Sydney's question at Yahoo Answers regarding root approximation

In summary, to approximate the solution to the equation x^(4) + x - 3 = 0 that satisfies the given condition, x is greater than or equal to 0, we can use Newton's method. Starting with an initial guess of x = 1, we can use the recursion x_{n+1}=(3(x_n^4+1))/(4x_n^3+1) to find an approximation of x that satisfies the equation. After multiple iterations, we reach a final approximation of x = 1.16403514029, which satisfies the given condition.
  • #1
MarkFL
Gold Member
MHB
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Here is the question:

CALCULUS HELP PLEASE!?

Approximate the solution to the following equation that satisfies the given condition: x^(4) + x - 3 = 0 ; x is greater than or equal to 0Please help I'm having so much trouble!

Here is a link to the question:

CALCULUS HELP PLEASE!? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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  • #2
Hello Sydney,

First, let's take a look at a plot of the function:

View attachment 582

We see the positive root is near $x=1$, so that will be a good initial guess.

Newton's method gives us the recursion:

$\displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$

In our case, this is:

$\displaystyle x_{n+1}=x_n-\frac{x_n^4+x_n-3}{4x_n^3+1}=\frac{3(x_n^4+1)}{4x_n^3+1}$

where $x_0=1$.

Now, I have a TI-89 Titanium graphing calculator, so what I do is enter the following:

1 [ENTER] Result: 1
(3(ans(1)^4+1))/(4ans(1)^3+1) ♦[ENTER] Result: 1.2
♦[ENTER] Result: 1.16541961577
♦[ENTER] Result: 1.16403726916
♦[ENTER] Result: 1.16403514029
♦[ENTER] Result: 1.16403514029

Since the last two approximations are the same, we have exceeded the limit of accuracy of the calculator, and we know the required root, to 12 digits, is:

$x\approx1.16403514029$
 

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Related to Sydney's question at Yahoo Answers regarding root approximation

1. What is root approximation?

Root approximation is a mathematical method used to find an approximate solution to an equation for which there is no exact solution. It involves using a series of calculations to get closer and closer to the actual solution.

2. Why is root approximation important?

Root approximation is important because it allows us to find solutions to equations that cannot be solved using traditional algebraic methods. It is also used in many fields, including engineering, physics, and finance, to make estimates and predictions.

3. How does root approximation work?

Root approximation works by starting with an initial guess for the solution and then using a series of calculations to refine the approximation. This is done by plugging the current approximation into the equation and using the resulting value to make a better guess for the solution.

4. What is the difference between root approximation and root finding?

Root approximation and root finding are closely related, but there is a subtle difference. Root approximation involves finding an approximate solution to an equation, while root finding involves finding the exact solution. Root approximation is often used when an exact solution is not possible.

5. Can root approximation be used for all equations?

No, root approximation cannot be used for all equations. It is most commonly used for polynomial equations, but can also be used for some transcendental equations. However, there are some equations for which root approximation is not a suitable method.

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