Difference between revisions of "Course notes/Numerical integration"
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Numerical integration is the process of going background from a given derivative of a function to the function itself using computational tools. The simplest form of numerical integration takes advantage of the approximation to the derivative that we get but stopping short of letting $h$ go all the way to zero. That is, | Numerical integration is the process of going background from a given derivative of a function to the function itself using computational tools. The simplest form of numerical integration takes advantage of the approximation to the derivative that we get but stopping short of letting $h$ go all the way to zero. That is, | ||
| − | :$f'(x)\approx \frac{f(x+h)-f(x)}{h}$ | + | :$\displaystyle f'(x)\approx \frac{f(x+h)-f(x)}{h}$ |
for a sufficiently small value of $h$. We'll use $h=0.1$ for now. | for a sufficiently small value of $h$. We'll use $h=0.1$ for now. | ||
Suppose we are given a velocity function $v(t)$ but not the position function $x(t)$ and are told that | Suppose we are given a velocity function $v(t)$ but not the position function $x(t)$ and are told that | ||
| − | :$\frac{dx}{dt} = v(t)= e^{-t^2}$. | + | :$\displaystyle\frac{dx}{dt} = v(t)= e^{-t^2}$. |
How do we find the function $x(t)$ if we don't recognize $v(t)$ as the derivative of some known function? | How do we find the function $x(t)$ if we don't recognize $v(t)$ as the derivative of some known function? | ||
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Let's find the position at a time $h$ later, which we'll call $t_1=t_0+h$. Using the approximation | Let's find the position at a time $h$ later, which we'll call $t_1=t_0+h$. Using the approximation | ||
| − | :$v(t_0)=x'(t_0) \approx \frac{x(t_0+h)-x(t_0)}{h}=\frac{x(t_1)-x(t_0)}{h}$ | + | :$\displaystyle v(t_0)=x'(t_0) \approx \frac{x(t_0+h)-x(t_0)}{h}=\frac{x(t_1)-x(t_0)}{h}$ |
Solving this equation for \(x(t_1)\) we | Solving this equation for \(x(t_1)\) we | ||
find that | find that | ||
| − | :\(x(t_1) = x(t_0)+h v(t_0)\). | + | :\(x(t_1) = x(t_0)+h \ v(t_0)\). |
Replacing all the letters but numbers, we get | Replacing all the letters but numbers, we get | ||
| − | : | + | :$x(2.1) = x(2)+0.1 \ v(2) = 3+0.1 e^{-4}$. |
Repeating this process, replacing $t_0$ with $t_1$, we find that | Repeating this process, replacing $t_0$ with $t_1$, we find that | ||
| − | \(x(t_2) = x(t_1)+h x(t_1)\) | + | \(x(t_2) = x(t_1)+h \ x(t_1)\) |
where \(t_2=t_1+h\). | where \(t_2=t_1+h\). | ||
| + | :$x(2.2) = x(2.1)+0.1 \ v(2.1) = (3+0.1 e^{-4}) + 0.1 e^{-(2.1)^2}$. | ||
| − | We can repeat this as long as we want, stopping when we get to some desired final time. If we start at $t=2$ and want the position at $t=5$, using $h=0.1$, we'll have to repeat the process | + | :$x(2.3) = x(2.2)+0.1 \ v(2.2)$. |
| + | :$x(2.4) = x(2.3)+0.1 \ v(2.3)$. | ||
| + | |||
| + | |||
| + | We can repeat this as long as we want, stopping when we get to some desired final time. If we start at $t=2$ and want the position at $t=5$, using $h=0.1$, we'll have to repeat the process 30 times. | ||
===Choosing $h$=== | ===Choosing $h$=== | ||
There is still the issue of choosing a value of $h$. We know that the | There is still the issue of choosing a value of $h$. We know that the | ||
derivative approximation is better for smaller \(h\) so we can expect that our estimate of $x(5)$ will be better for smaller $h$. However, the smaller we choose it, the more steps it will take to get to $t=5$. To achieve a desired accuracy, start with a convenient (big) value like | derivative approximation is better for smaller \(h\) so we can expect that our estimate of $x(5)$ will be better for smaller $h$. However, the smaller we choose it, the more steps it will take to get to $t=5$. To achieve a desired accuracy, start with a convenient (big) value like | ||
| − | $h = 0. | + | $h = 0.4$, record the resulting $x(5)$, and repeat the process for |
| − | $h$ | + | $h=0.2$. Repeat again with $h=0.1$. Repeat again with an $h$ yet again half the size. After three or four repetitions, you will see less |
and less change in the successive values of $x(5)$. When the change is less | and less change in the successive values of $x(5)$. When the change is less | ||
| − | than 0.01%, | + | than 0.01%, the accuracy of your approximate $x(5)$ value should be sufficient for WeBWorK. |
Revision as of 09:45, 16 September 2014
Numerical integration is the process of going background from a given derivative of a function to the function itself using computational tools. The simplest form of numerical integration takes advantage of the approximation to the derivative that we get but stopping short of letting $h$ go all the way to zero. That is,
- $\displaystyle f'(x)\approx \frac{f(x+h)-f(x)}{h}$
for a sufficiently small value of $h$. We'll use $h=0.1$ for now.
Suppose we are given a velocity function $v(t)$ but not the position function $x(t)$ and are told that
- $\displaystyle\frac{dx}{dt} = v(t)= e^{-t^2}$.
How do we find the function $x(t)$ if we don't recognize $v(t)$ as the derivative of some known function?
To do this using a computational approach, we need to know the position at some starting time $t_0$. Let's suppose we know that $x(2)=3$. From there on, we can use the approximation above. Let's try to determine the position at $t=5$.
Let's find the position at a time $h$ later, which we'll call $t_1=t_0+h$. Using the approximation
- $\displaystyle v(t_0)=x'(t_0) \approx \frac{x(t_0+h)-x(t_0)}{h}=\frac{x(t_1)-x(t_0)}{h}$
Solving this equation for \(x(t_1)\) we find that
- \(x(t_1) = x(t_0)+h \ v(t_0)\).
Replacing all the letters but numbers, we get
- $x(2.1) = x(2)+0.1 \ v(2) = 3+0.1 e^{-4}$.
Repeating this process, replacing $t_0$ with $t_1$, we find that \(x(t_2) = x(t_1)+h \ x(t_1)\) where \(t_2=t_1+h\).
- $x(2.2) = x(2.1)+0.1 \ v(2.1) = (3+0.1 e^{-4}) + 0.1 e^{-(2.1)^2}$.
- $x(2.3) = x(2.2)+0.1 \ v(2.2)$.
- $x(2.4) = x(2.3)+0.1 \ v(2.3)$.
We can repeat this as long as we want, stopping when we get to some desired final time. If we start at $t=2$ and want the position at $t=5$, using $h=0.1$, we'll have to repeat the process 30 times.
Choosing $h$
There is still the issue of choosing a value of $h$. We know that the derivative approximation is better for smaller \(h\) so we can expect that our estimate of $x(5)$ will be better for smaller $h$. However, the smaller we choose it, the more steps it will take to get to $t=5$. To achieve a desired accuracy, start with a convenient (big) value like $h = 0.4$, record the resulting $x(5)$, and repeat the process for $h=0.2$. Repeat again with $h=0.1$. Repeat again with an $h$ yet again half the size. After three or four repetitions, you will see less and less change in the successive values of $x(5)$. When the change is less than 0.01%, the accuracy of your approximate $x(5)$ value should be sufficient for WeBWorK.
