Question

e^x>1+x+(x^2)/2 (x>0)
prove the inequalities.

Answer 1

are you looking at power series in class?

Answer 2

do you know the taylor series expansion of e^x?

Answer 3

(First prove that : \(e^x=1+x+(x^2)/2!+(x^3)/3!+... \) by using power series)

i.e. \(e^x = 1 + x +\cfrac{x^2}{2} + (\text{ something positive } ) \)

Thus \(e^x > 1 + x + \cfrac{x^2 }{ 2} \)

Answer 4

Try to prove the given equation : \(e^x = 1 + x + \cfrac{x^2 }{ 2! } + \cfrac{x^3 }{ 3!} + ... \)

Answer 5

@ni21 have you already learned about Taylor series?

Answer 6

dear all, actually i have the answer but i don't understand..
in the solution , it shows using f' and f''...
does this related to taylor series?

Answer 7

related, yes, but it would be hard to explain that way...
how about post the answer and then I can explain it for you

Answer 8

f' and f'' are related to taylor series

Answer 9

try looking at 4.2 and 4.3 here
http://www.math.smith.edu/~rhaas/m114-00/chp4taylor.pdf

Answer 10

and just plug in 0 for a

Answer 11

\[f(x)=e^x-[1+x+(x^2)/2]\]\[f'(x)=e^x-1-x\]\[f''(x)=e^x-1>e^0-1=0\]\[(x>0)\]
from f''(x)>0, f'(x)>f'(0)
since

\[f'(0)=e^0-1-0=0, f(x)>0\]
from f'(x)>0, f(x)>f(0)
since f(0)=e^0-(1+0+0)=0, f(x)>0

therefore., \[e^x>1+x+(x^2)/2 when x>0\]

Answer 12

oh. It is defining a function as the difference between e^x and 1+x+(x^2)/2, then finding the behavior of that function by finding the first and second derivatives.

Answer 13

so if f(x) is positive, the statement is true.

Answer 14

@Peter14 ... does this mean the solution given is using mathematical induction?
i mean how can i know that i need to find until f'' or f'?

Answer 15

there is another question that actually find up to f' only..
here is the question..
\[\ln (1+x)>\frac{ x }{ 1+x} (x>0)\]
prove the inequalities

Answer 16

the solution doesn't use mathematical induction, or at least not that I can see.

Answer 17

if that so, then how can i know the solution needs f' or f''?
million thanks in advance, @Peter14:)
other members, i need your opinions, too:)

Answer 18

here. http://en.wikipedia.org/wiki/Racetrack_principle do you see how this applies?

Answer 19

i think i can see how it links... but if let say i stuck in the middle, can i ask you again?:)

Answer 20

sure, i'll be here

Answer 21

so for this question, ln(1+x)>x 1+x (x>0) prove the inequalities.. the method is same right?

Answer 22

yes, the method is the same.

Answer 23

thanks,@Peter14

Answer 24

hi @Peter14 , i try to solve for ln(1+x)>x 1+x (x>0), i mean try to prove this inequality but when i check the solution, it shows that i dont have to find f''. so now, how do i know whether i need to find up to f'' or f'? is there any condition or am i missed out anything that is related to?

Answer 25

let's do it together...

Answer 26

what's f'(x)?

Answer 27

(sorry, my latex isn't working)

Answer 28

\[\frac{ x }{ (1+x)^2 } \]

Answer 29

i simplified from this \[\frac{ 1 }{ 1+x }-\frac{ 1 }{ (1+x)^2 }\] to \frac{ x }{ (1+x)^2 }

Answer 30

so when x is positive, is x/((1+x)^2) always positive?

Answer 31

yes. f'(x) always positive

Answer 32

so then you don't need to find f''(x) because you can see from this that the derivative of ln (1+x) > the derivative of the other thing

Answer 33

if we were to look back at the first question that i've posted, when \[f'(x)=e^x-1-x\] means it is still not enough to prove that it is always positive? is it?

Answer 34

it's because e^x is sorta hard to evaluate without a calculator, so it's hard to prove just from that that e^x>1+x,
when you go to f''(x) it's easy because you know e^x is always greater than 1 for x>0

Answer 35

@ni21 you see?

Answer 36

i'm still trying to digest :)
so to put in other example, if i manage to f'(x)=sinx-xcosx, i still cannot prove that it is always positive and need to derive 1 more time, right?

Answer 37

yes, I think so

Answer 38

what was f(x) for that?

Answer 39

-2 cos x - x sin x ?

Answer 40

\[2(1-cosx)-xsinx\]

Answer 41

more or less the same.. without constant 2 yet the basic is still there:) that's impressive!

Answer 42

that's indefinite integrals

Answer 43

except I left out the C as is my common mistake

Answer 44

but please note that the condition for x i given as such (0<x<=pi/2)

Answer 45

ok, that's nice to know

Answer 46

yes, take the derivative of sin x - x cos x and all will be clear

Answer 47

glad to know that i'm in the right path.
just to double confirm with regarding the reason of f'(x) and f"(x); we only need to derive up to f''(x) because to assure that when x is positive then the derivative is always positive, am i right? or is there any concrete reason?

Answer 48

yes, you're right, though the english there is a bit awkward.

Answer 49

haha:) i know my english really poor.

Answer 50

it works, it's enough to get the message across, and I hadn't noticed before now.

Answer 51

you mean my english?

Answer 52

yes

Answer 53

anything else?

Answer 54

that should be enough for this time.
arigatou gozaimasu!
nice to learn from you.

Answer 55

nice to teach you