<p>marcus you subtract by 1 before dividing so if</p>
<p>e^(x-1)^2 = 1 + (x-1)^2 + (x-1)^4/2! +… e^(x-1)^2n/n!
then e^(x -1)^2 -1 = (x-1)^2 + (x-1)^4/2! + … e^(x-1)^2n/(n+1)!</p>
<p>Actually, sorry, I had it wrong. Isn’t it ((((x-1)^2) -1)^n) / n! That’s plugging in (x-1)^2 for every x then subtracting 1 cuz centered at 1, then to the n power</p>
<p>That’s part A</p>
<p>the whole e^(x-1)^2 is simply the coefficient. whatever the coefficient is there will only be (x-1)^2n as the x part of the equation b/c that is centered around 1, your answer has a squared x, and you only want a linear equation involving x.</p>
<p>Wait, how is that the coefficient? And what did you do with the centered at x = 1 part?</p>
<p>marcus you don’t need to worry about being centered at 1 for anything other than the interval of convergence part</p>
<p>and the coefficient is just the non-(x-1)^2n part which is 1/(n+1)!</p>
<p>^ Yea I get the coefficient, but don’t you have to plug in - 1 where there is an x for every term? Since that’s how you get the Taylor.</p>
<p>i remember proving in my class that if the maclaurin series for e^x^2 was plugging in x^2 for all the x’s. the reason why it is still a maclaurin is that it has just the x. when you substitute an (x-1) or (x-1)^2 for all x’s, now you are making it centered at the same time.</p>
<p>f = e^(x-1)^2.</p>
<p>taylor is just f^(n) (a) (x -a)^n all over (n!)</p>
<p>It doesn’t matter what the function is, since it will just provide a value for f. </p>
<p>It’s still going to be (x - a)^n or in this case (x - 1)^n.</p>
<p>O whattt, I didn’t know the function would automatically center itself. How do you know that?</p>
<p>exactly, that is why you can’t use a squared x inside of the final ^n</p>
<p>YEs, Xav, thats what I’m saying</p>
<p>i actually did not know on the test, but looking back it does make sense. all you need is just an (x-1)^n and it is centered at 1</p>
<p>That’s why I said ((((x-1)^2) -1)^n) / n!</p>
<p>x = (x-1)^2</p>
<p>but if you expand out you get (x^2-2x)^n/n!</p>
<p>you don’t have the x-1 and you have a square of x. once you simplify you need to have the x-1</p>
<p>marcusf329, we did the same thing haha, though i was a bit suspicious that it might be self centered at 1 already ~</p>
<p>hawk, yea i see how you get that, but how does that simplify to (x-1)^2n? I just left it as is cuz i know we don’t have to simplify on the AP tests, like you could leave a term as 2(x)^2 / 2!, you wouldn’t have to simplify to x^2</p>
<p>If I like bombed the FRQ but did decently on the MCs, how hard would it be for me to get a 3 overall, or just on AB?</p>
<p>to be completely honest, i am not 100% sure you just make the sub into the e^x, there may be additional coefficients, but if you do just sub in then it goes as follows:</p>
<p>e^x=1+x+x^2/2+x^3/3!</p>
<p>so </p>
<p>e^(x-1)^2 = 1 + (x-1)^2 +(x-1)^4/2+(x-1)^6/3!</p>
<p>since the first term is n=0, the second is n=1 so you need (x-1)^2n/n!</p>
<p>for aigiquinf, it will be close. if you got 60% of MC you get around 30 raw points. if you got 20% of FR you get about 11 points which is a 41 raw which i think will put you at a 3, but it will be close, depending on the scale.</p>
<p>^ Yea, I’m pretty sure you just plug in, don’t worry about about coefficients I’m pretty sure</p>
<p>and you still need to center it…</p>