1. ## prove these innequalities

how am i supposed to prove that the function is bigger then its tailor series??

the taylor series are written in endless way.
its should be equal

how to prove that its bigger??

2. Let $f_n(x) = e^x - \left( 1 + x + ... + \frac{x^n}{n!} \right)$.
You need to prove that $f_n(x) > 0$.
You can prove this by showing $f ' (x) > 0\text{ for }x>0$ and $f(0) \geq 0$.

Prove this by induction.
What you need to realize is that $f_{n+1}(x) ' = f_n(x)$.

3. regarding the first:
the base case
$
f_1(x) = e^x - \left( 1 + x\right)\\
$

i know that x>0
but it doesnt prove that
$
f_1(x)>0
$

??

4. Originally Posted by transgalactic
regarding the first:
the base case
$
f_1(x) = e^x - \left( 1 + x\right)\\
$

i know that x>0
but it doesnt prove that
$
f_1(x)>0
$

??
First $f_1(0) = e^0 - (1 + 0) = 0$. Second, $f_1'(x) = e^x - 1 > 0$ for $x>0$. Thus, the function $f_1$ is increasing for $x>0$. But at $x=0$ we have $f_1(0) = 0$ which means if $x>0$ then $f_1(x) > f_1(0) = 0$. Therefore, $e^x - (1+x) > 0 \text{ for }x>0$.

5. i wasnt given that x>0
so my general function are bigger then their taylor series by the remainder of Rn

this is the logic
how to formulate it into equation??

6. Originally Posted by transgalactic
i wasnt given that x>0
so my general function are bigger then their taylor series by the remainder of Rn

this is the logic
how to formulate it into equation??
This inequality is not true for $x<0$.
Just graph $y=e^x$ and $y=1 + x + \tfrac{1}{2}x^2$ and see what happens.

7. Originally Posted by transgalactic

how am i supposed to prove that the function is bigger then its tailor series??

the taylor series are written in endless way.
its should be equal

how to prove that its bigger??
Look at the sign of the remainder term for the taylor series?

CB

8. how to prove that??
drawing a graph is not considered as prove

i got the idea that i need to consider two cases x>0 then the function is smaller by the remainder
and x<0 then the function is bigger by the remainder.

how to describe it mathematicly
??

9. Originally Posted by transgalactic
how to prove that??
drawing a graph is not considered as prove

i got the idea that i need to consider two cases x>0 then the function is smaller by the remainder
and x<0 then the function is bigger by the remainder.

how to describe it mathematicly
??
Once again, consider the sign of the remainder term in Taylors theorem!

CB

10. if the sign is positive then the function is bigger then the series
if the sign is negative then the function is smaller then the series

??

11. How many boards are you posting this on? I feel like I am chasing you around the internet! I explained this to you on "Physics Forums" but since you might confuse others:

You are NOT asked to prove that "this function is larger than its Taylor series". You are asked to prove that it is larger than its Taylor polynomial of degree n, for any n.

12. i got your idea that its the function is bigger then the polynomial by the members with powers higher then "n".

but its only a logic
how to formulate it into mathematical equations??

13. i should prove it by induction??

14. Originally Posted by transgalactic
if the sign is positive then the function is bigger then the series
if the sign is negative then the function is smaller then the series

??
Yes, now look up one or more of the forms of remainder and determine the sign/s

CB

15. the remainder formula is
0<c<x
$
R_n(x)=\frac{f^{n+1}(c)}{(n+1)!}x^{n+1}
$

so i guess that for x>0 its:
$
R_n(x)=\frac{f^{n+1}(c)}{(n+1)!}x^{n+1}
$

for x<0
$
R_n(x)=-\frac{f^{n+1}(c)}{(n+1)!}x^{n+1}
$

should i prove it by induction??
if so then if n=1 then
$
e^x=1+x+ \frac{f^{1+1}(c)}{(1+1)!}x^{1+1}
$

but i dont know whats the value of c??
and still it depends on the values of x
??