anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@amistre64 i opened this one
anonymous
  • anonymous
i don't quite understand what I'm proving
amistre64
  • amistre64
the indication that t not= 0 suggests that we have to divide by t, to get to whatever we have to get to ...

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anonymous
  • anonymous
ok
anonymous
  • anonymous
i'm still kind of confused as to where to start
amistre64
  • amistre64
im thinking its 2 problems kind of combined
amistre64
  • amistre64
and you have to induce it to n, and n+1
anonymous
  • anonymous
it's actually under "Lemma" and it tells me to prove this lemma
anonymous
  • anonymous
ok so i'll prove it by induction the way we did it before?
amistre64
  • amistre64
\[(1+t)^n \ge 1+nt~:~for~n>1\] \[(1+t)^n \ge 1+nt+\frac{1}{2}n(n-1)t^2~:~for~n>2\] yeah, its pretty much the same concept
amistre64
  • amistre64
let n=2 for the first one, and prove it; then set it up for an n+1 run
anonymous
  • anonymous
ok and then let n=3 for the second one?
amistre64
  • amistre64
we know that (1+x)^n is equal to some binomial exapnsion ... so that might be needed as well
anonymous
  • anonymous
ok
amistre64
  • amistre64
im thinking this plays into the proof that we just covered :)
amistre64
  • amistre64
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
anonymous
  • anonymous
i'm kind of lost... should i start with the left side first? it looks less complicated
amistre64
  • amistre64
im thinking we can replace the left sides by the summation stuff, and pull out a few terms for comparison
anonymous
  • anonymous
i was using playtex and it crashed on me
amistre64
  • amistre64
\[(1+t)^n \ge 1+nt\] \[\sum_{k=0}^{n}\binom nk x^k \ge 1+nt~:~let~n=2\] \[\sum_{k=0}^{2}\binom 2k x^k \ge 1+2t\] \[1+2t+t^2\ge 1+2t\]for t\(\ne\)0 \[t^2\ge0\]
amistre64
  • amistre64
since t not=0, the equal bar underneath is moot
amistre64
  • amistre64
we know the left summation is good for an n+1 .... or whatever variable you want to redefine it as
amistre64
  • amistre64
multiply both sides by (1+t) like we did before, and it should be a cake walk
anonymous
  • anonymous
ok i got that part. but what i don't get is what does that prove?
anonymous
  • anonymous
i got the t^2>0 part. i don't know what to do with that information though
amistre64
  • amistre64
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
anonymous
  • anonymous
ok
amistre64
  • amistre64
thats just the basis step, its true for n=2; let n=s i guess; and show that its true for some s+1
amistre64
  • amistre64
n=\(\mu\) if you wan to get all greecey
anonymous
  • anonymous
what should i expect for the result of n=s then? that (1+t)^s>=1+st?
amistre64
  • amistre64
we should that:\[(1+t)\sum\binom \mu kx^k = \sum \binom{\mu+1}{k}x^k\]
amistre64
  • amistre64
we *showed .... that is
anonymous
  • anonymous
ok i'm condused. sorry i'm kind of slow
amistre64
  • amistre64
multiplying both sides by (1+t) is what we need to demonstrate right?
amistre64
  • amistre64
crashed again ...
anonymous
  • anonymous
yes from the last proof right?
amistre64
  • amistre64
\[(1+t)^n>1+nt\] \[(1+t)(1+t)^n>(1+nt)(1+t)\] \[(1+t)^{n+1}>(1+nt)(1+t)\] \[\sum_{k=0}^{\mu+1}\binom{\mu+1}{k}t^k~>~1+(\mu+1)t+\mu t^2\]
amistre64
  • amistre64
drops some expansions out the left side ... what do start to get?
anonymous
  • anonymous
ok i understand this part
anonymous
  • anonymous
do we just ignore the equal under the > tho?
amistre64
  • amistre64
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
anonymous
  • anonymous
ok!
amistre64
  • amistre64
we know that its stricly > for n=2, we want to show that it is strictly true for numbers bigger than 2 as well
anonymous
  • anonymous
i see
anonymous
  • anonymous
how do i get the st^2 to go away on the right?
amistre64
  • amistre64
\[\binom{\mu+1}{0}+\binom{\mu+1}{1}t+\binom{\mu+1}{2}t^2+\sum_{k=0}^{\mu+1}\binom{\mu+1}{k}t^k~>~1+(\mu+1)t+\mu t^2\] \[1+{\mu+1}t+\binom{\mu+1}{2}t^2+\sum_{k=0}^{\mu+1}\binom{\mu+1}{k}t^k~>~1+(\mu+1)t+\mu t^2\] \[\binom{\mu+1}{2}t^2+\sum_{k=0}^{\mu+1}\binom{\mu+1}{k}t^k~>~\mu t^2\] whats the right up for the (u+1 2), and how does it relate to u?
amistre64
  • amistre64
*the write up
anonymous
  • anonymous
(u+1,2)=(u+1)!/(2*(u-1)!) right?
amistre64
  • amistre64
\[\frac{(\mu+1)!}{(\mu+1-2)!}\] \[\frac{(\mu+1)!}{(\mu-1)!}\] \[\frac{(\mu+1)(\mu)\cancel{(\mu-1)!}}{\cancel{(\mu-1)!}}\] is... \[\mu(\mu+1)>\mu\]
anonymous
  • anonymous
i thought there would be a 2 at the bottom because of the k!(n-k)! at the bottom?
amistre64
  • amistre64
\[\binom {n}{k}=\frac{n!}{(n-k)!}\] \[\binom {\mu+1}{2}=\frac{(\mu+1)!}{(\mu+1-2)!}\]
amistre64
  • amistre64
\[\mu(\mu+1)>\mu\] \[\mu(\mu+1)-\mu>0\] \[\mu(\mu+1-1)>0\] \[\mu^2>0~:~for~all~\mu>1\]
anonymous
  • anonymous
\[\left(\begin{matrix}n \\ k\end{matrix}\right)\] is actually \[\frac{n!}{k!(n-k)!}\]in my textbook
amistre64
  • amistre64
.... 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
amistre64
  • amistre64
\[\frac12\mu(\mu+1)>\mu\]happier :)
anonymous
  • anonymous
ok i get that and what should i do next?
anonymous
  • anonymous
i got \[s^2-s>0\]
amistre64
  • amistre64
or written another way: s^2 > s isnt this true for all s>2 ?
anonymous
  • anonymous
yes so that means we prove the first part by induction?
amistre64
  • amistre64
it shows that the left side is bigger than the right side: (Bigger) t^2 + (more) > (smaller) t^2
amistre64
  • amistre64
yes, the first part is proved
anonymous
  • anonymous
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
amistre64
  • amistre64
at least I believe it is
amistre64
  • amistre64
the second one would follow the same concepts yes
anonymous
  • anonymous
thank you so much!
amistre64
  • amistre64
youre welcome :)

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