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OpenStudy (anonymous):
whats up?
OpenStudy (anonymous):
just give me a moment let me screen shot it
OpenStudy (anonymous):
OpenStudy (anonymous):
question 3 and 4
OpenStudy (anonymous):
o man this will be hard to show on this haha
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OpenStudy (anonymous):
@ganeshie8 @zepdrix
OpenStudy (anonymous):
@dashdg oh man just try
OpenStudy (anonymous):
3. have you proved doing induction?
OpenStudy (anonymous):
i haven tried anything as of now
OpenStudy (anonymous):
what class is this for?
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OpenStudy (anonymous):
advanced cacl bro
OpenStudy (anonymous):
is that after multivariable? i am a senior math major ha
OpenStudy (anonymous):
its level 534 and it comes after calc 3
OpenStudy (anonymous):
advanced calculus 1
OpenStudy (anonymous):
is it like real analysis?
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OpenStudy (anonymous):
have you dealt with Cauchy squeneces?
OpenStudy (anonymous):
am not a major in math , i major in electrical .. yes its like real analysis
OpenStudy (anonymous):
yes we have
OpenStudy (anonymous):
Do you know bernoulli's inequality by any chance?
OpenStudy (anonymous):
no not off hand
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OpenStudy (anonymous):
whats it say?
OpenStudy (anonymous):
no bernoullis equality bro
OpenStudy (anonymous):
If $$r>1,x\ge -1,$$ then: $$(1+x)^r>1+rx$$
OpenStudy (anonymous):
ahhhh, BOOOOO. Your problem is an immediate consequence =/
OpenStudy (anonymous):
i am going to use a form of proof called induction:
base case a^1>1+1(a-1)
a^1>1+a-1
a^1> a checks for all a>1 so it checks
Suppose it works for k^n>1+n(k-1)
prove for k+1....
(K+1)^n>1+n(k+1-1)
K+1^n>1+n*k
because of the inequality it checks
qed
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OpenStudy (anonymous):
thanks my genius boys..
OpenStudy (anonymous):
i have plenty more questions as wel
OpenStudy (anonymous):
im sorry i am going to need to get some sleep i have an 8am class tomorrow
OpenStudy (anonymous):
alright thanks @dashdg
OpenStudy (anonymous):
@hartnn @campbell_st @mathstudent55
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