Continuation of composite function problem problematic

[valmancer]

I asked about the first part of this problem in https://www.physicsforums.com/showthread.php?t=592408. I thought the best idea was to start another thread for the second part.

Homework Statement

Part a) Given f(x)=ax+b, and f⁽³⁾(x)=64x+21, find the values of the constants a and b.

(note: f⁽³⁾(x) means fff(x))

I figured out a=4 and b=1

Part b) Suggest a rule for f⁽ⁿ⁾(x).

Homework Equations

f⁽³⁾(x)=a³x+b(a²+a+1)

The Attempt at a Solution

From the equation I got in the first part (above) I reasoned that the genereal rule would be:

f⁽ⁿ⁾(x)=aⁿx+b(a^n-1+a^n-2...a+1)

a=4 and b=1 so,

f⁽ⁿ⁾(x)=4ⁿx+(4^n-1+4^^n-2...4+1)

However that isn't a very good general rule and the answer given in the book is:

4ⁿx+(4ⁿ-1)/3

I have no idea how to reach that from what I've got now.

 

[Bearded Man]

4^{n-1} + 4^{n-2} + \cdots + 4 + 1 is also
4^{n-1} + 4^{n-2} + \cdots + 4^1 + 4^0
Can you think of those terms in another way now?

 

[valmancer]

Thanks for the answer,

I thought of them like that almost immediately after writing them down, but still can't get them into any other form. I understand that the exponent is always one more until it reaches n-1. I don't know how I can simplify a^n+a^m, which I think is needed here.

 

[jing2178]

Have you covered geometric sequences and series yet?

 

[Bearded Man]

Are you familiar with geometric sequences and their sums?

Looks like he jing ^ beat me to it!

 

[valmancer]

I'm afraid we haven't done anything with sequences or series yet. That's why I think I'm doing something fundamentally wrong :). Could there be another way of doing this question?

It would be great if someone could point me to someplace where I can learn about sequences and series and whatever else I need to know for this question.

Thanks for the responses

 

[Bearded Man]

A geometric sequence is simply a sequence of terms where each successive term is multiplied by some r. Take this, for example:
a_n = 4^{n-1}
1, 4, 16, 64, ...
It is clear the r is 4. The sum of a geometric sequence is given by
S_n = \dfrac{a_1(1 - r^{n+1})}{1 - r}
The authors note that the expression in parenthesis may defined as the sum of a geometric sequence of n-1 terms, specifically:
\dfrac{1 \ast (1 - 4^{(n-1)+1})}{-3}
Which simplifies to the expression given.

 

[valmancer]

Thank you for the answer Bearded Man,

I understand everything else but the a_n term you use. What does a₁ mean? Why is it equal to 1 in my problem?

 

[Bearded Man]

Thank you for the answer Bearded Man,

I understand everything else but the a_n term you use. What does a₁ mean? Why does is it equal to 1 in my problem?

a_n is an expression for a given term in the sequence. so a₁ is the first term in the sequence, 1, and a₂ is the second term in the sequence, 4, etc. If you wanted to find a₅, you would find 4^5-1.

 

[valmancer]

Thank you once again Bearded Man,

I finished the question. The only thing missing is some justification for the formula used (sum of series=...), but I understand if you don't have time to explain why it works.

 

[Bearded Man]

Thank you once again Bearded Man,

I finished the question. The only thing missing is some justification for the formula used (sum of series=...), but I understand if you don't have time to explain why it works.

Recall that we can add two equations together.
S_n = a + ar + ar^2 ... + ar^n
rS_n = ar + ar^2 + ar^3 ... + ar^{n+1}
S_n - rS_n = a - ar^{n+1}
S_n (1 - r) = a(1-r^{n+1})
S_n = \dfrac{a(1-r^{n+1})}{1-r}

That is the sum of a geometric sequence. You can derive the sum of an arithmetic sequence, like 1,4,7,10,13... in a similar fashion, try it. An arithmetic sequence is a sequence of terms where each following term is d greater than the previous term (3 in the previous example). You can also derive a formula for a_n of an arithmetic sequence as well (it's actually really easy).

 

[SammyS]

...

a=4 and b=1 so,

f⁽ⁿ⁾(x)=4ⁿx+(4^n-1+4^^n-2...4+1)

However that isn't a very good general rule and the answer given in the book is:

4ⁿx+(4ⁿ-1)/3

I have no idea how to reach that from what I've got now.

If you want to show that (4^n-1+4^^n-2...4+1) is the same as (4ⁿ-1)/3 try the following:

\displaystyle \text{Let } S=4^{n-1}+4^{n-2}+\dots+4^{2}+4+1\ .

\displaystyle \text{Then multiply }S\text{ by 4: }\quad 4S=4^n+4^{n-1}+4^{n-2}+\dots+4^{2}+4\ .

Now take 4S - S, which is of course 3S, and solve for S.