- #1
psholtz
- 136
- 0
Is there a simple way to show that when we differentiate the following expression (call this equation 1):
[tex]Y(x) = \frac{1}{n!} \int_0^x (x-t)^n f(t)dt[/tex]
that we will get the following expression (call this equation 2):
[tex]Y'(x) = \frac{1}{(n-1)!}\int_0^x (x-t)^{n-1}f(t)dt[/tex]
It's simple enough to prove "by hand", for low-order polynomials. For instance, for n=1 we would have:
[tex]Y(x) = \int_0^x (x-t)f(t)dt = x\int_0^x f(t)dt - \int_0^x tf(t)dt[/tex]
[tex]Y'(x) = \int_0^x f(t)dt + xf(x) - xf(x) = \int_0^x f(t)dt[/tex]
and it can be proved just as easily for n=2, 3, etc.. But is there a straightforward way to prove it for general n? By induction perhaps? I'm reading a text where the author just goes from Equation 1 to Equation 2, and keeps rolling, w/o giving much in the way of explanation.
[tex]Y(x) = \frac{1}{n!} \int_0^x (x-t)^n f(t)dt[/tex]
that we will get the following expression (call this equation 2):
[tex]Y'(x) = \frac{1}{(n-1)!}\int_0^x (x-t)^{n-1}f(t)dt[/tex]
It's simple enough to prove "by hand", for low-order polynomials. For instance, for n=1 we would have:
[tex]Y(x) = \int_0^x (x-t)f(t)dt = x\int_0^x f(t)dt - \int_0^x tf(t)dt[/tex]
[tex]Y'(x) = \int_0^x f(t)dt + xf(x) - xf(x) = \int_0^x f(t)dt[/tex]
and it can be proved just as easily for n=2, 3, etc.. But is there a straightforward way to prove it for general n? By induction perhaps? I'm reading a text where the author just goes from Equation 1 to Equation 2, and keeps rolling, w/o giving much in the way of explanation.