Taylor's Series For One Variable Engineering Maths, Btech first year

Taylor Series

For One Variable :

Taylor Series is defined by it's usage. It is used to represent a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point.
Suppose we have a function $f(x)$ to be represented in powers of $(x-a)$, then: $$f(x)=f(a)+\frac{(x-a{)}^1}{1!}{f}^{\prime }(a)+\frac{(x-a{)}^2}{2!}{f}^{″}(a)+\frac{(x-a{)}^3}{3!}{f}^{‴}(a)+.....$$ The formula is quite easy to memorize if you observe the pattern here. Ofcourse you have to cram some part of it (after all, everything falls down on cramming something!) here $f(a)$ and $\frac{(x-a)}{1!}{f}^{\prime }(a)$, after that, in the next terms, the power of $(x-a)$ keeps increasing by 1, similarly the factorial in the denominator keeps increasing by 1, and the derivatives of the function at the point $a$ also keeps increasing by 1.

Sample Problem 1 :

Expand $\sin x$ in powers of $(x-\pi /2)$ by Taylor Series.

Solution :

$$\text{Let }f(x)=\sin x$$ $$x-a=x-\pi /2$$ $$\Rightarrow a=\pi /2$$ By Taylor Series, $$f(x)=f(a)+\frac{(x-a{)}^1}{1!}{f}^{\prime }(a)+\frac{(x-a{)}^2}{2!}{f}^{″}(a)+\frac{(x-a{)}^3}{3!}{f}^{‴}(a)+.....$$ $$\text{Here, }f(\pi /2)=1$$ $$f^{\prime }(x)=\cos x;f^{\prime }(\pi /2)=0$$ $$f^{″}(x)=-\sin x;f^{″}(\pi /2)=-1$$ $$f^{‴}(x)=-\cos x;f^{‴}(\pi /2)=0$$ $$f^{⁗}(x)=\sin x;f^{⁗}(\pi /2)=1$$ Putting values, $$f(x)=1+\frac{(x-\pi /2)}{1}(0)+\frac{(x-\pi /2{)}^2}{2}(-1)$$ $$+\frac{(x-\pi /2)}{3\times 2}(0)+\frac{(x-\pi /2{)}^4}{4\times 3\times 2}(1)+...$$ $$=1-\frac{(x-\pi /2{)}^2}{2}+\frac{(x-\pi /2{)}^4}{24}+...$$ You don't need to simplify it further, just leave it in the powers of $(x-\pi /2)$.