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From MF26,
① sin x = x - x³/3! + x^5 / 5! -...+ (-1)^r x^(2r + 1)/(2r + 1)! + ...
② f(x) = f(0) + x f'(0) + x²/2! f"(0) + xⁿ/n! f(ⁿ)(0) +...
From ① we can say that :
③ sin(x³) = x³ - (x³)³/3! + (x³)^5 / 5! + ...
Since we want f^(15) (0) , the required term in ② will be x^15 /15! f^15 (0)
We want the equivalent term in ③. So we look for the power of x to be 15. We should pick the term in ③ that has it. This term is (x³)^5 / 5! = x^15 / 5!
Comparing these two terms' coefficients,
1/15! f^15 (0) = 1/5!
f^15 (0) = 15! / 5!
= 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6
= 10897286400
f^20 (0) will be shown in the comments section
① sin x = x - x³/3! + x^5 / 5! -...+ (-1)^r x^(2r + 1)/(2r + 1)! + ...
② f(x) = f(0) + x f'(0) + x²/2! f"(0) + xⁿ/n! f(ⁿ)(0) +...
From ① we can say that :
③ sin(x³) = x³ - (x³)³/3! + (x³)^5 / 5! + ...
Since we want f^(15) (0) , the required term in ② will be x^15 /15! f^15 (0)
We want the equivalent term in ③. So we look for the power of x to be 15. We should pick the term in ③ that has it. This term is (x³)^5 / 5! = x^15 / 5!
Comparing these two terms' coefficients,
1/15! f^15 (0) = 1/5!
f^15 (0) = 15! / 5!
= 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6
= 10897286400
f^20 (0) will be shown in the comments section
For f^20 (0) ,
We extract the following term from ① :
x^20 / 20! f^20 (0)
But what we realise from ③ is that, all the powers of x are odd.
There is no term with an even power of x.
The closest power to x^20 that we can get is :
(-1)³ (x³)^(2(3)+1) / (2(3) + 1)! = -x^21 / 7!
So the coefficient of x^20 is 0.
Which means that :
1/20! f^20 (0) = 0
f^20 (0) = 20! × 0 = 0
We extract the following term from ① :
x^20 / 20! f^20 (0)
But what we realise from ③ is that, all the powers of x are odd.
There is no term with an even power of x.
The closest power to x^20 that we can get is :
(-1)³ (x³)^(2(3)+1) / (2(3) + 1)! = -x^21 / 7!
So the coefficient of x^20 is 0.
Which means that :
1/20! f^20 (0) = 0
f^20 (0) = 20! × 0 = 0
thanks