Power rule proof

[Callumnc1]

Homework Statement:

Pls see below

Relevant Equations:

Pls see below

For this proof,

I am unsure how they got from line 3 to line 4.

If I simplify and collect like terms for line 3 I get $f'(a) = 4a^{n-1}$

Would some please be able to help?

Many thanks!

 

[fresh_42]

What does $a^n$ mean? It is $\underbrace{a\cdot a\cdot \ldots\cdot a}_{n\text{ times}}.$ Hence for $n=i+j$ we get $$a^n=\underbrace{a\cdot a\cdot \ldots\cdot a}_{n\text{ times}}=\underbrace{a\cdot a\cdot \ldots\cdot a}_{i \text{ times}} \cdot \underbrace{a\cdot a\cdot \ldots\cdot a}_{j\text{ times}}=a^i \cdot a^j =a^{i+j}$$

 

[Callumnc1]

What does $a^n$ mean? It is $\underbrace{a\cdot a\cdot \ldots\cdot a}_{n\text{ times}}.$ Hence for $n=i+j$ we get $$a^n=\underbrace{a\cdot a\cdot \ldots\cdot a}_{n\text{ times}}=\underbrace{a\cdot a\cdot \ldots\cdot a}_{i \text{ times}} \cdot \underbrace{a\cdot a\cdot \ldots\cdot a}_{j\text{ times}}=a^i \cdot a^j =a^{i+j}$$

Thank you for your reply @fresh_42!

Did I not add the exponents correctly? For example, $a^{n-2}a^1 = a^{n-1}$

Many thanks!

 

[Mark44]

If I simplify and collect like terms for line 3 I get
$f'(a) = 4a^{n-1}$

Would some please be able to help?

Did I not add the exponents correctly?

That's not at all what you did wrong. You missed the significance of the ... in the first of the expressions you circled.
The expression $a^{n-1} + a^{n-2}a + \dots + aa^{n-2} + a^{n-1}$ does NOT have just four terms in it. The ... in the middle is called an ellipsis, and means "continuing the same pattern." In fact, this pattern indicates that there are n terms in all, thus leading to the conclusion that $f'(a) = na^{n - 1}$.

 

[Callumnc1]

That's not at all what you did wrong. You missed the significance of the ... in the first of the expressions you circled.
The expression $a^{n-1} + a^{n-2}a + \dots + aa^{n-2} + a^{n-1}$ does NOT have just four terms in it. The ... in the middle is called an ellipsis, and means "continuing the same pattern." In fact, this pattern indicates that there are n terms in all, thus leading to the conclusion that $f'(a) = na^{n - 1}$.

Thank you for your reply @Mark44 !

I see what you mean about me missing the dots.

However how dose,

$a^{n-1} + a^{n-2}a + \dots + aa^{n-2} + a^{n-1} = na^{n-1}$?

Many thanks!

 

[Frabjous]

Thank you for your reply @Mark44 !

I see what you mean about me missing the dots.

However how dose,

$a^{n-1} + a^{n-2}a + \dots + aa^{n-2} + a^{n-1} = na^{n-1}$?

Many thanks!

Try writing it out explicitly for n=5.

 

[pasmith]

Thank you for your reply @Mark44 !

I see what you mean about me missing the dots.

However how dose,

$a^{n-1} + a^{n-2}a + \dots + aa^{n-2} + a^{n-1} = na^{n-1}$?

Many thanks!

$$\begin{multline*} \overbrace{a^{n-1} + aa^{n-2} + \dots + a^{k}a^{n-1-k} + \dots + a^{n-2}a + a^{n-1}}^{\mbox{n terms}} \\ = \overbrace{a^{n-1} + a^{n-1} + \dots + a^{n-1} + \dots + a^{n-1} + a^{n-1}}^{\mbox{n terms}} = na^{n-1}\end{multline*}$$

 

[Frabjous]

@Callumnc1 I worry about how much you are learning given the large number of problems you are posting. I would suggest going back and looking at the problems you have posted and seeing if you can solve them without help. If you have issues with many of them, then all PF is doing is helping you solve problems, not learn the material.

 

[Callumnc1]

$$\begin{multline*} \overbrace{a^{n-1} + aa^{n-2} + \dots + a^{k}a^{n-1-k} + \dots + a^{n-2}a + a^{n-1}}^{\mbox{n terms}} \\ = \overbrace{a^{n-1} + a^{n-1} + \dots + a^{n-1} + \dots + a^{n-1} + a^{n-1}}^{\mbox{n terms}} = na^{n-1}\end{multline*}$$

Thank you for your reply @pasmith ! Sorry, how did you get you last result:

$$\begin{multline*} \\ = \overbrace{a^{n-1} + a^{n-1} + \dots + a^{n-1} + \dots + a^{n-1} + a^{n-1}}^{\mbox{n terms}} = na^{n-1}\end{multline*}$$

Many thanks!

