<p>i believe that by plugging in (x-1)^2 it is already centered.</p>
<p>yeah - basically let “x = (x-1)^2”</p>
<p>or think of it this way:</p>
<p>let g(x) = e^x = 1 + x + x^2/2! + … + x^n/n! + …
so therefore g((x-1)^2) = e^(x-1)^2 = 1 + (x-1)^2 + (x-1)^4/2! + … + (x-1)^2n/n! + …</p>
<p>yup, stimulus we have the same idea</p>
<p>Thank god!!! I did it that way wahhooo</p>
<p>does anyone remember what c) asked?/</p>
<p>a) asked for e^(x-1)^2
b) asked for the actual f(x)
c)???
d) asked for POI…</p>
<p>If a series equals the same number for no matter what value you enter, does that make it convergent?</p>
<p>^ not sure, smart way of thinking though if teh steps are too complicated lol, guess and check</p>
<p>C) asked for interval of convergence</p>
<p>IOC for C) was (-oo,oo) right??</p>
<p>yes, unfortunately i put the radius of convergence as infinity, so i may lose one point. i may have explained it though and gotten full credit.</p>
<p>^radius is infinity, isn’t it? Just the interval is (-infinity, infinity)?</p>
<p>Vis a vis the evil FR number 6 (which will be the death of my score) I am 99% sure that you have to redevelop the series around 1 for part a. The original series was centered around 0, so if the second series was also centered around 0, then it would be fine to just plug in. Remember that when you do a taylor series, the coefficient is f^n(c)/n! If e^(x^2-1) (or any of its successive derivatives) have different values at 1 than at 0, then the Taylor polynomials of e^(x^2-1) at 0 and at 1 are not equal.</p>
<p>I don’t know anyone who got FR 6 correct. It was crazy evil. I’ve decided that calculus doesn’t exist and I’m not going to worry about it. I have an A in my BC class for the semester, my GPA is going up, whatevs to the college board.</p>
<p>If I put the interval of convergence as all real numbers, will I get credit?</p>
<p>I got -inf to inf with 1/n+1 as one of the factors, but i thought it had to be wrong so i just thought i messed it up and thus left it at 0<x<2.</p>
<p>i think i got d right though still, we’ve all agreed on no inflection points, right?</p>
<p>i can’t believe i stupidly did that, but i’m happy i somehow got all of part a and b haha, which is about 4 points.</p>
<p>If I put the interval as convergence as All Real Numbers, will I get credit? Or did I have to state from negative infinity to positive infinity?</p>
<p>i think all real numbers should give you full credit</p>
<p>e^x=1+x+ x^2/2 + x^3/3! + x^4/4!</p>
<p>a)find e^(x-1)^2 first 4 non 0 terms and general term</p>
<p>e^x around x=1 is </p>
<p>e^x=1+(x-1)+(x-1)^2/2 +(x-1)^3/3!</p>
<p>e^(x-1)^2 = 1 + [(x-1)^2-1] + [(x-1)^2-1]^2/2 + [(x-1)^2-1]^3/3! +….+ [(x-1)^2-1]^(n-1)/(n-1)!</p>
<p>b)f(x)=[e^(x-1)^2 -1]/(x-1)^2 four none zero and general term</p>
<p>e^(x-1)^2 = 1 + [(x-1)^2-1] + [(x-1)^2-1]^2/2 + [(x-1)^2-1]^3/3! +….+ [(x-1)^2-1]^n/n!</p>
<p>e^(x-1)^2-1 = [(x-1)^2-1] + [(x-1)^2-1]^2/2 + [(x-1)^2-1]^3/3! + [(x-1)^2-1]^4/4!</p>
<p>[e^(x-1)^2-1]/(x-1)^2 = 1/(x-1)^2 <[(x-1)^2-1] + [(x-1)^2-1]^2/2 + [(x-1)^2-1]^3/3! + [(x-1)^2-1]^4/4!></p>
<p>f(x)= [e^(x-1)^2-1]/(x-1)^2 = [(x-1)^2-1]/(x-1)^2 + [(x-1)^2-1]^2/2(x-1)^2 + [(x-1)^2-1]^3/3!(x-1)^2 + [(x-1)^2-1]^4/4!(x-1)^2 +…+ [(x-1)^2-1]^n/n!(x-1)^2</p>
<p>c) ratio test for radius of convergence</p>
<p>p= lim(n->inf) )ratio test<br>
p=lim(n -> inf) ratio test (a (n+1)/ a (n) =0 < 1 always
==> -inf<x<+inf</p>
<p>d) take 2nd derivative of f(x)</p>
<p>no inflection point because it always greater than 0</p>
<p>im goin to curse all of u who got FR#6 all correct
you are tremendously making the curve not as fun!</p>
<p>ctqctq is this your answer, or is this gained from an official source? </p>
<p>several students discussed this a couple of days ago and concluded that you substitute (x-1)^2 for x. </p>
<p>if you simplify your inside than you get (x^2 - 2x) which is not centered around 1 and it has x^2 which i have never seen. </p>
<p>by just sub. (x-1)^2 for x it automatically re-centers. </p>
<p>additionally, someone said they did it all out again and go the sub. answer.</p>
<p>Yeah, ctqctq’s solution is not correct for the taylor series expansion of e^(x-1)^2 because like hawkswim09 said, it automatically re-centers when you just sub in for x. I just confirmed it on my TI-89 which can do taylor series expansions. Just sub in (x-1)^2 for x. Yay for series manipulation.</p>
<p>Seeing all this makes me realize I should probably go with my first answer on every math test I take from here on out, because I put down my original thoughts for every single one, and then had to force myself not to change some of them when I checked over my work. </p>
<p>And I STILL changed the last one anyway because I TOTALLY had no POI (showed the ratio test et al, interval was neg. infinity to infinity) and then said it was at x=1. Grrrrrrrrrrrrr.</p>