Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 !

share|cite|improve this question
no $u \in C^\infty$ @Amzoti – Theorem Feb 24 '13 at 18:40
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 $$\sum_{j=0}^{j=1} \frac{1}{j!} \phi^j(0)$$

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.