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On this post, will be explained about analytic method versus numerical method. Well get on with it, this is it his discussion.

Analytic method is also known as a true method because he gives us a true solution (exact solution), that is a solution that has zero error.

$\quad$ Unfortunately analytic method just superior for a number of limited issues that is the issue that has a simple geometry interpretations and are low. In fact the question that appears in the real world is often nirlanjar as well as involving shapes and a complicated process. As a result practical value completion of the analytic method becomes limited.

$\quad$ When analytic method can no longer applied, then the question solution is still searchable by using a numerical method. Numerical method is a technique used to formulate the mathematical issue so it can be solved by a regular calculation or arithmetic operations (add, less, times and division). Method does that mean way, while numerical means numbers. So, numerical method literally means how to numeracy by using the numbers.

The main difference between numerical methods with the analytic method lies in the two things. First, the solution by using numerical methods always shaped numbers. Compare it to the analytic method that usually produce solutions in the form of mathematical function which further the mathematical function can be evaluated to generate the value in the form of numbers.

$\quad$ Secondly, with numerical methods, we just get a solution that approached or close to a true solution so that numerical solution is also called approximations solutions, but approximations solutions can be made seteliti we want. Approximations solutions is certainly not exactly the same with true solution, so there is a difference between the two. This is what difference called with error.

$\quad$ For example illustration completion with numerical method, look into an understatement integration-of course follows:

$I= \int \limits^1_{-1} (4-x^2)dx$

With analytical methods, we can find true solutions easily. In the integral calculus-of course we know integration techniques for this simple function, namely:

$\int ax^n dx = \frac {1}{n+1} ax^{n+1} + C$

Then, we can do integration tribes of it, then calculate the result as follows an integral function:

$I=\int \limits^1_{-1} (4-x^2) dx $

$= [4x-x^3/3]^{x=1}_{x=-1}$

$ = [4(1)-(1)/3]-[4(-1)-(-1)/3] = 22/3$

$\quad$ Consider that $4x-x^3/3$ is analytic solution from form mathematic function, while $22/3$ adalah numeric value Integrals – certainly that is obtained by how to evaluate the mathematic function to the limits of integration on the x = 1 and x =-1.

$\quad$ Compare the completion of the above when the integration issues resolved by numerical method as follows. Once again, in calculus we certainly still remember that the interpretation of the integral geometry of integrals $f(x)$, axis-$x$, line $x=a$ and line $x=b$. Area of the area can be approximated in the following way. Divide the area integration of a number of trapezoid on interval $[-1, 1]$ With a width of 0.5, note (that picture) provided. The spacious integration area approached with four broad fruit trapezoid, or

$I \approx p+q+r+s$

$\approx [(f(-1)+f(-1/2)).(0,5/2))$

$ + (f(-1/2)+f(0)).(0,5/2))$

$ + (f(0)+f(1/2)).(0,5/2))$

$+ (f(1/2)+f(1)).(0,5/2))]$

$\approx 0,5/2 .[f(-1)+2.f(-1/2)+2.f(0)$

$+2.f(1/2)+f(1)]$

$\approx 0,5/2 .[3+7,5+8+7,5+3]$

$\approx 7,25$

$\quad$ Which is a approximations solution (sign "$\approx $" meaning "about") towards a true solution (22/3). Approximations solution error true solution is:

$Error = |7,25-22/3| $

$= |7,25-7,33..| = 0,08333...$

$\quad$ Of course we can minimize this error by making a smaller trapezoid wide (which means the amount of trapezoid more and more, which means the number of computing more and more). This example also shows that even though solutions with a numerical method is approximation, but the results can be made seteliti as possible by changing computing parameters (in the example above, width trapezoid are reduced) Integrals.

On this post, will be explained about analytic method versus numerical method. Well get on with it, this is it his discussion.

Analytic method is also known as a true method because he gives us a true solution (exact solution), that is a solution that has zero error.

$\quad$ Unfortunately analytic method just superior for a number of limited issues that is the issue that has a simple geometry interpretations and are low. In fact the question that appears in the real world is often nirlanjar as well as involving shapes and a complicated process. As a result practical value completion of the analytic method becomes limited.

$\quad$ When analytic method can no longer applied, then the question solution is still searchable by using a numerical method. Numerical method is a technique used to formulate the mathematical issue so it can be solved by a regular calculation or arithmetic operations (add, less, times and division). Method does that mean way, while numerical means numbers. So, numerical method literally means how to numeracy by using the numbers.

The main difference between numerical methods with the analytic method lies in the two things. First, the solution by using numerical methods always shaped numbers. Compare it to the analytic method that usually produce solutions in the form of mathematical function which further the mathematical function can be evaluated to generate the value in the form of numbers.

$\quad$ Secondly, with numerical methods, we just get a solution that approached or close to a true solution so that numerical solution is also called approximations solutions, but approximations solutions can be made seteliti we want. Approximations solutions is certainly not exactly the same with true solution, so there is a difference between the two. This is what difference called with error.

$\quad$ For example illustration completion with numerical method, look into an understatement integration-of course follows:

$I= \int \limits^1_{-1} (4-x^2)dx$

With analytical methods, we can find true solutions easily. In the integral calculus-of course we know integration techniques for this simple function, namely:

$\int ax^n dx = \frac {1}{n+1} ax^{n+1} + C$

Then, we can do integration tribes of it, then calculate the result as follows an integral function:

$I=\int \limits^1_{-1} (4-x^2) dx $

$= [4x-x^3/3]^{x=1}_{x=-1}$

$ = [4(1)-(1)/3]-[4(-1)-(-1)/3] = 22/3$

$\quad$ Consider that $4x-x^3/3$ is analytic solution from form mathematic function, while $22/3$ adalah numeric value Integrals – certainly that is obtained by how to evaluate the mathematic function to the limits of integration on the x = 1 and x =-1.

$\quad$ Compare the completion of the above when the integration issues resolved by numerical method as follows. Once again, in calculus we certainly still remember that the interpretation of the integral geometry of integrals $f(x)$, axis-$x$, line $x=a$ and line $x=b$. Area of the area can be approximated in the following way. Divide the area integration of a number of trapezoid on interval $[-1, 1]$ With a width of 0.5, note (that picture) provided. The spacious integration area approached with four broad fruit trapezoid, or

$I \approx p+q+r+s$

$\approx [(f(-1)+f(-1/2)).(0,5/2))$

$ + (f(-1/2)+f(0)).(0,5/2))$

$ + (f(0)+f(1/2)).(0,5/2))$

$+ (f(1/2)+f(1)).(0,5/2))]$

$\approx 0,5/2 .[f(-1)+2.f(-1/2)+2.f(0)$

$+2.f(1/2)+f(1)]$

$\approx 0,5/2 .[3+7,5+8+7,5+3]$

$\approx 7,25$

$\quad$ Which is a approximations solution (sign "$\approx $" meaning "about") towards a true solution (22/3). Approximations solution error true solution is:

$Error = |7,25-22/3| $

$= |7,25-7,33..| = 0,08333...$

$\quad$ Of course we can minimize this error by making a smaller trapezoid wide (which means the amount of trapezoid more and more, which means the number of computing more and more). This example also shows that even though solutions with a numerical method is approximation, but the results can be made seteliti as possible by changing computing parameters (in the example above, width trapezoid are reduced) Integrals.

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