How do you use Newton's Method to approximate the positive root of the equation #sin(x)=x^2# ?

Answer 1

The answer is 0.8767262154.

Recall that Newton's Method uses the formula:

#x_(n+1)=x_n-(f(x_n))/(f'(x_n))#

So we need to change the equation into a function. This is done by moving all terms to one side:

#f(x)=sin x-x^2#

And we need the derivative:

#f'(x)=cos x - 2x#
The easiest way to iterate is to program your calculator. Enter #f(x)# into #Y_1# and #f'(x)# into #Y_2#. Then enter a very short program that does this:
#A-(Y_1(A))/(Y_2(A))->A#

You can go to my website for specific instructions for the TI-83 or the Casio fx-9750 .

Finally, you need a starting value, #x_1#. Since the question is asking for a positive root, we know that #sin x# has a maximum value of #1# at #x=pi/2# and #x^2# is #1# at #x=1#. So it is safe to say that the root is less than #1# and therefore will be the starting value. Note, we don't want a starting value close to #0# because #0# is also a root but not positive. So enter the following into your calculator,
#1->A#

Then execute the program until you get the desired accuracy:

#0.8913959953# #0.8769848448# #0.8767262985# #0.8767262154# #0.8767262154#

We get 3 digits of accuracy after 2 iterations, 7 after 3 iterations, and 10 after 4 iterations. So the answer converges very quickly for this root.

If you aren't getting these values or you are finding that the convergence is very slow for another function, it is most likely that your calculator is in DEGREE mode rather than RADIAN mode.

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Answer 2

To use Newton's Method to approximate the positive root of the equation ( \sin(x) = x^2 ), you start by guessing a value, say ( x_0 ), close to the root. Then, you iterate using the formula:

[ x_{n+1} = x_n - \frac{\sin(x_n) - x_n^2}{\cos(x_n) - 2x_n} ]

Repeat this process until you reach a desired level of accuracy.

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Answer 3

To use Newton's Method to approximate the positive root of the equation ( \sin(x) = x^2 ):

  1. Start with an initial guess ( x_0 ) for the positive root.

  2. Set up the iteration formula: [ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} ]

  3. Calculate ( f(x_n) ), which represents ( \sin(x_n) - x_n^2 ).

  4. Calculate ( f'(x_n) ), which represents the derivative of ( f(x) ), ( \frac{d}{dx}(\sin(x) - x^2) ).

  5. Substitute ( x_n ), ( f(x_n) ), and ( f'(x_n) ) into the iteration formula to find ( x_{n+1} ).

  6. Repeat steps 3-5 until the value of ( x_{n+1} ) converges to the desired accuracy. This is typically done until ( |x_{n+1} - x_n| ) is smaller than a predetermined tolerance level.

  7. The final value of ( x_{n+1} ) is an approximation of the positive root of the equation ( \sin(x) = x^2 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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