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can some one help me figure out those 2 problems ^
i think ur suppose to convert the summation to a natural log of a factorial
<p>well</p>
<p>i'm a high school student taking BC Calc.</p>
<p>i dont know how to solve it, i ahvent learned enough for this hardcore 5hit</p>
<p>well, at least i tried</p>
<p>BTW which college do u go to</p>
<p>hmm id try figuring out the top first and if the whole equation still turns out to be 0/0 use L'hopitals rule i think</p>
<p>i wanna say the first one goes to 1/2...?</p>
<p>well, maybe the top goes to 1/2....so then the whole thing goes to infinity? </p>
<p>gah, i'm tired and sick! my apologies haha</p>
<p>What calculus is this? It seems more like something off of the Putnam.</p>
<p>I get 1/2 for the first one and 1 for the second one...</p>
<p>I think we figured out the first one, so here is my explanation for number 2:</p>
<p>Since 2^0 = 1, I substitute 2^n in for k and eliminate the sigma notation. That leaves you with (log2 (sqrt(2*2^n-1)))/n</p>
<p>simplify to (log2(2^(n+1)-1))/n</p>
<p>plugging in zero results in 0/0, so use l'hopitals rule</p>
<p>the n on bottom goes to 1, and the top derives (according to my ti-89) to 2^n/(2*2^n-1)</p>
<p>Plugging in zero gives you one.</p>
<p>Edit: here is the first one</p>
<p>plug in 2^n for k and eliminate the sigma notation</p>
<p>this leaves you with (log2 (2^(n/2)))/n</p>
<p>using log properties simplify to (n/2 *log2(2))/n</p>
<p>log2(2) goes to 1 and the rest simplifies to 1/2</p>
<p>I hope those are right, who knows, i could be way off base.</p>
<p>Great explaination Kirbus. </p>
<p>I am trying to do the derivative on part B with my TI-89. How do you input a log base 2? I tried just to type in log2(8), but it doesn't equal 3 as expected. Instead the whole expression is just displayed for the answer.</p>
<p>Change the base to one the calculator can use eg. log2(8) = lg 8/lg 2 = 3.</p>
<p>edit: ^Beat me to it...</p>
<p>use the change of base formula to make it log(sqrt(2^(n+1)-1)/log (2)</p>
<p>NB. You can differentiate it by hand, in fact it may be quicker that way than inputting into the calculator.</p>
<p>And the 2nd one can be proven by integration, I think.</p>
<p>omfg i hope that's not on the bc exam</p>
<p>Edit: it cannot be proven by integration. The easiest way is the summation.</p>
<p>Alright guys thanks. It has been two years since I used that change of base formula. I guess you start to forget things if you don't use them.</p>