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#math
<Dacicus> Seems pretty easy, so I must be overlooking something
<andrew_> prove sum(i=0..n, (2i + 1)) = (n+1)^2
<andrew_> by induction
<andrew_> i get what im supposed to do here, but its not adding up
<TRWBW> andrew_: what's (n+1)^2?
<andrew_> and i know that; sum(i=0..k+1, (2i+1)) = sum(i=0..k, (2i+1)) + (k+1)
<TRWBW> andrew_: nope
<TRWBW> andrew_: it's + (2(k+1)+1)
<andrew_> ok, thats what the book says in front of me
<andrew_> with k+1 it doesnt touch the i
<TRWBW> andrew_: forget the book. do you understand induction?
<andrew_> basic induction, yes
<TRWBW> andrew_: and can you expand out (n+1)^2?
<andrew_> 2i+1 is part of the sum
<cyclicFifths> andrew_: just use the rule for the sum, you don't need the book to tell you that
<andrew_> n^2 + 2n + 1
<andrew_> easy enough
<TRWBW> andrew_: so (n+1)^2-n^2=?
<TRWBW> andrew_: kinda looks like 2i+1, doesn't it?
<andrew_> where did you derrive that from?
<TRWBW> andrew_: a little birdy told me.
<andrew_> :) cute
<andrew_> i cant see the correlation :/
<Olathe> Those birdies are annoying, always telling people things when they're trying to work.
<diseaser> "Use the definition of continuity and the properties of limits to show that the function f(x) = x*sqrt(16-x^2) is continuous on the interval [-4, 4]" ... I have used limit laws to rearrange it into: x * lim[x->a] 16 - lim[x->a] x^2 .. if I substitute a in there (kinda following along with the book) I end up with -a^2 + 16a .. but i'm not sure how to show that its actually continuous on that interval.. any suggestions?
<TRWBW> andrew_: i don't know how to make this any simpler. maybe try this, sum i=0..n 1 = (n+1), but iduction
<kaizoku> Any suggestions?
<Olathe> "...the same at any age" is a better way of putting it.
<kaizoku> I just can't seem to get it to be a regular exponential or logarithmic function so I can determine the Domain and Range.
<Olathe> You want to find the domain and range of f ?
<kaizoku> I want to know how to do that.
<Olathe> f(x) = log_3(x + 5) + log_3(-x)
<JIMJONESBALLIN> kaizoku what are you trying to do?
<Olathe> log_any accepts any and only positive arguments.
<kaizoku> I am trying to determine the domain and range of the functions.
<Olathe> So, when is x + 5 positive and when is -x positive ?
<Olathe> When they're both positive, f can work.
<JIMJONESBALLIN> So log is defined for what?
<kaizoku> So -5=< x <0?
<JuanTelez> hi everyone
<Olathe> kaizoku: Yep.
<Olathe> Only -5 < x
<kaizoku> Damnit!! It was that simple!!!
<Olathe> If it's -5 exactly, log gets a zero as input.
<kaizoku> Alright
<kaizoku> Thank you!!
<Olathe> No problem.
<kaizoku> I feel like a total idiot now..
<kaizoku> Oh, x can't be equal to -1 or -4, right?
<kaizoku> Because there is no log(1)?
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