<p>The entire math department can't come up with the correct solution to this BC Calc question.</p>
<p>Let f be the function given by f(x)= e^(-2x^2) (read as e raised to the negative 2 x squared)</p>
<p>(a) Find the first four nonzero terms and the general term of the power seriews for f(x) about x=0.</p>
<p>(b) Find the interval of convergence of the power seriews for f(x) about x = 0. Show the analysis that leads to your conclusion.</p>
<p>(c) Let g be the function given by the sum of the first four nonzero terms of the power seriews for f(x) about x=0. Show that l f(x) - g(x) l < 0.02 for -0.6 less than or equal to x less than or equal to 0.6.</p>
<p>recall the power series expansion for e^x is 1 + x + x^2/2! etc etc etc</p>
<p>knowing this, and since we know d/dx e^x = e^x, simply substitute in -2x^2 for x in the expansion for e^x:</p>
<p>1 + (-2x^2) + (-2x^2)^2/2!...-(2x^2)^n/n!</p>
<p>PART B:</p>
<p>we know the general term of the series is sum{from 0 to inf}[-(2x^2)^n/n!]. since in the exponent, we have n, we can use the ratio test, where we find the limit of a<em>n+1/a</em>n as n-->inf:</p>
<p>((-2x^2)^n+1)/(n+1)! * n!/((-2x^2)^n) = (2x^2)/(n+1), the limit of which (as n--> inf) is 0. from the defenition of the ratio test, we know that if the limit = 0, the interval of convergence is all real numbers.</p>
<p>PART C:</p>
<p>we know the maximum error must be less than the first unused term of the expansion, or (-2x^2)^4/4!, which with simplification is equivalent to (2x^8)/3. knowing the max bounds for x are -.6 and .6, and since our power is even, we can substitute .6 in for x, and we get .0119 as the difference. since .0119 < .02, we have shown that |f(x) - g(x)| < .02.</p>
<p>hope this helps:) (and if anyone spots a mistake, please let me know/correct it)</p>
<p>For the record, it's pretty bad form to post an essentially trivial question with the grandiose subject line you used. Many people who would otherwise have been glad to help are turned off by this sort of silliness.</p>
<p>It's also really bad form to spam multiple forums with off-topic posts. CC is not a homework help site at all, so the post is really off-topic everywhere, but especially on the MIT and Caltech forums, where most readers are able to do such problems. :)</p>
<p>freebird raises a good point. if the entire math faculty couldn't do this problem, i'd be worried about more than just a simple series problem...</p>
<p>still...series is a relatively simple calc topic (simple compared to say, stuff that comes later), and i'd assume all math teachers have encountered it before...</p>
<p>sure thing, freeburd...i was just amazed since the math teachers i've had have all been excellent, both in terms of quality of instruction as well as training.:)</p>