Notes on Exponential and Logarithmic Function and Series,Expansion of e^x,a^x and log(1+x) | Grade 12 > Mathematics > Binomial Theorem | KULLABS.COM

Exponential and Logarithmic Function and Series,Expansion of e^x,a^x and log(1+x)

  • Note
  • Things to remember

xx

Graph for Exponential function Source:www.shelovesmath.com
Graph for exponential function Source:www.sheloves math.com

Exponential Function:The function which is in the form of $$\;y=f(x)=a^x,\;\;\;\;\;a>0$$ is called an exponential function in which the base a is constant and the power or index x is a variable. The given figure shows us the type of graph the exponential function portrays when the value of a is >1 or 0<a<1.

Logarithmic Function: The function that is the inverse of an exponential function is called a logarithmic function which is denoted by logbx.If we have an exponential function y=ax, then the logarithmic function is given by x=logay .

Euler's number:Let us assume that we have an exponential function y=(1 + 1/n )n. When we increase the value of n in the given function, the value of y tends to reach a value close to 2.71828... Thus the limit of the function (1 + 1/n )n denoted by e which is the Euler's number.The number e is a famous irrational number and is one of the most important numbers in mathematics. The first few digits are: 2.7182818284590452353602874713527 (and more ...) It is called Euler's number after Leonhard Euler. Thus, $$\;e=\lim_{n\to\infty}\bigg(1+\frac{1}{n}\bigg)^n$$The function which has the base e(Euler's number ) i.e. f(x)=ex is known as natural exponential function. The logarithmic function having base e is known as natural logarithmic function. It is written as logx or logex.

Expansion of ex:

We know,

$$e^x\;=1\;+\;\frac{x}{1!}\;+\;\frac{x^2}{2!}\;+\;\frac{x^3}{3!}\;+\;.\;.\;.\;.\;.\;.\;.\;+\frac{x^r}{r!}\;+\;.\;.\;.\;.\;.\;to\;\infty\;,\;x\in\;R$$

Proof:To prove,$$e^x\;=1\;+\;\frac{x}{1!}\;+\;\frac{x^2}{2!}\;+\;\frac{x^3}{3!}\;+\;.\;.\;.\;.\;.\;.\;.\;+\frac{x^r}{r!}\;+\;.\;.\;.\;.\;.\;to\;\infty\;,\;x\in\;R$$

we will use,$$\;e=\lim_{n\to\infty}\bigg(1+\frac{1}{n}\bigg)^n$$

$$\;Let\;n>1\;,\;then\;\frac{1}{n}<1\;$$

Now using the binomial theorem, we have

$$\bigg(1+\frac{1}{n}\bigg)^n\;=\;1+(nx)\frac{1}{n}\;+\frac{(nx)(nx-1)}{2!}.\frac{1}{n^2}$$

$$+\frac{(nx)(nx-1)(nx-3)}{3!}.\frac{1}{n^3}+\;.\;.\;.\;.\;.$$

$$\;=1+x+\frac{x}{2}\bigg(x-\frac{1}{n}\bigg)\;+\;\frac{x\bigg(x-\frac{1}{n}\bigg)\bigg(x-\frac{2}{n}\bigg)}{3!}\;+\;.\;.\;.\;.\;.\;.\;.$$

$$\;In\;the\;limit\;,\;when\;n\;\rightarrow\;\infty\;then\;\frac{1}{n}\rightarrow\;0.\frac{2}{n}\rightarrow\;0,\;and\;so\;on\;,\;we\;have\;$$

$$\;=\;=\lim_{n\to\infty}\bigg[\;=1+x+\frac{x}{2}\bigg(x-\frac{1}{n}\bigg)^{nx}\;+\;\frac{x\bigg(x-\frac{1}{n}\bigg)\bigg(x-\frac{2}{n}\bigg)}{3!}\;+\;.\;.\;.\;.\;.\;.\;.\bigg]$$

$$\rightarrow\;\big[\lim_{n\to\infty}\bigg[\;=1+x+\frac{x}{2}\bigg(x-\frac{1}{n}\bigg)^{nx}\bigg]\;=\;e^x\;$$

