- #1
mathboy
- 182
- 0
Notation: I = integral sign from 0 to 1, D= partial derivative symbol.
Please help me prove that for any smooth function f:R^n -> R defined on a neighbourhood of a in R^n,
f(x) = f(a) + I{(D/Dt)f(a+t(x-a))dt}
Here's my attempt:
(D/Dt)f(a+t(x-a))dt = d[f(a+t(x-a)] (justification needed?)
so
I [(D/Dt)f(a+t(x-a))dt] = I d[f(a+t(x-a)]
= f(a+1(x-a)) - f(a+0(x-a)) (Fundamental theorem of calculus, right?)
= f(x)-f(a).
Am I right, or am I making many unjustified steps here?
Please help me prove that for any smooth function f:R^n -> R defined on a neighbourhood of a in R^n,
f(x) = f(a) + I{(D/Dt)f(a+t(x-a))dt}
Here's my attempt:
(D/Dt)f(a+t(x-a))dt = d[f(a+t(x-a)] (justification needed?)
so
I [(D/Dt)f(a+t(x-a))dt] = I d[f(a+t(x-a)]
= f(a+1(x-a)) - f(a+0(x-a)) (Fundamental theorem of calculus, right?)
= f(x)-f(a).
Am I right, or am I making many unjustified steps here?
Last edited: