Question

It is known that the Maclaurin's series for
f(x) is 1 + 3x + 5x² + 7x³ + ... .
Find the value of f'(0), f''(0) and f'''(0).

Answer 1

I differentiate 3 times and can get the values of f'(0) and f''(0) but not f'''(0).
Why is this so ? :(

Answer 2

Because that is what the formula says you should do :)

Answer 3

The first time you differentiate f(x) you get f'(x), right @HatcrewS ?

Answer 4

\[\Large\rm f(x)=1+3x+5x^2+7x^3+...\]
\[\Large\rm f(x)=\sum_{0}^{\infty}\frac{f^n(0)}{n!}x^n\]
Is this what you're having trouble with? The third derivative?
Maybe I misunderstood your question though -_-
\(\Large\rm \implies\quad \frac{f^{(3)}(0)}{3!}=7x^3\)

Answer 5

3=f'(0)
using the idea that a1=f'(0) for functions where x exists at 0.

Answer 6

Woops I left out the x^3 on the left lol.

Answer 7

@zepdrix i differentiate til i get d³y/dx³ and sub in x = 0

Answer 8

Oh you're not using the formula? :3 my bad.

Answer 9

And f''(0)= a2*2! which would be 5*2=10

Answer 10

f'(x) = 3 + 10x + 21x²
f''(x) = 10 + 42x
f'''(x) = 42

Answer 11

Looks good c:

Answer 12

thank you :D

Answer 13

f'''(x) is the constant 42 :)

Answer 14

thanks everyone for helping! :DD

Answer 15

Thanks for asking :)

Answer 16

Don't ignore the remaining terms :

f'(x) = 3 + 10x + 21x² + ...
f''(x) = 10 + 42x + ...
f'''(x) = 42 + ....

they vanish olny after plugging in x=0
(just nitpicking :P)

Answer 17

Good point @ganeshie8