Question

the nth term in the expansion of f(a+h) is ​

Answer 1

Answer:

The nth term in the expansion of $f(a+h)$ is $f(a) + \frac{h}{1!}f^{'}(a) + \frac{h^{2} }{2!}f^{''}(a) + ....... + \frac{h^{(n-1)} }{(n-1)!} f^{(n-1)}(a) + \frac{h^{n} }{n!} f^{n}(a).$  

Step-by-step explanation:

As per Taylor's theorem, "if $f$ is a continuous function and n times differentiable in an interval $[x, x + h]$, then there exists some point in this interval, denoted by $x +$λ$h$, for some λ ∈ [0, 1], such that

$f(a) + \frac{h}{1!}f^{'}(x) + \frac{h^{2} }{2!}f^{''}(x) + ....... + \frac{h^{(n-1)} }{(n-1)!} f^{(n-1)}(x) + \frac{h^{n} }{n!} f^{n}(x +$λ$h)$ "

Thus, to get the nth term in the expansion of $f(a+h)$, $x$ is substituted with $a$.

Therefore, the nth term in the expansion of $f(a+h)$ is $f(a) + \frac{h}{1!}f^{'}(a) + \frac{h^{2} }{2!}f^{''}(a) + ....... + \frac{h^{(n-1)} }{(n-1)!} f^{(n-1)}(a) + \frac{h^{n} }{n!} f^{n}(a).$

Answer 2

$\displaystyle \sf The \: nth \: term \: in \: the \: expansion \: of \: f(a+h) \: is \: \: \frac{ {h}^{n - 1} }{(n - 1)!} {f}^{(n - 1)}(a)$

Given :

The function f(a + h)

To find :

The nth term in the expansion of f(a + h)

Solution :

Step 1 of 2 :

Find Taylor's Series expansion

If f(x + h) can be expanded as an infinite series then

$\displaystyle \sf f(x + h) = f(x) + hf'(x) + \frac{ {h}^{2} }{2!} f''(x) + \frac{ {h}^{3} }{3!} f'''(x) + . . . + \frac{ {h}^{n - 1} }{(n - 1)!} {f}^{(n - 1)}(x) + . . . . \infty$

f(x) possesses derivatives of all orders

Step 2 of 2 :

Find nth term in the expansion of f(a+h)

Putting x = a in the expansion of f(x + h) we get

$\displaystyle \sf f(a + h) = f(a) + hf'(a) + \frac{ {h}^{2} }{2!} f''(a) + \frac{ {h}^{3} }{3!} f'''(a) + . . . + \frac{ {h}^{n - 1} }{(n - 1)!} {f}^{(n - 1)}(a) + . . . . \infty$

Hence the nth term in the expansion of f(a+h)

$\displaystyle \sf{ = \frac{ {h}^{n - 1} }{(n - 1)!} {f}^{(n - 1)}(a) }$

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