From the above two equations,

$$e^x\;=1\;+\;\frac{x}{1!}\;+\;\frac{x^2}{2!}\;+\;\frac{x^3}{3!}\;+\;.\;.\;.\;.\;.\;.\;.\;+\frac{x^r}{r!}\;+\;.\;.\;.\;.\;.\;to\;\infty\;,\;x\in\;R$$

Result 1: When is replaced by -x,

$$e^{-x}\;=1\;-\;\frac{x}{1!}\;+\;\frac{x^2}{2!}\;-\;\frac{x^3}{3!}\;+\;.\;.\;.\;.\;.\;.\;.\;+(-1)^r\frac{x^r}{r!}\;+\;.\;.\;.\;.\;.\;to\;\infty$$

Result 2:When x is replaced by 1 or -1,

when x=1,$$e\;=1\;+\;\frac{1}{1!}\;+\;\frac{1}{2!}\;+\;\frac{1}{3!}\;+\;.\;.\;.\;.\;.\;.\;.\;$$

When x=-1,$$e^{-1}\;=1\;-\;\frac{1}{1!}\;+\;\frac{1}{2!}\;-\;\frac{1}{3!}\;+\;.\;.\;.\;.\;.\;.\;.\;$$

The Value of e:

We know that,$$e\;=1\;+\;\frac{1}{1!}\;+\;\frac{1}{2!}\;+\;\frac{1}{3!}\;+\;.\;.\;.\;.\;.\;.\;.\;$$

$$\;or\;e\;=2\;+\;\frac{1}{2!}\;+\;\frac{1}{3!}\;+\;.\;.\;.\;.\;.\;.\;.\;$$

$$\;e>2\;as\;all\;the\;terms\;are\;positive\;$$

Since, r!=1.2.3.4. . . . . . . . . .r

r!>1.2.2.2. . . . . . . . . .r

>2r-1

$$\therefore\;\frac{1}{r!}<\frac{1}{2^{r-1}}$$

$$\;\therefore\;e\;\;=\;1\;+\;\frac{1}{1!}\;+\;\frac{1}{2!}\;+\;\frac{1}{3!}\;+\;\frac{1}{4!}\;+\;.\;.\;.\;.\;.\;.\;.\;<\;1\;+\;\frac{1}{2}\;+\;\frac{1}{2^2}\;+\;\frac{1}{2^3}\;+\;\frac{1}{2^4}\;+\;.\;.\;.\;.\;.\;.\;.\;$$

$$\;<\;1+\frac{1}{1-\frac{1}{2}}\;=\;1+2\;=\;3\;$$

$$\;\therefore\;e<3$$

So, 2<e<3.

Expansion of ax

If a is a positive number then,

$$\;a^x\;=\;1\;+\;\frac{x}{1!},(log_ea)\;+\;\frac{x^2}{2!},(log_ea)^2\;+\;.\;.\;.\;.\;.\;.\;.+\;\frac{x^r}{r!},(log_ea)^r\;+\;.\;.\;.\;to\infty$$

Proof:To prove,$$\;a^x\;=\;1\;+\;\frac{x}{1!},(log_ea)\;+\;\frac{x^2}{2!},(log_ea)^2\;+\;.\;.\;.\;.\;.\;.\;.+\;\frac{x^r}{r!},(log_ea)^r\;+\;.\;.\;.\;to\infty$$

Let a be any positive number and kbe any number such that,

$$\;a=e\;\;\;\;\;\;\;\;\;\;\;\therefore\;k=\;log_ea\;$$

Now,

$$\;a^x=(e^k)^x=e^kx$$

$$e^{kx}\;=1\;+\;\frac{kx}{1!}\;+\;\frac{kx^2}{2!}\;+\;\frac{kx^3}{3!}\;+\;.\;.\;.\;.\;.\;.\;.\;+\frac{(kx)^r}{r!}\;+\;.\;.\;.\;.\;.\;to\;\infty\;,\;x\in\;R$$

$$\;a^x\;=\;1\;+\;\frac{x}{1!},(log_ea)\;+\;\frac{x^2}{2!},(log_ea)^2\;+\;.\;.\;.\;.\;.\;.\;.+\;\frac{x^r}{r!},(log_ea)^r\;+\;.\;.\;.\;to\infty$$

Note:The expansion of ax is an exponential series.

