To analyze Simpson’s method of numerical integration to create a natural log table for numbers 1, 1.1, 1.2, 1.3, …… , 2 that is accurate to 5 decimal places (error less than .000005).
Simpson’s Rule:
Simpson’s method rule is to approximate ∫_a^b▒〖f(x)dx,use〗 in the method we use the parabolas to calculate the value, S = ∆x/3(y_0+4y_1+2y_2+4y_3+..+2y_(n-2)+4y_(n-1)+y_n)

The Y’s are the value of f at the partition points x_0=a,〖 x〗_1=a+∆x, x_2=a+2 ∆x, …… , x_(n-1)= a+(n-1)∆x, 〖 x〗_n=b
Always the number n is even, and ∆x= ((b-a))/n
Definition of Natural Logarithm:
To find the function, we will use the definition of the Natural logarithm which is the function given by ln⁡x= ∫_1^x▒〖1/x〗 dx

Moreover, before we use the Simpson’s rule we will calculate the n value for all the natural log of the following numbers: 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, and 2.
To find n value for all functions, we will use the error estimates in the Simpson’s Rule:
It states that if f^((4)) is the continuous and M is any upper bound for the value of |f^((4)) | on [a, b], then the error E_s in the Simpson’s Rule approximation o the integral of f from a to b for n steps satisfies the inequality, the formula is
|E_s |≤(M 〖(b-a)〗^5)/(180 n^4 )

Let’s find the n value, calculate the n value for all the natural log of the following numbers: 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, and 2

Functions N value ln⁡1= ∫_1^1▒〖1/x〗 dx 0 ln⁡〖1.1〗= ∫_1^1.1▒〖1/x〗 dx 8 ln⁡〖1.2〗=∫_1^1.2▒〖1/x〗 dx ln⁡〖1.3〗= ∫_1^1.3▒〖1/x〗 dx ln⁡〖1.4〗= ∫_1^1.4▒〖1/x〗 dx ln⁡〖1.5〗= ∫_1^1.5▒〖1/x〗 dx ln⁡〖1.6〗= ∫_1^1.6▒〖1/x〗 dx ln⁡〖1.7〗= ∫_1^1.7▒〖1/x〗 dx ln⁡〖1.8〗= ∫_1^1.8▒〖1/x〗 dx ln⁡〖1.9〗= ∫_1^1.9▒〖1/x〗 dx ln⁡2= ∫_1^2▒〖1/x〗 dx 14

First, we will find the fourth derivative of the function ln⁡1=∫_1^1▒1/x dx

f(x)=1/x=x^(-1) f^1 (x)=〖-x〗^(-2) f^2 (x)=2x^(-3) f^3 (x)= -6x^(-4) f^4 (x)=24x^(-5)

On the interval [1, 1] the upper bound of the derivative is |f^((4) ) (1) |=24〖(1)〗^(-5)= 24 = k
Now use the error bound formula, which is

E_n^S≤(k(b-a))/(180n^4 )

Here, k = 24, b=1, a=1

0.000005 ≤ (24(1-1))/(180n^4 )
24(0)/180(0.000005) ≤ n^4
∜((24 (0))/(180(0.000005)))≤n

0 = n

Now let’s find then value for the function of ln⁡〖1.1〗= ∫_1^1.1▒〖1/x〗 dx

The fourth derivative of the function is f^4 (x)=24x^(-5)
On the interval [1, 1.1] the upper bound of the derivative is |f^((4) ) (1) |=24〖(1)〗^(-5)= 24 = k

E_n^S≤(k(b-a))/(180n^4 )

Here, k = 24, b=1.1, a=1

0.000005 ≤ (24(1.1-1))/(180n^4 )
24(0.1)/180(0.000005) ≤ n^4
∜((24 (0.1))/(180(0.000005)))≤n

7.1861≅n , even n value 8 = n However, according to the Simpson’s rule n is even, nearest even value of the n is 8.

Next,
Let’s find n that guarantees an error less than .000005 for the function of ln⁡2= ∫_1^2▒〖1/x〗 dx

On the interval [1, 2] the upper bound of the derivative is |f^((4) ) (1) |=24〖(1)〗^(-5)= 24 = k

E_n^S≤(k(b-a))/(180n^4 )

Here E_n^S= 0.000005, a= 1, b= 2

0.000005 ≤ (24(2-1))/(180n^4 )
24/180(0.000005) ≤ n^4
∜(24/(180(0.000005)))≤n

12.7789 ≅ n
14 = n