Taylor Series Equation Formula Latex Code and Summary

Taylor Series

Tags: #math #taylor series

Equation

$$y(x)=y(a+u)=y(a)+u\frac{\mathrm{d} y}{\mathrm{d} x}+\frac{1}{2!}u^{2}\frac{\mathrm{d}^2 y}{\mathrm{d} x^{2}}+\frac{1}{3!}u^{3}\frac{\mathrm{d}^3 y}{\mathrm{d} x^{3}}+...$$

Latex Code

                                 y(x)=y(a+u)=y(a)+u\frac{\mathrm{d} y}{\mathrm{d} x}+\frac{1}{2!}u^{2}\frac{\mathrm{d}^2 y}{\mathrm{d} x^{2}}+\frac{1}{3!}u^{3}\frac{\mathrm{d}^3 y}{\mathrm{d} x^{3}}+...

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Introduction

$y(x)=y(a+u)=y(a)+u\frac{\mathrm{d} y}{\mathrm{d} x}+\frac{1}{2!}u^{2}\frac{\mathrm{d}^2 y}{\mathrm{d} x^{2}}+\frac{1}{3!}u^{3}\frac{\mathrm{d}^3 y}{\mathrm{d} x^{3}}+...$

Latex code for Taylor Series Expansion

$n!$ : Factorials of n
$\frac{\mathrm{d}^{n} y}{\mathrm{d} x^{n}}$ : Derivative of y over x evaluated at point x=a

If y(x) is well-behaved in the vicinity of x = a then it has a Taylor series as above equation.

Discussion

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Phyllis Price

All my studying is towards passing this exam.

2024-03-23 00:00
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Olivia Evans reply to Phyllis Price

Nice~

2024-03-28 00:00:00.0
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Ronald Walker

Oh, how I wish to conquer this exam.

2023-05-22 00:00
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Cheryl Mitchell reply to Ronald Walker

Gooood Luck, Man!

2023-05-25 00:00:00.0
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Joe Lane

Striving to pass this upcoming test.

2023-07-07 00:00
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Melissa Clark reply to Joe Lane

You can make it...

2023-07-28 00:00:00.0
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