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Newton's Method on Linear Approximation - Examples Part 2: Where it fails to Converge by mes

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· @mes · (edited)
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Newton's Method on Linear Approximation - Examples Part 2: Where it fails to Converge
https://youtu.be/mDSyBPsiv1Q

In this video I recap on Newton's Method on Linear Approximation and go through 4 very useful examples illustrating when Newton's Method either fails to converge or converges very slowly depending on the initial starting approximation you select.

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# Newton's Method on Linear Approximation - Examples Part 2: Where it fails to Converge

![Newton's Method Examples Part 2.jpeg](https://files.peakd.com/file/peakd-hive/mes/23vspBV1NicctLZ956B4N6zLFNkubANbcPkGEvRg6NVmmHjh737x3AB7NjyQvsBLSAP4S.jpeg)

## Recap on Linear Approximation

![image.png](https://files.peakd.com/file/peakd-hive/mes/23t8CfKPvXW7EWXop2Gti83gXXpfuUM3tXeg7Au36B3ygxxmNaV2y2XHPWUSDR9K7S3G1.png)

## Example 1

Explain why Newton’s method doesn’t work for finding the root of the equation x<sup>3</sup> – 3x + 6 = 0 if the initial approximation is chosen to be x<sub>1</sub> = 1.

**Solution:**

![image.png](https://files.peakd.com/file/peakd-hive/mes/23t8DBnQEtNKs13eoSJ1A8rWLjTHMYvADXL4grsL9xjXhJMmqFogSfYYvq3cFjYoaC4ix.png)

## Example 2

Use Newton’s Method to find the root of the equation x<sup>3</sup> – x = 1 for the following initial approximations:

>a) x<sub>1</sub> = 1
b) x<sub>1</sub> = 0.6
c) x<sub>1</sub> = 0.57

Graph f(x) = x<sup>3</sup> – x – 1 and it’s tangent lines at x<sub>1</sub> = 1, 0.6, and 0.57 to explain why Newton’s Method is so sensitive to the value of the initial approximation.

**Solution:**

![image.png](https://files.peakd.com/file/peakd-hive/mes/23t8CSPFBxuJYQWHeoynRu3GoSjAiJxDXd8dDWqbweepB97eXwFA2exLz8M2pvpwinRBg.png)

![image.png](https://files.peakd.com/file/peakd-hive/mes/Eo2BpKHNZbth42UeRJsbCJkv4JevbCvCnUhrhm5NDtKLwP1ZdtvTohsok491wbUq5JB.png)

## Example 3

Explain why Newton’s method fails when applied to the equation ∛x = 0 with any initial approximation x<sub>1</sub> ≠ 0. 

Illustrate your explanation with a sketch.

**Solution:**

![image.png](https://files.peakd.com/file/peakd-hive/mes/23t8D6MYf9SWog7DV4FL5DXDD91JGAyuttiofsr2eWe1nLNazZ7NqjnExMtPhDom7odGq.png)

![image.png](https://files.peakd.com/file/peakd-hive/mes/23t8EjqhGAA8YzbqA6fKtvetbxpM9HduGC9UwK7z4jL81ujqBWbW6qJih2bSn169fM4qm.png)

## Example 4

If:

![image.png](https://files.peakd.com/file/peakd-hive/mes/23wr9i58hzJCXfzMiCdWugwu3h7t8vjj5fesNEmgSEcoEfJP5L7U9YSo8hoqWpE1KJ94D.png)

Then the root of the equation f(x) = 0 is x = 0. 

Explain why Newton’s method fails to find the root no matter which initial approximation x<sub>1</sub> ≠ 0 is used. Illustrate your explanation with a sketch.

![image.png](https://files.peakd.com/file/peakd-hive/mes/23t8DBnQF2rBQBSoSaRkWVgZssHrx9GvPZ2TaHGgQc5ESo27J8T1irmNv8arm2mNrnpkZ.png)

![image.png](https://files.peakd.com/file/peakd-hive/mes/48HhKJZ3D9hrTJtQ35a7wHYDMuMhymrvPNjhkAJdHXaRpRox6yWXEKFVs1vCPEbYfX.png)

![image.png](https://files.peakd.com/file/peakd-hive/mes/23t8DBnQFDCAT1GsZLbuULrSJMsbHcA2in5B7kumWxiwFfrnRkija4G9azsWGCHbcEArG.png)
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