Use induction to prove that, if f(x) = xe^{ax}, then f^n(x) = (n a^{n-1} + a^n x) e^{ax}

[KamCh]

 

[MarkFL]

Hello and welcome to FMH!

Let's look at the base case $P_0$:

[MATH]f^{(0)}(x)=(0+x)e^{ax}=f(x)\quad\checkmark[/MATH]
The base case is true, and so we state $P_n$

[MATH]f^{(n)}(x)=\left(na^{n-1}+a^nx\right)e^{ax}[/MATH]
As our inductive step, we can differentiate both sides with respect to $x$...what do you get?

 

[Steven G]

What have you tried? Did you find f' (x)? Does it match the formula? We can't help you unless you tell us where you are stuck. Please reply back.

 

[KamCh]

I know that statement holds true for n=0, I didn't know I had to differentiate. Here's what I have after diff.
I know I have to sub k and then k+1 but I don't know what sequence have to equate to in order to show that left side is equal to the right side. Thank you.

 

[Steven G]

What is f' (x), just to see what you know? Hint: there are no n's in it

 

[MarkFL]

When I take $P_n$ as I stated it, and differentiate both sides, I get:

[MATH]f^{(n+1)}(x)=a^ne^{ax}+\left(na^{n-1}+a^nx\right)ae^{ax}[/MATH]
[MATH]f^{(n+1)}(x)=\left(a^n+\left(na^{n-1}+a^nx\right)a\right)e^{ax}[/MATH]
[MATH]f^{(n+1)}(x)=\left(a^n+na^{n}+a^{n+1}x\right)e^{ax}[/MATH]
[MATH]f^{(n+1)}(x)=\left((n+1)a^{(n+1)-1}+a^{n+1}x\right)e^{ax}[/MATH]
And so, we have derived $P_{n+1}$ from $P_n$, thereby completing the proof by induction. Does that make sense?

 

[KamCh]

I understand it now, thank you very much


I found F'(x), but I didn't know that you can derive f^(n+1) from f(n) by diff.

 

[HallsofIvy]

KamCh wrote "I know that statement holds true for n=0, I didn't know I had to differentiate."

Then you seem to be completely misunderstanding the problem! $\displaystyle f^{(n)}$ means the nth derivative of f. Notice the parentheses about n. $\displaystyle f^n$ without the parentheses is the nth power, $\displaystyle f^{(n)}$ is the nth derivative. The problem asks you to use induction to show that the nth derivative of $\displaystyle f(x)= xe^{ax}$ is $\displaystyle f^{(n)}= (nax^{n-1}+ a^nx)e^{ax}$.

 

[KamCh]

That makes it so clear, thank you!

 

[HallsofIvy]

Actually, what I wrote was incorrect- I miscopied. What you want to prove is that the nth derivative of $\displaystyle f(x)= xe^{ax}$ is $\displaystyle f^{(n)}(x)= (a^{n-1}+ a^nx)e^{ax}$.