Modified Euler Method

Added: 2026-05-05

[00:00:02]Welcome to another video.

[00:00:05]I recently made a video on Oiler method.

[00:00:09]That's a numerical method of evaluating a function. If you don't know the function, but you know the derivative of the function and you know the value of the function at one point, but you want to find the value at another point. So, it's more like a solution to a differential equation. And in one of the comments I read that um I should talk about the modified oiler method. So well you shouldn't watch this video if you don't know what the oiler method is about. So you can actually watch the oiler method video and then come back and watch the modified oiler method.

[00:00:48]Let's get into the video.

[00:00:53][music] So this is what we do for the oiler method. We get the next point by using the previous point and using the derivative. So this f of x subn y subn is just the value of the derivative. So this is the derivative function multiplied by the step. Okay, which is just h. But usually this assumption is that you're taking the slope at the start of a curve to be the slope all through the curve. You know that is not true. Okay, that's why we have to take steps and we say, okay, we're taking small steps. You're not too far. So your answer is more likely to be accurate.

[00:01:43]But we know you don't take the the slope at the beginning of a function and assume it is the same slope throughout the function for however many steps you're going to take. So the modified oiler function says no, we're going to make that we're not going to make that assumption. The slope you get at the beginning, you're going to look at the slope you're going to get at the end and take an average of both of them. That gives you better accuracy. And that's what we do. So what we do first is we perform the usual oiler method. We get our y n + one but then we compute the slope at the end. So what we do is we actually take an average. Watch the modified. So look at what the modified is the modified modified oiler.

[00:02:38]Okay.

[00:02:42]So the modified oiler method says we are going to instead of saying h times just the value of the derivative at the start remember this is the beginning the initial value we're going to say it's going to be the value of the slope at x subn y subn which is what you do for your normal oiler method.

[00:03:10]And then we're going to add it to the value of the slope at the end. x sub n + 1 comma y sub n + 1 and then we're going to divide it by two.

[00:03:29]That's all.

[00:03:32]That's the only difference. We're saying the slope at the beginning is not the slope at the end, but they're different.

[00:03:38]So instead of us just assuming that the slope at the beginning is the slope at the end, we might as well find find the two slopes at the beginning and at the end. When I say end, I mean where you stop after your step and then average the two together. Okay? And you still do your normal h here. And then you still have your y subn plus and then you still have your y n + 1 equals. So this is the same as this except that instead of using a single slope, you're using an average of two slopes. One at the start and one at the end. And that's the only modification that you do. But in the formula, sometimes this two goes under this h. So it looks as if the step is actually have and then you're having two slopes added together. But really really you're taking the average of the slopes.

[00:04:37]Another explanation is you see this value you got that I say at the end. It is usually not called at the end because look it looks like y n +1 is the same thing as where we're going. But that's not true. So the first time you perform the oiler method if you're planning to do the modified oiler method you call this the predictor of the final point. So this would have a superscript of P which means predictor.

[00:05:11]Okay, make sure you put that notation there. This is helping us predict and that's the same value you're going to put here. It is the predictor.

[00:05:24]A superscript telling you that that's not the final answer. The final answer is the one without the P as a as a superscript. Do you see that? Okay, let's actually do it. Okay, so the first time you do Oiler method, you're getting the predictor. You then use the predictor to actually get your final answer. And that's what it's going to look like. So this problem is asking us to estimate the value of y at one at x= 1/2 starting from x= 0. So you can see when x is 0, y is zero. So we're starting from the origin and they want us to use step size of 0.25 or 1 over 4. Okay.

[00:06:10]So how do you go from zero using steps of 1 over 4 to 1/2? You have to take two steps. The first step goes from 0 to 1/4 and the second step will take you to where you have to be. So we only need to perform two operations each of them with steps of 1/4. Okay. Now, so I'm going to use the modified Oiler method. But the first thing you want to do is use the Oiler method first. So remember our formula is we're going to get the predictor by starting from the initial value of y multiplying by the step size and evaluating the function at the original point which is n. Okay. So here I'm just going to go straight into it. I know that the predictor of the value of y at n + one, which is the next step, the next step for us is going to be at 1 over4 because we're going from zero to 1 over4. From 1 over4 to 1/2. So it's going to be y at 1 over4. Let me put 1 over4 here. Okay.

[00:07:11]And this is the predictor of what the value will be at 1 over4. It's going to be equal to the initial value of y which is zero.

[00:07:21]Okay. Plus the the step size is 1 over 4 or 0.25 whichever like but I want to do the fraction. And then we're going to evaluate the function. The value of the function is going to be um here's our value the value of the function. Now dy dx is going to be x^2. What was the original value of x? It was zero. is going to be 0^ 2 minus the original value of y is also 0. So um this predictor is going to give us 0 + 1 / 4 of 0 which is zero. So actually that is what we get as our predictor value now but it is not the final value. If this was the normal oiler, the regular, the simple oiler method, that's our answer.

