Video: Pack 5 β€’ Paper 2 β€’ Question 20

Pack 5 β€’ Paper 2 β€’ Question 20

02:49

Video Transcript

Using π‘₯ 𝑛 plus one is equal to seven minus two over π‘₯𝑛 plus five over π‘₯𝑛 cubed with π‘₯ nought equal to 4.5, find the values of π‘₯ one, π‘₯ two, and π‘₯ three to three decimal places.

This is an example of an iterative formula. It looks much scarier than it is. All it’s saying is that to find the next value of π‘₯, take the current value, and substitute it in the formula. The starting value π‘₯ nought is 4.5. In this case, to find π‘₯ one, we’ll substitute π‘₯ nought into the formula. π‘₯ one is equal to seven minus two over 4.5 add five over 4.5 cubed. Correct to three decimal places, π‘₯ one is equal to 6.610.

To find π‘₯ two, we substitute π‘₯ one into our formula. Notice it’s helpful to use the previous answer button on your calculator to save time in the next few steps and to prevent any errors from rounding to early. Correct to three decimal places, π‘₯ two is 6.715.

Finally, to find π‘₯ three, we substitute π‘₯ two into our formula or we use the previous answer button to find π‘₯ two. We should therefore just be able to press equals to get the value for π‘₯ three. Correct to three decimal places, π‘₯ three is 6.719. And there we go, we found the values for π‘₯ one, π‘₯ two, and π‘₯ three.

Describe the relationship between π‘₯ one, π‘₯ two, and π‘₯ three and the equation π‘₯ to the power of four minus seven π‘₯ cubed plus two π‘₯ squared minus five is equal to zero.

Now, this equation isn’t plucked out of nowhere. This is all part of the same question. So there should be some relationship between this equation and our iterative formula. In fact, if we start by adding seven π‘₯ cubed to both sides of this equation then subtract two π‘₯ squared and then add five. We have something very close to the original iterative formula. We can divide through by π‘₯ cubed to get π‘₯ is equal to seven minus two over π‘₯ plus five over π‘₯ cubed.

This means that the values of π‘₯ one, π‘₯ two, and π‘₯ three are estimates to the solution of the equation π‘₯ to the power of four minus seven π‘₯ cubed plus two π‘₯ squared minus five is equal to zero.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.