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Why is the following taylor expansion right , of a function $\phi(t)=u(tx+(1-t)x)$

my book says $$\phi(1)=\sum_{uj=0}^{j=m-1} \frac{1}{j!} \phi(t)^j + \frac{1}{(m-1)!} \int_0^1 (1-t)^{m-1} \phi^m(t)dt$$

can some one explain me ? I don't quite get it . I am not even sure if its right .

Thanks !

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no $u \in C^\infty$ @Amzoti –  Theorem Feb 24 '13 at 18:40
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up vote 0 down vote accepted

Here its just application of Taylor expansion . Using the fundamental theorem of calculus , $$f(x)= f(a)+\int_a^x f'(t)dt$$

Repeatedly applying this we get taylor expansion in integral form .

Now in case of multivariable its quite easy to go with parametrization and use taylor expansion for single variable . So, lets put for $x,y \in \Omega$ , here we particularly need $\Omega$ to be convex. so let $\phi (t)=u(xt+(1-t)y$

hence $\phi (1) =u(x)$

$$\phi(1) =\phi(0)+ \sum_{j=0}^{j=1} \frac{1}{j!} \phi^j(0) + \int_0^1 \frac{(1-t)^k}{k!}\phi^{k+1}(t)$$

To find $$\sum_{j=0}^{j=1} \frac{1}{j!} \phi^j(0)$$

$$\frac {d^j}{dt^j}\phi(t) =\frac{d^j}{dt^j}u(xt+(1-t)y) =.........$$

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