Question

obtain the Taylor's series for sec x at x=π/4​

Answer 1

Answer:

The final answer is $sec(\pi/4) = \sqrt{2}+\sqrt{2}(x-\pi/4)+\sqrt{2}(x-\pi/4)^2/2!....$

Step-by-step explanation:

Given we have,

$f(x)$ $=$ $secx$

We need to find the Taylor series for $secx$ at $\pi /4$. The Taylor Series expansion is basically an expansion of a function infinite sum of terms which are derived based on the given function. Here we have been given the function secx. We will be deriving the first couple of terms so that the pattern for the expansion can be understood.

$f(\pi/4)$ =$sec(\pi/4)$=$\sqrt{2} }$

$f'(x)$ = $secx*tanx$

$f'(\pi/4)$ = $\sqrt{2} * 1$= $\sqrt{2}$

$f"(x)$ = $(secx*tanx)$$*$ $tanx$ + $secx*sec^{2}x$ = $secx$ $( tan^{2}x + sec^{2}x )$= $secx$

$f"(\pi/4)$ $=$ $\sqrt{2}$

$f'''(x)$ $= sec x*tanx$

$f'''(π/4)$ $= \sqrt{2}$

The Taylor series expansion is as given below

$f(x) =$ ∑$f^n(a)/n!(x-a) =$ $f(x) + (x-a)f'x + (x-a)^2f''(x)/2!+(x-a)^3f'''(x)/3!...$

Now we will apply the answers we found out earlier to the Taylor Series Expansion.

$sec(\pi/4) = \sqrt{2}+\sqrt{2}(x-\pi/4)+\sqrt{2}(x-\pi/4)^2/2!....$

Hence this is the final answer.

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