Use induction to prove that, if f(x) = xe^{ax}, then f^n(x) = (n a^{n-1} + a^n x) e^{ax}
- Thread starter KamCh
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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?
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?
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.
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.
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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?
[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?
HallsofIvy
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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 \(\displaystyle f^{(n)}= (nax^{n-1}+ a^nx)e^{ax}\).
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 \(\displaystyle f^{(n)}= (nax^{n-1}+ a^nx)e^{ax}\).
HallsofIvy
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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 \(\displaystyle f^{(n)}(x)= (a^{n-1}+ a^nx)e^{ax}\).