# Exponential and Logarithmic Series Formulas

Not Everyone feels comfortable to solve problems related to the Exponential And Logarithmic Series. To help such people we have listed the Exponential And Logarithmic Series Formulas that will save them from doing lengthy calculations. Check out the Exponential and Logarithmic Series Formulae provided to solve the problems related with ease. Besides, you will also learn the concept better after solving problems by applying Formulae.

## Exponential and Logarithmic Series Formulae Sheet & Tables

The Concept of Exponential and Logarithmic Series is not going to be horror again for you with the list of formulas provided concerning it. Try to recall the Exponential And Logarithmic Series Formulas regularly instead of worrying about how to solve the related problems. Thus, you can overcome the burden of doing calculations and get the results quickly.

1. The Number ‘e’

The sum of the series $$1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\ldots \ldots \ldots . .+\infty$$ is denoted by the number e i.e.
e = $$\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n}$$

• The number e lies between 2 and 3. Approximate value of e = 2.718281828.
• e is an irrational number.

2. Some standard deduction from Exponential Series

(i) ex = $$1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots \ldots . \frac{x^{n}}{n !}+\ldots \ldots \infty$$

(ii) e-x = $$1-\frac{x}{1 !}+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\ldots \ldots \frac{(-1)^{n}}{n !} x^{n}+\ldots \ldots \infty$$ (Replace x by -x)

(iii) e = $$1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\ldots \ldots . \infty$$ {Putting x = 1 in (i)}

(iv) e-1 = $$1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\ldots \ldots . \infty$$ {Putting x = 1 in (ii)}

(v) $$\frac{e^{x}+e^{-x}}{2}=1+\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}+\frac{x^{6}}{6 !}+\ldots . . \infty$$

(vi) $$\frac{e^{x}-e^{-x}}{2}=x+\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\ldots \ldots \ldots \infty$$

(vii) ax = 1 + x(logea) + $$\frac{x^{2}}{2 !}$$(logea)2 + $$\frac{x^{3}}{3 !}$$ (logea)3

3. Logarithmic Series

If -1 < x ≤ 1
(i) loge(1 + x) = x – $$\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\ldots \ldots \infty$$

(ii) log(1 – x) = – x – $$\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}+\ldots \ldots \infty$$

(iii) log(1 + x) – l0g(1 – x) = log$$\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\ldots . .\right)$$

(iv) log(1 + x) + l0g(1 – x) = log (1 – x2) = -2 log$$\left(\frac{x^{2}}{2}+\frac{x^{4}}{4}+\frac{x^{6}}{6}+\ldots . .\right)$$

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