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Integration

A set of curves that each pass through the origin, have equations y=f(sub1) (x), y=f(sub2) (x), y=f (sub3)(x)... where f ' (sub n) (x) = f (sub n-1) (x)and f(sub1)(x)=x^2
This is a question in Exercise 8E Q 4 (C1 modular book).
Could somebody please explain what the subscripts mean? I believe there is some sort of relation...

Also, would anybody know how to solve this... Given that root y= (x^1/3)+3
show that y=(x^2/3) + Ax^1/3 +B where A and B are integers to be found.
I see how to find B which is 9 but cannot solve for A. I know I should integrate (that is the chapter) but do not know what to do exactly... Please help.
Original post by MathMeister
A set of curves that each pass through the origin, have equations y=f(sub1) (x), y=f(sub2) (x), y=f (sub3)(x)... where f ' (sub n) (x) = f (sub n-1) (x)and f(sub1)(x)=x^2
This is a question in Exercise 8E Q 4 (C1 modular book).
Could somebody please explain what the subscripts mean? I believe there is some sort of relation...


y=f1(x)y = f_1(x)

etc

Where fn′(x)=fn−1(x)f'_n(x) = f_{n-1}(x)

andf1(x)=x2f_1(x) = x^2

Is that the question?

If so - the middle line tells you the relationship
Original post by MathMeister

Also, would anybody know how to solve this... Given that root y= (x^1/3)+3
show that y=(x^2/3) + Ax^1/3 +B where A and B are integers to be found.
I see how to find B which is 9 but cannot solve for A. I know I should integrate (that is the chapter) but do not know what to do exactly... Please help.


If root y = something then
Y = the something squared




Posted from TSR Mobile
Original post by gdunne42

I done that.. :smile:
Original post by TenOfThem
y=f1(x)y = f_1(x)

etc

Where fn′(x)=fn−1(x)f'_n(x) = f_{n-1}(x)

andf1(x)=x2f_1(x) = x^2

Is that the question?

If so - the middle line tells you the relationship

I get that I should integrate and knew how to answer them. But I just do not understand it... Is the n as in x^n?
Original post by MathMeister
I done that.. :smile:


Then you know what A and B are ?

Posted from TSR Mobile
(edited 9 years ago)
Original post by MathMeister
I get that I should integrate and knew how to answer them. But I just do not understand it... Is the n as in x^n?


No

The n is n

like the nth term
Original post by gdunne42
Then you know what A and B are ?

Posted from TSR Mobile

Still don't know A. :/
Original post by TenOfThem
No

The n is n

like the nth term

I guessed that at one point. Uhm. I still don't quite get how integrating gets f2(x)
Original post by MathMeister
I guessed that at one point. Uhm. I still don't quite get how integrating gets f2(x)


fn′(x)=fn−1(x)f'_n(x) = f_{n-1}(x)

So, if n=2

f2′(x)=f1(x)f'_2(x) = f_1(x)
Original post by MathMeister
Still don't know A. :/


Because, I suspect, you didn't square it properly
If root y = a + b
y = (a + b)^2 = (a + b)(a + b)
Not a^2 + b^2


Posted from TSR Mobile
(edited 9 years ago)
Original post by gdunne42
Because, I suspect, you didn't square it properly
If root y = a + b
y = (a + b)(a + b)
Not a^2 + b^2


Posted from TSR Mobile

Thank you, uh yes... I cannot believe I missed that.
Original post by TenOfThem
fn′(x)=fn−1(x)f'_n(x) = f_{n-1}(x)

So, if n=2

f2′(x)=f1(x)f'_2(x) = f_1(x)

fn′(x)=fn−1(x)f'_n(x) = f_{n-1}(x)
If the nth derivative = n-1th function of x
.. then surely the function (before differentiated to equal fn−1(x) f_{n-1}(x)
is the integral of fn−1(x) f_{n-1}(x) ??
So that's why you integrate...?
Original post by MathMeister
fn′(x)=fn−1(x)f'_n(x) = f_{n-1}(x)
If the nth derivative = n-1th function of x
.. then surely the function (before differentiated to equal fn−1(x) f_{n-1}(x)
is the integral of fn−1(x) f_{n-1}(x) ??
So that's why you integrate...?


yes
Original post by TenOfThem
yes

Eureka!

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