[00:08:11]But we're not going to accept it because this is just the predictor. So what we're going to do is go back and say y at 1 / 4, which is the real value we need is going to use this formula. It's going to be y subn the original value plus the predictor 1 / 4 * now we're going to average two values. The first value is going to be using the initial value 0 0. So we're going to be getting 0^ 2 - 0. That's the first one. Plus, we're going to be using the final value, which is going to be the value of x after you've taken one step, which is 1 / 4 plus the value of y after you've taken the step. Sorry, it's minus 1 / 4^ 2 rather minus the value we just got now 0 0 based on this relationship.

[00:09:08]So, what does that give us? That gives us 0 + 1 over 4. Oh, divided by two.

[00:09:16]This is all divided by two. Don't forget we're taking the average of it.

[00:09:26]Or I can rewrite this formula because this is what you actually see. It's y subn + h over 2. So it's like having the step. So I think this is better. x subn y subn plus f of x sub n + 1 y sub n + one predictor. This is the better uh way just do h over two. So I think that's what I'm going to do here. I'm going to do h over 2 which is going to be 1 over8. So we can just do 1 over 8 here.

[00:10:01]Okay. So it's 1 over8. Then I don't have to deal with that. So this is 1 over 8 of this is 0 + 1 over 16 1 over 16. Ah what is 8 * 16? That's going to be 64 * 2 that's 128. Okay. So that's going to be 1 over 128. Okay. So our y so at 1 over4 = 1 over 128. This is what the modified oiler method is going to give you as answer. If it were in um the simple oiler method, the predictor that you got would have been your final answer zero.

[00:10:55]So now we have gotten the first step.

[00:10:58]We're going to take one more step a quarter and move on to get the second value. Okay. So let's do that again.

[00:11:06]We're going to get the predictor y the predictor value at 1/2 which is where we're going uh for this function is going to be the initial value which is this one. um 1 over 128.

[00:11:25]Remember you always perform the simple oiler method first to get the predictor.

[00:11:30]Okay, so it's going to be times plus h which is 1 / 4 * the value of the function at these the the old value here x is 1 / 4 and y is 1 over 128. So, it's going to be 1 over 4^ 2ar - 1 over 128. Yeah, I haven't done this math before. So, 1 over 128 + 1 over 4 * 7 over 128. So, now I need this to also have a four down here. So, it's going to be four times. So, it's going to be 4 + 7, that's 11.

[00:12:17]That's going to be 11 over 4 * 128 is 512.

[00:12:23]So this is the predicted value or this is the predictor value for what we're going to get ultimately. So y predictor at the end. Remember we're going to 1/2.

[00:12:34]I'm not going to one this time. You can continue it to one just following the steps. Okay? And compare it to what the ordinary oiler method will show you. at n + 1 which is going to be y at 1/2. So this is no longer the predictor is going to be the initial value of y for us here which is 1 over 128 1 over 128 um plus and then we're going to have half of the step just as I did it in this formula half of the step will now be 1 over8 and then we're going to have the two um values I know that initially we got 7 over 128 here when we did it here. So, I'm I might as well just write 7 over 128.

[00:13:24]We add what the value of the slope is going to be at the end. So, we're going to go here and use this right here. So, it's going to be x^2, but our x is 12. So it's going to be 12 squared minus the predictor squared minus a predictor rather 11 over 512.

[00:13:52]So this is what we need to do. We want everybody to be 512.

[00:13:58]So what I'm going to do is I'm instead of writing this as 7 over 128 I'm going to write it as what do I need to four that's 28 over 1 over 512 plus this is 1 over 4 when you square it what do I need to multiply that's 128 + 128 over 512 - 11.

[00:14:31]Nice.

[00:14:32]This is going to give me 1 over 128 + 1 8 times um in fact just subtract this from this.

[00:14:47]You're going to get 17. Add 17 to 128.

[00:14:51]That's going to be 140.

[00:15:01]Five. Yeah. 145. That was correct.

[00:15:06]145 over 5002.

[00:15:11]128. Well, how do I make this 128 multiply by 4? And this that's 32. So, it's going this is going to be 32 over um 8 * 512 is going to be 1,024,048 4,096.

[00:15:33]Oh my.

[00:15:37]Yes, that's going to be 4,096 + 145 over 4,096.

[00:15:50]So this is instead of saying equal to we just do the approximation sign and it would be 177 over 4096.

[00:16:01]Is there any number that divides the top and the bottom? Does three divide this?

[00:16:06]No. So that's it. This is the approximate value using the modified oilers the modified oiler method. Okay. Yeah, it's a lot of calculator work or you can just deal with fractions because I don't want to use a calculator. That's why I'm here. And um never stop learning. Those who are stop learning, stop living.

[00:16:29]Bye-bye.