Basically what Euler's method does is approximate the
value of a point that's close to a point that you have, by using a small
value and the derivative of the function. Other than questions like these, VCE doesn't go into it too much, but later on it can be used to help approximate solutions to certain problems that involve a differential equation that doesn't have an explicit solution.
So basically in VCE, all this is, is manipulating the following formula:
\approx f(x)+hf'(x))
, the smaller
is the better the approximation will be.
So what the above formula is doing is taking an initial point, and then by extending the tangent of the curve at that point a short distance, we can approximate a point close to it. This can be repeated with a small step size to get an approximate solution further along the curve, as below.
So we take the first point, and add to the
value the product of the step size
and the value of the derivative at that point.
This will give out another
value that is an approximation of
)
. Then we use that point and repeat, approximating the next point until we get to
.
 & \approx f\left(x\right)+hf'\left(x\right)<br />\\ f\left(0\right)=1<br />\\ f\left(0+0.5\right) & \approx1+0.5\left(0+1\right)<br />\\ & =1.5<br />\\ \implies f\left(0.5\right) & =1.5<br />\\ f\left(0.5+0.5\right) & \approx1.5+0.5\left(0.5+1.5\right)<br />\\ & =2.5<br />\\ f\left(1+0.5\right) & \approx2.5+0.5\left(1+2.5\right)<br />\\ & =4.25<br />\\ f\left(1.5+0.5\right) & \approx4.25+0.5\left(1.5+4.25\right)<br />\\ & =7.125<br />\\ \implies f\left(2\right) & \approx7.125<br />\end{alignedat})
EDIT: Beaten, was typing an explanation/Matlabing
Home
Help
Search
Login
