Assignment 04

Details: For this assignment, you are to use the switch-case statement to branch to the various choices for the first menu. You are forbidden to use <cmath> (which is the same as <math.h>) for this assignment! If you want to confer with Professor Frink, that's ok. Your code had better make sure that allowable values are entered upon prompts. And as far as we are concerned, you can only find the roots of positive numbers. Also, think about what kind of precision you might want in your computations for the formulas above.

When option #1 is chosen, the user is to be prompted for the value of x that the user wishes to exponentiate. Accept only values (real numbers) between and including 0 and 7. You are to use this formula for the computation:

e^x = 1 + x + x²/2 + x³/6 + x⁴/24 + x⁵/120 + x⁶/720.

That's an approximation that should be pretty good. Output the computed value.

When option #2 is chosen, the user is to be prompted for the value of x (any real value) for which to compute the sin(x). But then, they are to next be prompted for a positive integer between and including 1 and 5 representing the accuracy of the computation. This value will determine the number of terms to be included in the truncated Taylor series which calculates an approximation to the sin(x). This is that formula:

sin(x) = x¹/1! - x³/3! + x⁵/5! - ... + x^2k-1/(2k-1)! k = 1, 2, 3, ...

Thus, if one wants 2 terms, that corresponds to k = 2 in this formula and sin(x) = x¹/1! - x³/3!. Of course, it's not really equal (=), but we'll pretend. By the way, you will learn in calc II that the larger k is, the better the approximation is. And if you add on terms forever, the equality is indeed guaranteed. Try it! It's definitely worth the time. Note: k! is "k factorial" is k * (k-1) * (k-2) * ... * 3 * 2 * 1 (e.g. 4! = 4*3*2 *1 = 24). And, of course, output the value.

When option #3 is chosen, you will present the user with another menu that looks like this:

Roots

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1. enter x

2. square root

3. cube root

Here, options 2 and 3 should be denied until option 1 is first chosen. Once option 1 has been chosen, options 2 and 3 will compute and output the square root of x and the cube root of x, respectively. After the value is output, return to the first menu. Now, how are you to compute the roots? You will code an iterative method (Newton's method) to compute the roots of x. What that means is that you compute a new x value based on the old x value. Here's the formula for the square root:

x_n+1 = (x_n + A/x_n) / 2 for n = 0, 1, 2, ...

So, A is the number you are trying to find the square root of (e.g. square root of 16 is 4 since 4² is 16), x₀ is an initial guess, and

x₁ = (x₀+ A/x₀) / 2,

x₂ = (x₁+ A/x₁) / 2,

x₃ = (x₂+ A/x₂) / 2,

etc.

If you keep doing this forever, the last x value you compute is the square root of A. Cooool? Well, yeeaaahhhh, sort of. You don't have infinite time, hence you cannot iterate forever. But, Newton's method "converges" to a value very close to the answer really quickly no matter what your first guess is. So, you will code it to repeat this computation just 6 times and use an initial guess (that's x₀) of A, that's just the value you are trying to find the square root of. (And to be sure you understand, A is the value you prompt for in option 1 of option 3.) Try this with your calculator: in the formula above, let A be 16 and start with 16 as an initial guess at the square root. You'll see that the x_n's get really close to 4 pretty fast. And for

In fact, your program should present this menu until the user chooses the quit option.