Question

Completely Lost Please Help!!
If f(x) = sin x and g(x) = cos x determine the following.
f(4n+1)(x) g(4n+1)(x)
f(4n+2)(x) g(4n+2)(x)
f(4n+3)(x) g(4n+3)(x)
f(4n+4)(x) g(4n+4)(x)
Now, use your answers to calculate F(157)(x) when F(x) = 4 sin x − 3 cos x.

Answer 1

f' = cos
f'' = -sin
f''' = -cos
f'''' = sin = f

Likewise for g, there is a cycle four long.

Hence
\[ F^{(157)}(x) = F^{(4 \cdot 39 + 1)}(x) = F'(x) \]

Answer 2

See what to do now?

Answer 3

Talk to me

Answer 4

so 4(cos)+1
4(-sin)+1)
and so on

Answer 5

What that means is the (4n+1)th derivative. E.g.,
\[ f'''' =f ^{(4)}, \ \ \ f''''' = f^{(5)} = f^{(4 \cdot 1 + 1)} \]

Answer 6

\[ f' = f^{(1)}, \ \ f'' = f^{(2)}, \ \ f''' = f^{(3)}, \ ... \]

Answer 7

oh so what is n?

Answer 8

n is an integer. Hence

if n = 0
4n + 1

if n = 1,
4n + 1 = 5

if n = 2,
4n + 1 = 9
etc.

The idea is the 1st derivative is equal to the 5th derivative is equal to the ninth derivative etc.

I.e.,
\[ f' = f^{(1)} = f^{(5)} = f^{(9)} = ... \]
Likewise for 4n + 2:
\[ f'' = f^{(2)} = f^{(6)} = f^{(10)} = ... \]
and for 4n +3 and 4n + 4

Answer 9

*correction: if n = 0, then 4n + 1 = 1

Answer 10

so im going to have to find 9 derivatives? or more?

Answer 11

No, the whole idea is you see the pattern so you only need to calculate 4 derivatives. For example, with sin
f = sin
f' = cos
f'' = -sin
f''' = -cos
f'''' = sin = f

I.e., the 4th derivative equals the original function. Hence the fifth derivative
\[ f''''' = f^{(5)} \] is equal to the derivative of the 4th derivative. And the fourth derivative is the original function. Hence the 5th derivative is just the first derivative!
\[ f^{(5)} = (f^{(4)})' = f' = f^{(1)} \]

Answer 12

Likewise the 6th derivative is the second derivative, because
\[ f^{(6)} = (f^{(5)})' = (f')' = f^{(2)} \]

Answer 13

oh gotcha, so the answer will be one of the four sin cos, -sin or -cos since it just has x beside it and you plug in them for x?

Answer 14

sort of. But can you see now that
\[ f^{(1)} = f^{(5)} = f^{(9)} = f^{(13)} = ... \]
i.e.,
\[ f^{(4n+1)} \] is the same for integers \( n \geq 0 \).

Answer 15

yes i see, so sin(x) then would be for 4n+1

Answer 16

This is what you have to show.

And then show that \( f^{(2n+2)} \) is the same for all integers n.

And show that \( f^{(2n+3)} \) is the same for all integers n

And show that \( f^{(2n+4)} \) is the same for all integers n.

===

Then show the same relationships for g(x) = cos x

Answer 17

*correction: 4n + 1, 4n + 2, 4n + 3, 4n + 4. Not 2n + ...

Answer 18

0_o so since f(x) = sin x = = sin^2(x) for 4n+1)

Answer 19

No. Let's go back to this:
f = sin
f' = cos
f'' = -sin
f''' = -cos
f'''' = sin = f

What is the 5th derivative?

Answer 20

cos? or si

Answer 21

remember that \( f^{(n)} \) means the nth derivative of f. It does NOT mean the nth power of f.

Answer 22

Yes. The fifth derivative of f is written, in notation as
\[ f^{(5)} \] That's just a definition of a piece of notation in mathematics. Now, by definition of the 5th derivative in itself, it is the derivative of the fourth derivative: i.e.,
\[ f^{(5)} = (f^{(4)})' \]

Now, what is the fourth derivative of f equal to?

Answer 23

sin

Answer 24

sin

Answer 25

right. Hence
\[ f^{(5)} = (sin)' = cos \]

Answer 26

that is, the 5th derivative is equal to the first derivative. In notation:
\[ f''''' = f' \] or in the new compact notation of this question:
\[ f^{(5)} = f^{(1)} \]

Again, this does NOT mean the 5th power of f; it's the fifth derivative. This second formula is just a way of writing the first formula.

Answer 27

So the question is asking you to prove several things. The first thing it asking you to prove is that
\[ f^{(4n+1)} \] is the same for all integers \( n \geq 0 \). I.e.,
\[ f^{(1)} = f^{(5)} = f^{(9)} = f^{(13)} = ... \]
Make sense?

Answer 28

sobut isnt it asking for more than just f(1), f95) and f(9)?

Answer 29

Yes. It's asking you to establish 8 formulae, at the top of your question:
f(4n+1)(x) g(4n+1)(x)
f(4n+2)(x) g(4n+2)(x)
f(4n+3)(x) g(4n+3)(x)
f(4n+4)(x) g(4n+4)(x)

Where the first number here in brackets means the (4n+j)th derivative

Answer 30

Then use that for the second part of the question as I indicated in my first answer.

Anyway. Play around with this for a bit. I need to help someone else for a while.

Answer 31

do you know what the anser to the very last one is i have all 8 but i need the last one