EPTFree Math Help - Page 3
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Complete
United States1864 Posts
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krndandaman
Mozambique16569 Posts
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Sigh
Canada2433 Posts
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Dalguno
United States2446 Posts
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Trezeguet
United States2656 Posts
On September 03 2011 07:25 krndandaman wrote: im so tempted to give up on alot of my summer homework problems and just ask you math geniuses but i should legitly try at least an hour on a problem before giving up... will you guys be back tomorrow? i might have some questions i cant solve that i want to ask T.T Sure, post in this thread or skype me, or PM me and we can always call out for help to 343 if we need it. On September 03 2011 07:29 Sigh wrote: I'll definitely contact you if I have problems. I took calculus last year (gr 12) and i struggled heavily and i have to take one calculus course in my first year of university. Thanks a lot for this! I would be glad to help! On September 03 2011 08:51 Dalguno wrote: Oooh, this could help me greatly. I'm taking AP Calculus this year, and am not the best at math. Probably will be hitting you up That's what this is for, bring it on! | ||
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Trezeguet
United States2656 Posts
On September 03 2011 07:11 Complete wrote: Any statistics background? I can do intro stats stuff, it mostly has to do with confidence that I am giving good help that I don't really want to do more than intro. | ||
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Vlare
748 Posts
Pm me if you can do it ^_^ | ||
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boon2537
United States905 Posts
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krndandaman
Mozambique16569 Posts
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blah_blah
346 Posts
On September 03 2011 01:36 n.DieJokes wrote:I don't want the answer but if you have a hint or idea I'd like to see it. 3s are better than 2s, since 3+3 = 6 and 3^2=9, whereas 2+2+2=6 and 2^3=8. So you want to use as many 3s as you can. Using 333 3s leaves a 1 left over, which is no good. So use 332 3s (summing to 996) and either a 4 or two 2s. As for a proof, note that if you have any number N which is 5 or bigger, you can replace it by either two copies of N/2 (if N is even), or (N+1)/2 and (N-1)/2 (if N is odd). This leaves the sum invariant but increases the product (why?). So you have only 1s, 2s, 3s, or 4s. It's easy to see that 1s are bad (why?) Now use the above observation that 3s are better than 2s. If you have more than 3 2s or more than 2 4s, you can split them up into 3*3>2^3 or 3*3*2>4^2, which increases the product. So you can have at most 2 2s or 1 4. From here the answer (and indeed the general answer for any sum, not just 1000) follows immediately. e:f,b -- I guess I should read the thread through next time! | ||
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pyaar
United States423 Posts
Find an exact value for the sum of (-1)^n * 3n/2^(3n-1) from 1 to infinity. | ||
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blah_blah
346 Posts
On September 04 2011 10:43 pyaar wrote: I don't need help, but some of you mathy people might find this fun (or trivial?). This is a problem my BC teacher put on the sequence and series test last year that threw everyone off. I thought it was pretty tricky, here it is: Find an exact value for the sum of (-1)^n * 3n/2^(3n-1) from 1 to infinity. It's trivial if you know the the trick, at least. Let f(x) := \sum_{n\geq 0} x^n = 1/(1-x). Differentiating gives g(x) := f'(x) := \sum_{n\geq 1} nx^{n-1} = 1/(1-x)^2. Now, after a bit of algebra, your series is \sum_{n\geq 1} (-3/4)n(-1/8)^{n-1}, or (-3/4)g(-1/8), or -16/27. | ||
infinitestory
United States4053 Posts
On September 04 2011 10:04 krndandaman wrote: internet is not helping me at all with this... Given: f(x) = -x^2 -2x +3 find: f^1(-3) the inverse function thing I solved it, but idk if my answer is right. internet is not helping me at all. would the answer be -3,1 ? also, find: f^1(x) ; there is no inverse function at all right? lol summer hw is just basic calc intro review but I forgot everything You found f^-1(0). (you set -x^2 -2x +3 = 0 and took f^-1 of both sides) | ||
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krndandaman
Mozambique16569 Posts
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infinitestory
United States4053 Posts
On September 04 2011 13:25 krndandaman wrote: ah i did that because f(-3) = 0 did i make a huge mistake here? i totally forgot how to do this lol... The values of f^-1(-3) are the values of x such that f(x) = -3. To see why, just take x= f^-1(f(x)) = f^-1(-3). | ||
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krndandaman
Mozambique16569 Posts
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See.Blue
United States2673 Posts
On September 04 2011 01:12 Vlare wrote: How are you with partial derivatives? Pm me if you can do it ^_^ Post the problem! | ||
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See.Blue
United States2673 Posts
On September 04 2011 14:41 krndandaman wrote: I think I understood what you explained... I got an answer of 3 now. Is this right? If not I'm seriously lost and I'm surprised nothing on google explains this when I search "finding value of inverse function" Think of a function as 'mapping' point along the x-axis to a point along the y-axis. An inverse function is just the opposite, it takes points from along the y-axis and puts them somewhere along the x-axis. So finding f-inverse(-3) is just feeding y=-3 (because the inverse takes in y values) and seeing which points on along the x-axis they end up on (ie where the graph of f(x) intersects y = -3) which is all infinitestory said. | ||
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targ
Malaysia445 Posts
How do I get the integral of 5(1 + x^2)^1/2 ? I checked with my friend and Wolfram Alpha, and they both used trigonometric substitution. However since this question appeared the first chapter of integration, where trigonometric substitution was not taught yet, I feel that there might be an easier way to get the answer. Any ideas? | ||
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Adeny
Norway1233 Posts
I.E. how would you go about solving it step by step? | ||
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