Answer to Integral Calculus problem. Here is a photo

deepa Varakala

Answered question

2022-05-01

alenahelenash

Expert2023-05-02Added 556 answers

We can write the Maclaurin series for

f (x) = \sin (x^{4})

using the formula:

\sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}

where

f^{(n)} (x)

denotes the

n

th derivative of

f (x)

.
To find the derivatives of

f (x)

, we can use the chain rule and the derivative of

\sin (x)

f^{'} (x) = \cos (x^{4}) \cdot 4 x^{3}

f^{″} (x) = (- \sin (x^{4})) \cdot 16 x^{6} + (\cos (x^{4})) \cdot 12 x^{2}

{f^{″}}^{'} (x) = (- \cos (x^{4})) \cdot 64 x^{9} + (- \sin (x^{4})) \cdot 36 x^{5}

f^{(4)} (x) = (\sin (x^{4})) \cdot 256 x^{12} + (- \cos (x^{4})) \cdot 540 x^{8} + (\sin (x^{4})) \cdot 90 x^{4}

f^{(5)} (x) = (\cos (x^{4})) \cdot 1024 x^{15} + (\sin (x^{4})) \cdot 2160 x^{11} + (- \cos (x^{4})) \cdot 900 x^{7}

Then, we can evaluate each derivative at

x = 0

f (0) = \sin (0) = 0

f^{'} (0) = \cos (0) \cdot 4 \cdot 0^{3} = 0

f^{″} (0) = - \sin (0) \cdot 16 \cdot 0^{6} + \cos (0) \cdot 12 \cdot 0^{2} = 0

{f^{″}}^{'} (0) = - \cos (0) \cdot 64 \cdot 0^{9} - \sin (0) \cdot 36 \cdot 0^{5} = 0

f^{(4)} (0) = \sin (0) \cdot 256 \cdot 0^{12} - \cos (0) \cdot 540 \cdot 0^{8} + \sin (0) \cdot 90 \cdot 0^{4} = 0

f^{(5)} (0) = \cos (0) \cdot 1024 \cdot 0^{15} + \sin (0) \cdot 2160 \cdot 0^{11} - \cos (0) \cdot 900 \cdot 0^{7} = 0

Therefore, the Maclaurin series for

f (x) = \sin (x^{4})

is:

f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} = 0 + 0 x + 0 x^{2} + 0 x^{3} + 0 x^{4} + 0 x^{5} + \dots = 0

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