Asukers Posted June 1, 2011 Report Share Posted June 1, 2011 Just wanted to say a big THANKS to everyone here, I got 18/20 on this IA with your help! 1 Reply Link to post Share on other sites More sharing options...
dessskris Posted June 10, 2011 Report Share Posted June 10, 2011 what do you mean? the IA of this task or of an old task other than this one? Reply Link to post Share on other sites More sharing options...
sdshjsjasj Posted June 10, 2011 Report Share Posted June 10, 2011 the ia of this task. the same task has been given 2 years in a row. so the people who graduated this year would have already done the same task hat IB 1s are getting this year. Reply Link to post Share on other sites More sharing options...
dessskris Posted June 10, 2011 Report Share Posted June 10, 2011 oh, sorry you cannot get it then. because it = cheating. Reply Link to post Share on other sites More sharing options...
Ezeh Posted June 10, 2011 Report Share Posted June 10, 2011 we have the same IA that was given last year. if anyone couullld pleeaassseeeeeee give me a sample IA that they did last year it would be awesome thxx There's a fine line between asking for help and asking for the answers. 1 Reply Link to post Share on other sites More sharing options...
dessskris Posted June 10, 2011 Report Share Posted June 10, 2011 I can help you but not give you answers. Reply Link to post Share on other sites More sharing options...
Guest 1529 Posted June 10, 2011 Report Share Posted June 10, 2011 Just wanted to say a big THANKS to everyone here, I got 18/20 on this IA with your help! You're welcome (LOOOL) Reply Link to post Share on other sites More sharing options...
snmatg Posted June 14, 2011 Report Share Posted June 14, 2011 Sn=a^x meaning if a=3 and x=3 the sum is 27. You get this from the property of logarithms where you pull out the exponent. So from t1=(x ln a)/1, the x is just pulled out from it's actual exponential spot. 3 ln 3= ln 3^3 can you explaın how can you fınd this? ı dont understand.please help me Reply Link to post Share on other sites More sharing options...
dessskris Posted June 15, 2011 Report Share Posted June 15, 2011 read it carefully. it's basicly saying that3ln(3) = ln(3^3)and you can use this to "manipulate" t and S Reply Link to post Share on other sites More sharing options...
sdshjsjasj Posted June 16, 2011 Report Share Posted June 16, 2011 umm.. for the first task when x=1 and we are meant to find S10 for various positive values of a. i tried a=0.02. the values i get for Sn are alternating positive and negative values. for eg. for S2 i get..2.343 but then for S3 i get something like -0.288. the graph is not exponential but zigzag however the values eventually do lead up to 0.02. is that normal? or am i doing something wrong? Reply Link to post Share on other sites More sharing options...
dessskris Posted June 17, 2011 Report Share Posted June 17, 2011 smart. this is actually your limitation. when a=0.02 you don't find a pattern similar to previous, which means a cannot be 0.02 in this task. then generalise the limitation if you know what I mean. do you know what I mean? Reply Link to post Share on other sites More sharing options...
sdshjsjasj Posted June 17, 2011 Report Share Posted June 17, 2011 no, i dont get what you mean by generalize the limitation :/alsoo for x=1 and a=2, sor S10 i get exactly 2. but isnt it not meant to touch 2 since 2 is an asymptote?? and for S11 and 12 it goes above 2. like 2.0000004 ..etc. Reply Link to post Share on other sites More sharing options...
dessskris Posted June 17, 2011 Report Share Posted June 17, 2011 -_____- haha if it doesn't work with 0.02, does it work with 0.56, 3.25, 10.68, etc etc? you know what I mean now?and actually it's meant to be 2. it's not really an asymptote :/ Reply Link to post Share on other sites More sharing options...
sdshjsjasj Posted June 17, 2011 Report Share Posted June 17, 2011 yeahh i get it thxx umm some people were actually saying that that is not a limitation because it still complies with the general statement that we are trying to find. that as n approaches infinity the value of Sn = a. s is it still a limitation?? smart. this is actually your limitation. when a=0.02 you don't find a pattern similar to previous, which means a cannot be 0.02 in this task. then generalise the limitation if you know what I mean. do you know what I mean? If one works out T9(0.2,1). T9=0.2, which proves the general statement of a^x. It is true that it has a different pattern but it does approach the general statement. so why is it than a limitation? and any ideas about the scope of the general statement? what does that even mean? Reply Link to post Share on other sites More sharing options...
dessskris Posted June 18, 2011 Report Share Posted June 18, 2011 ughh you were the one who told me that it didn't work with 0.02 fine then it's not the limitation hahahaha sorry! I think the scope is actually similar to a limitation... Reply Link to post Share on other sites More sharing options...
BAWAWI Posted June 18, 2011 Report Share Posted June 18, 2011 I need some help with defining Tn (a,x) as the sum of the first n terms? Reply Link to post Share on other sites More sharing options...
dessskris Posted June 18, 2011 Report Share Posted June 18, 2011 nah it's just saying that the notation Tn means the sum of the first n terms Reply Link to post Share on other sites More sharing options...
omgpop Posted June 24, 2011 Report Share Posted June 24, 2011 OK, so I've read through this thread trying to figure out how to find the both general statements required in this portfolio. I know what they both are, but I did it through just examining my data... It was fairly obvious when I looked at it. I keep reading about the Taylor Series and derivatives and stuff, but we haven't done calculus yet, so it makes no sense to me (at least I think that stuff is calculus). Like, for the first general statement on the first page of the portfolio, when x has been kept at 1, I know the general statement (not 100% clear if I'm just allowed to post the general statements or not...), but I did it because I just looked at the asympototes the graph produced, and what number I kept consistently getting when I calculated S10 I knew what the general statemtent was, but how do I mathematically prove it without just reasoning it or calculus... If I can post the general statements, let me know so my question can be more clear... Reply Link to post Share on other sites More sharing options...
dessskris Posted June 24, 2011 Report Share Posted June 24, 2011 hmmm I've tried to work out so many IAs that I forgot which one needs what method, but if I remember correctly, this task should not require calculus :/ actually you are not allowed to post any answer (which includes general statements) here but some people still did haha if you like you can PM me or anybody whom you think could help you. actually there are various methods, depending on how you see it. I'd say the easiest is the graphical method (plotting your data and finding the best regression). or, as you might have guessed, using that Taylor thing. Reply Link to post Share on other sites More sharing options...
omgpop Posted June 24, 2011 Report Share Posted June 24, 2011 OK, so I can just use the Taylor Series to calculate the first general statement... Can I just use the Taylor Series for the second one they ask on the second page? They ask for two; one for when x=1 and the other for when x can be anything. I see how I can use the Taylor Series to find the general statement when x=1, but I'm not so sure how to use it for the other one. Also, we were never actually taught the Taylor Series in our class; is it OK for me to just say the series given in this portfolio is an example of the Taylor Series, and then provide a footnote for where I got the Taylor Series from, and then solve for the general statements?Any help is much appreciated. Reply Link to post Share on other sites More sharing options...
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