 

[Callumnc1]

@Callumnc1 I worry about how much you are learning given the large number of problems you are posting. I would suggest going back and looking at the problems you have posted and seeing if you can solve them without help. If you have issues with many of them, then all PF is doing is helping you solve problems, not learn the material.

Thank you for your reply @Frabjous ! Yes I agree with you suggestion! I will definitely try to solve the old threads!

Many thanks!

 

[Callumnc1]

Try writing it out explicitly for n=5.

Thank you for your reply @Frabjous !

For n = 5, then

$a^{n-1} + a^{n-2}a + \dots + aa^{n-2} + a^{n-1} = a^4 + a^4 + a^4 + a^4 = 4a^4$ (I am unsure what the dots mean)

Many thanks!

 

[Mark44]

For n = 5, then $a^{n-1} + a^{n-2}a + \dots + aa^{n-2} + a^{n-1} = a^4 + a^4 + a^4 + a^4 = 4a^4$ (I am unsure what the dots mean)

No, that's wrong. You should not have the dots (ellipsis) here. For n = 5, and writing all of the terms in the first expression you circled, we have:
$a^4 + a^3a + a^2a^2 + aa^3 + a^0a^4 =$
$a^4 + a^4 + a^4 + a^4 + a^4 = 5a^4$
This means that if $f(x) = x^5$, then $f'(x) = 5x^4$ and $f'(a) = 5a^4$.
As for the meaning of the dots, please reread post #4 in which I explained what they mean.

 

[Callumnc1]

No, that's wrong. You should not have the dots (ellipsis) here. For n = 5, and writing all of the terms in the first expression you circled, we have:
$a^4 + a^3a + a^2a^2 + aa^3 + a^0a^4 =$
$a^4 + a^4 + a^4 + a^4 + a^4 = 5a^4$
This means that if $f(x) = x^5$, then $f'(x) = 5x^4$ and $f'(a) = 5a^4$.
As for the meaning of the dots, please reread post #4 in which I explained what they mean.

Thank you for your help @Mark44!

Sorry, so for n = 5, continuing the pattern will give $a^2a^{n-3} = a^4$, correct? Now I need to generalize that for n which is hard.

EDIT: That's what @pasmith has done, I understand now :)

Many thanks!

 

[Callumnc1]

$$\begin{multline*} \overbrace{a^{n-1} + aa^{n-2} + \dots + a^{k}a^{n-1-k} + \dots + a^{n-2}a + a^{n-1}}^{\mbox{n terms}} \\ = \overbrace{a^{n-1} + a^{n-1} + \dots + a^{n-1} + \dots + a^{n-1} + a^{n-1}}^{\mbox{n terms}} = na^{n-1}\end{multline*}$$

Thank you for help @pasmith ! I understand now :)

 

[Frabjous]

Thank you for your help @Mark44!

Sorry, so for n = 5, continuing the pattern will give $a^2a^{n-3} = a^4$, correct? Now I need to generalize that for n which is hard.

EDIT: That's what @pasmith has done, I understand now :)

Many thanks!

The thing to notice is that the exponents go from (n-1) to 0 and 0 to (n-1) which is n terms.

In general, you need to spend more time staring at and manipulating things. Be patient. Problems take time to solve, otherwise we would call them easies.

 

[Callumnc1]

The thing to notice is that the exponents go from (n-1) to 0 and 0 to (n-1) which is n terms.

In general, you need to spend more time staring at and manipulating things. Be patient. Problems take time to solve, otherwise we would call them easies.

Thank you for you reply @Frabjous!

I agree with your advice! How long should try solving the problem for before looking at the solution?

Many thanks!

 

[Frabjous]

Until you have run out of ideas. Ideally, you would go away for a while and try again before looking. Part of problem solving is to figure out different ways of going after a problem. Hopefully one of them will work. When you get a hint from PF, play with it for a while before asking for additional help.

For this problem, you should have tried to solve it for a specific value of n starting at the xⁿ-aⁿ stage. You would have at least learned what an ellipsis is.

 

[Callumnc1]

Until you have run out of ideas. Ideally, you would go away for a while and try again before looking. Part of problem solving is to figure out different ways of going after a problem. Hopefully one of them will work. When you get a hint from PF, play with it for a while before asking for additional help.

For this problem, you should have tried to solve it for a specific value of n starting at the xⁿ-aⁿ stage. You would have at least learned what an ellipsis is.

Thank your for your advice @Frabjous! I will try to do that more!