's method to find the positive value of which satisfies . Compute enough approximations so that your answer is within .05 of the exact answer.

The Attempt at a Solution

For the limit problem I simply tried taking the derivative of the two equations and subtracting them which gave me 0/0. I know the answer to the problem is infinity but I have no idea how one would arrive at that solution?

On the second problem I really don't have a clue. I don't know how to apply that to the normal Newtons method formula to a problem worded like this any help is appreciated.

 

[HallsofIvy]

$$\frac{1}{x^4}- \frac{4}{x^2}= \frac{1- 4x^2}{x^4}$$.
Apply L'Hopital's rule to that.

"Use Newton's method to find the positive value of which satisfies ."
I have no clue what that means. There seem to be words missing after "positive value of" and "satisfies".

 

[redbird555]

sorry it should read.
Use Newton's method to find the positive value of x which satisfies x=1.5cos(x) . Compute enough approximations so that your answer is within .05 of the exact answer. (You may use any starting point you deem appropriate.)

 

[redbird555]

$$\frac{1}{x^4}- \frac{4}{x^2}= \frac{1- 4x^2}{x^4}$$.
Apply L'Hopital's rule to that.

"Use Newton's method to find the positive value of which satisfies ."
I have no clue what that means. There seem to be words missing after "positive value of" and "satisfies".

The only thing is my problem is 3/(x^4) instead of 1/(x^4) I tried to apply hopitals rule to your example and I came out with just zero after taking the derivative of the result and then plugging in zero back into the derivative.

 

[SammyS]

$$\frac{1}{x^4}- \frac{4}{x^2}= \frac{1- 4x^2}{x^4}$$.
Apply L'Hopital's rule to that.
...

@ redbird555,

Does L'Hopital's rule apply to $\displaystyle \lim_{x\to 0}\,\frac{3- 4x^2}{x^4}\ ?$ Why not?

You should be able to do this without L'Hopital .

 

[redbird555]

It would and your right it wouldn't even really need to apply hereI could just plug in zero. However if I do that I get a divided by 0 answer and I know the answer to the problem is a negative infinity so I'm not sure what's up.

 

[SammyS]

Looks like +∞ to me.

 

[redbird555]

ugghhh sorry I had a brain fart for a second it is positive infinity. thanks for all the help guys does anyone know how to complete the 2nd one I'm completely lost?? I relisted the problem with the correct info.

 

[SammyS]

Do you know Newton's method ?

To get a starting value, graph:

y=x

and

y=cos(x)​

On the same set of axes & see approx where they intersect.

 

[redbird555]

yes i understand how Newtons method works but I think the y=1.5cosx is confusing me.
I should be able to take any value for x and plug it in so let's say I take 3. Now that per Newtons method 3-[1.5cos(3)/(-3sin(3)/2)] this gives me -4.01525 and that is not correct which is why I'm stumped

 

[SammyS]

x = 3 is a poor choice.

See the graph I suggested.

What are you using for your function?

What's the derivative?

 

[redbird555]

ok lol I am either very confused or there was some mis-communication. I'm using 1.5cos(x) as the function. so by that graph 1 would look like a logical choice therefor Newtons method reads x1-[f(x)/f'(x)] so I would get 1-[1.5cos(1)/(3sin(1)/2)] correct? If so that leaves me with a final value of 1.64209 which is not correct either. I've done quite a few Newtons method problems on this homework and was able to do them fine that's why I keep thinking its some organization mistake in the equation.

 

[SammyS]

From one of your earlier posts:

sorry it should read.
Use Newton's method to find the positive value of x which satisfies x=1.5cos(x) . Compute enough approximations so that your answer is within .05 of the exact answer. (You may use any starting point you deem appropriate.)

I thought you were solving x = 1.5 cos(x) .

 

[redbird555]

Well its asking for an x value that would satisfy that equation and as far as I could gather that's what I was trying to solve for in my calculations? I just need to figure it out before 1130 seeing as that's when my homework is due and this is the last problem I have

 

[SammyS]

Newton's method finds a root of a function. You need a function, f(x), such that f(x) = 0 for the value of x that solves your equation. such a function is:

f(x) = x - 1.5cos(x)

Find f ' (x) for that function.

Then $\displaystyle x_1 = x_0-\frac{f(x_0)}{f'(x_0)}\,.$

In general: $\displaystyle x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)}\,.$