Proving a conjecture in differentiation
How do I prove the conjecture that
d^(n)y
______ = k^(n) y
dx^(n)
The n are in the same position like 2 in the second derivative, d2y/dx2. And the n on the RHS is an exponent, but the y is not.
I know to use induction but I'm stuck at proving P(k+1),
d^(k+1)y
______
dx^(k+1)
Thank you very much!
Re: Proving a conjecture in differentiation
for n = 1
$\displaystyle \frac{dy}{dx} = ky$
assume true for $\displaystyle n$ ...
$\displaystyle \frac{d^ny}{dx^n} = k^n y$
$\displaystyle \frac{d}{dx} \left[\frac{d^ny}{dx^n} = k^n y\right] = \frac{d^{n+1}y}{dx^{n+1}} = k^n \cdot \frac{dy}{dx} = k^n \cdot ky = k^{n+1} y$
Re: Proving a conjecture in differentiation
Quote:
Originally Posted by
Tutu
How do I prove the conjecture that
d^(n)y
______ = k^(n) y
dx^(n)
skeeter does this using induction on n and assuming that dy/dx= ky is that true? You do not mention it in your post but without that conclusion is not true.
Quote:
The n are in the same position like 2 in the second derivative, d2y/dx2. And the n on the RHS is an exponent, but the y is not.
I know to use induction but I'm stuck at proving P(k+1),
d^(k+1)y
______
dx^(k+1)
Thank you very much!