The logarithmic series (Expansion of log(1+x))

$$\;log(1+x)\;=x-\frac{x^2}{2}+\frac{x^3}{3}-{x^4}{4}\;+\;.\;.\;.\;.\;.\;.$$

This series has been found to be valid for x=1, but not for x=-1.thus,

$$log_e(1+x)\;=\;x\;-\;\frac{x^2}{2}\;+\;\frac{x^3}{3}\;-\;\frac{x^4}{4}\;+\;.\;.\;.\;.\;.\;.\;.\;.$$

Which is a logarithmic series and is valid for -1<x<1. This series can be used to calculate the natural logarithmic of numbers. Thus, putting x on both sides we get,

$$\;ln2\;=\;1-\frac{1}{2}\;+\;frac{1}{3}\;-\frac{1}{4}\;+\;.\;.\;.\;.\;.\;.$$

Result 1: If we change x by –x in the equation,

$$log_e(1+x)\;=\;x\;-\;\frac{x^2}{2}\;+\;\frac{x^3}{3}\;-\;\frac{x^4}{4}\;+\;.\;.\;.\;.\;.\;.\;.\;.$$

We get,

$$log_e(1-x)\;=\;-x\;-\;\frac{x^2}{2}\;-\;\frac{x^3}{3}\;-\;\frac{x^4}{4}\;+\;.\;.\;.\;.\;.\;.\;.\;.$$

Now, Subtracting The two equations,

$$log_e(1+x)- log_e(1-x)\;=\;x\;-\;\frac{x^2}{2}\;+\;\frac{x^3}{3}\;-\;\frac{x^4}{4}\;+\;.\;.\;.\;.\;.\;.\;.\;-[ x\;-\;\frac{x^2}{2}\;-\;\frac{x^3}{3}\;-\;\frac{x^4}{4}\;+\;.\;.\;.\;.\;.\;.\;.\;.].$$

$$\;log_e\frac{(1+x)}{(1-x)}\;=\;2\bigg[\;x\;+\frac{x^3}{3}+\;\frac{x^5}{5}\;+\;.\;.\;.\;.\bigg]$$

Note : The exponential series is valid for all values of x. The logarithmic series is valid only when -1<x<1.

Taken reference from

( Basic mathematics Grade XII and A foundation of Mathematics Volume II and Wikipedia.com )



  • The function which is in the form of $$\;y=f(x)=a^x,\;\;\;\;\;a>0$$ is called an exponential function in which the base a is constant and the power or index x is a variable. The given figure shows us the type of graph the exponential function portrays when the value of a is >1 or 0<a<1.
  • The function that is the inverse of an exponential function is called a logarithmic function which is denoted by logbx.If we have an exponential function y=ax, then the logarithmic function is given by x=logay .
  • $$e^x\;=1\;+\;\frac{x}{1!}\;+\;\frac{x^2}{2!}\;+\;\frac{x^3}{3!}\;+\;.\;.\;.\;.\;.\;.\;.\;+\frac{x^r}{r!}\;+\;.\;.\;.\;.\;.\;to\;\infty\;,\;x\in\;R$$
  • $$\;a^x\;=\;1\;+\;\frac{x}{1!},(log_ea)\;+\;\frac{x^2}{2!},(log_ea)^2\;+\;.\;.\;.\;.\;.\;.\;.+\;\frac{x^r}{r!},(log_ea)^r\;+\;.\;.\;.\;to\infty$$
  • $$\;log(1+x)\;=x-\frac{x^2}{2}+\frac{x^3}{3}-{x^4}{4}\;+\;.\;.\;.\;.\;.\;.$$
.

Very Short Questions

0%

ASK ANY QUESTION ON Exponential and Logarithmic Function and Series,Expansion of e^x,a^x and log(1+x)

No discussion on this note yet. Be first to comment on this note