Question

For every real \(t>-1\), and every \(n \in N\), we have
\[(1+t)^n \ge 1+nt, (1+t)^n \ge 1+nt+\frac{1}{2}n(n-1)t^2;\]when \(t \neq 0\), the inequalities are strict for \(n>1\),\(n>2\), respectively.

Answer 1

@amistre64 i opened this one

Answer 2

i don't quite understand what I'm proving

Answer 3

the indication that t not= 0 suggests that we have to divide by t, to get to whatever we have to get to ...

Answer 4

ok

Answer 5

i'm still kind of confused as to where to start

Answer 6

im thinking its 2 problems kind of combined

Answer 7

and you have to induce it to n, and n+1

Answer 8

it's actually under "Lemma" and it tells me to prove this lemma

Answer 9

ok so i'll prove it by induction the way we did it before?

Answer 10

\[(1+t)^n \ge 1+nt~:~for~n>1\]
\[(1+t)^n \ge 1+nt+\frac{1}{2}n(n-1)t^2~:~for~n>2\]
yeah, its pretty much the same concept

Answer 11

let n=2 for the first one, and prove it; then set it up for an n+1 run

Answer 12

ok and then let n=3 for the second one?

Answer 13

we know that (1+x)^n is equal to some binomial exapnsion ... so that might be needed as well

Answer 14

ok

Answer 15

im thinking this plays into the proof that we just covered :)

Answer 16

the right sides are expansions, at least partial, and we can use the expansion of the binomial for the first few terms is a thought

Answer 17

i'm kind of lost... should i start with the left side first? it looks less complicated

Answer 18

im thinking we can replace the left sides by the summation stuff, and pull out a few terms for comparison

Answer 19

i was using playtex and it crashed on me

Answer 20

\[(1+t)^n \ge 1+nt\]
\[\sum_{k=0}^{n}\binom nk x^k \ge 1+nt~:~let~n=2\]
\[\sum_{k=0}^{2}\binom 2k x^k \ge 1+2t\]
\[1+2t+t^2\ge 1+2t\]for t\(\ne\)0

\[t^2\ge0\]

Answer 21

since t not=0, the equal bar underneath is moot

Answer 22

we know the left summation is good for an n+1 .... or whatever variable you want to redefine it as

Answer 23

multiply both sides by (1+t) like we did before, and it should be a cake walk

Answer 24

ok i got that part. but what i don't get is what does that prove?

Answer 25

i got the t^2>0 part. i don't know what to do with that information though

Answer 26

it proves that it is strictly greater than the right side for n=2, we need to show that it is true for 2+1, for 2+1+1, for 2+1+1+....+1

Answer 27

ok

Answer 28

thats just the basis step, its true for n=2; let n=s i guess; and show that its true for some s+1

Answer 29

n=\(\mu\) if you wan to get all greecey

Answer 30

what should i expect for the result of n=s then? that (1+t)^s>=1+st?

Answer 31

we should that:\[(1+t)\sum\binom \mu kx^k = \sum \binom{\mu+1}{k}x^k\]

Answer 32

we *showed .... that is

Answer 33

ok i'm condused. sorry i'm kind of slow

Answer 34

multiplying both sides by (1+t) is what we need to demonstrate right?

Answer 35

crashed again ...

Answer 36

yes
from the last proof right?

Answer 37

\[(1+t)^n>1+nt\]
\[(1+t)(1+t)^n>(1+nt)(1+t)\]
\[(1+t)^{n+1}>(1+nt)(1+t)\]
\[\sum_{k=0}^{\mu+1}\binom{\mu+1}{k}t^k~>~1+(\mu+1)t+\mu t^2\]

Answer 38

drops some expansions out the left side ... what do start to get?

Answer 39

ok i understand this part

Answer 40

do we just ignore the equal under the > tho?

Answer 41

since t has a domain of (-1,0), 1 - fraction is never negative so we aint got to bother worrying about a sign flip either

Answer 42

ok!

Answer 43

we know that its stricly > for n=2, we want to show that it is strictly true for numbers bigger than 2 as well

Answer 44

i see

Answer 45

how do i get the st^2 to go away on the right?

Answer 46

\[\binom{\mu+1}{0}+\binom{\mu+1}{1}t+\binom{\mu+1}{2}t^2+\sum_{k=0}^{\mu+1}\binom{\mu+1}{k}t^k~>~1+(\mu+1)t+\mu t^2\]

\[1+{\mu+1}t+\binom{\mu+1}{2}t^2+\sum_{k=0}^{\mu+1}\binom{\mu+1}{k}t^k~>~1+(\mu+1)t+\mu t^2\]

\[\binom{\mu+1}{2}t^2+\sum_{k=0}^{\mu+1}\binom{\mu+1}{k}t^k~>~\mu t^2\]

whats the right up for the (u+1 2), and how does it relate to u?

Answer 47

*the write up

Answer 48

(u+1,2)=(u+1)!/(2*(u-1)!) right?

Answer 49

\[\frac{(\mu+1)!}{(\mu+1-2)!}\]
\[\frac{(\mu+1)!}{(\mu-1)!}\]
\[\frac{(\mu+1)(\mu)\cancel{(\mu-1)!}}{\cancel{(\mu-1)!}}\]
is...
\[\mu(\mu+1)>\mu\]

Answer 50

i thought there would be a 2 at the bottom because of the k!(n-k)! at the bottom?

Answer 51

\[\binom {n}{k}=\frac{n!}{(n-k)!}\]
\[\binom {\mu+1}{2}=\frac{(\mu+1)!}{(\mu+1-2)!}\]

Answer 52

\[\mu(\mu+1)>\mu\]
\[\mu(\mu+1)-\mu>0\]
\[\mu(\mu+1-1)>0\]
\[\mu^2>0~:~for~all~\mu>1\]

Answer 53

\[\left(\begin{matrix}n \\ k\end{matrix}\right)\] is actually \[\frac{n!}{k!(n-k)!}\]in my textbook

Answer 54

.... hmm, you may be right about that :) write it up and see what we get ... i got too many things rolling around in me head to sort them

Answer 55

\[\frac12\mu(\mu+1)>\mu\]happier :)

Answer 56

ok i get that and what should i do next?

Answer 57

i got \[s^2-s>0\]

Answer 58

or written another way:

s^2 > s isnt this true for all s>2 ?

Answer 59

yes so that means we prove the first part by induction?

Answer 60

it shows that the left side is bigger than the right side:

(Bigger) t^2 + (more) > (smaller) t^2

Answer 61

yes, the first part is proved

Answer 62

yay ok so i think the second thing is actually the same because 1/2(n)(n+1) is actually (n+1,2) so i think we cancel that out so i can use the to prove k=3

Answer 63

at least I believe it is

Answer 64

the second one would follow the same concepts yes

Answer 65

thank you so much!

Answer 66

youre welcome :)