<p>hi, I need some help with several problems related to sequences and series. If you are really good at them, please help me. all of them are old fr questions (i think).
<em>POSSIBLE</em> rewards (depends on how much u help me):
I have:
AP Chem MC: (1984, 1989, 1994, 1999,)(<---with answers), 2004 (questions, no answers yet)
AP CHEM FR: 1961-1999 Q's and answers
Ap Calc: 1997-2002 FR , a lot of old FR questions, a few FR answers.</p>
<p>or u dont want any rewards and you just want to do it to help me. either way is fine.
The answers MUST be right, or MOST of them should be right. I need them by tonight. </p>
<p>1)The power series x+ ((x^2)/2) + ((x^3)/3) + ...+ ((x^n)/n) converges if and only if.
Answer: -1 less than/equal to X < 1
Please explain.</p>
<p>2) When we use (e^x) (approximate symbol)(1+x+ ((x^2)/2) to esimate the squareroot of e, the remainder is no greater than
a. 0.021 b. 0.034 c. 0.042 d. 0.067 e. 0.742</p>
<p>3) Left f be the function f(x)= 1/(x-1) and g be the function g(x)= (-1)/((x-1)^2)
a. find a power series, centered at o, for f
b. for what values of x does the series in part a converge?
c. Determine a power series for g
d. for what values of x does the series in part c converge?</p>
<p>4) Left f be the function f(x)= (x-5)/(xsquare -x-2)
a. write the function using partial functions
b. find a power series, centered at o, for f
c. for what values of x does the series in part b converge?</p>
<p>5) Left f be the function given by f(x)= e^(x/2)
a. write the first 4 nonzero terms and the general term for the Taylor series expansion of f(x) about x=0
b. use the result from part a to write the first 3 nonzero terms and the general term of the series expansion about x=0 for g(x)= ((e^(x/2))-1) / X</p>
<p>does anyone know how to do them?
please. im really bad at series.</p>
<p>Okay here are the methods I use to solve these problems. They may not be the fastest but they definitely work. Hopefully these aren?t real questions because they are really difficult. I would really appreciate it if you could send me all of those past papers you listed because I?m self-studying all of those courses (+more) at a school which only offers the IB programmer (so I basically have 0 resources). Please send them as soon as you can.
1. For the first problem (x^n/n): (Difficulty=V.Hard; requires complete mastery of series)
First you should recognize the series as (x^n)<em>(1/n). 1/n is divergent, and therefore you should realize that the series is either conditionally convergent or divergent.
To find the convergence values (or if the series is divergent) we must use the D'Alembert's Ratio Test which is: Abs( Un+1/Un)<1 if convergent ---ABS=Absolute Value of
Using this formula:
Abs<a href="write%20this%20out%20so%20you%20understand">((x^n<em>x)/(n+1))</em>(n/x^n)</a>
Cancelling out gives: Abs: [x</em>(n/n+2)]<1 (1)
Now we must find out whether or not n/(n+2) converges or diverges. We can use the ratio test again to find the convergence of n/(n+2):
Un+1/Un?=[(n+1)(n+2)]/[(n+3)(n)] ; canceling out and expanding gives: (n2+3n+2)/(n2+3n) (lim n->∞ ). As n tends to infinity n/(n+2) tends to 1.
Going back to equation (1), this means that Abs (x)<1. Which means -1<x<1. to="" find="" out="" if="" 1="" or="" -1="" converge="" we="" put="" them="" back="" into="" the="" original="" equation:="" x^n="" n.="" x="1" then="" series="" becomes="" n="" which="" is="" divergent="" (you="" should="" memorize="" this)="" by="" integral="" test.="" but="" an="" alternating="" harmonic="" (1+-1="" 2+1="" 3+-1="" 4?)="" you="" as="" convergent="" (the="" because="" an+1<a="" for="" all="" and="" lim="" n-="">∞ an =0).
So in the end we have -1 is less than or equal to x<1
2. For the second question you must use the Lagrange error bound theorem (this is difficult), and you should realize that if e^x≈ 1+x+ (x^2)/2? then:
e^.5x≈1+x^.5+x/(2^0.5)
In this case x=1 so the power series becomes: e^.5≈1+0.5+1/(2^0.5)
The Lagrange formula is too difficult to type out so you must look in your text book or ask your math teacher. But if you plug in the correct parameters you will have an equation that is: Rn<max abs(1="" 8e^0.5="" 3!)="" which="" is="" equal="" to="" 0.034348?.="" answer="" b.="" 3.a.="" basically="" just="" write="" the="" (maclaurin)="" power="" series="" for="" 1="" (x-1).="" f(x)="(1-x)^-1" f(0)="1" f?(x)="-(1-x)^-2" f??(x)="2(1-x)^-3" etc?="" then="" becomes:="" 1+x+x2+="" x3+="" x4+="" x5+="" ?xn="" 3.b)="" since="" you="" know="" that="" x^n="" can="" use="" ratio="" test="" find="" its="" convergence.="" i="" won?t="" go="" over="" minor="" steps="" of="" because="" already="" explained="" them="" in="" an="" earlier="" question.="" using="" you?ll="" end="" up="" with="" abs(xnx="" xn="" )<1="" simplifies="" abs(x)<1;="" therefore="" -1<x<1.="" but="" we="" must="" remember="" back="" and="" plug="" x="1" oscillates="" between="" -1="" diverges="" (like="" sine="" curve)="" when="" exponential="" so="" as="" well.="" final="" solution="" (excuse="" pun)="" 3.c.="" )="" same="" idea="" question="" 3a.="" maclaurin="" series:="" written:="" (-1)-(2x)-(3x^2)-(4x^3)-?-nx^(n-1)="" 3.d.="" here="" 3b.="" basic="" algebra="" simplify:="" this="" case="" with:="" abs[x2*(n+1="" n)]<1.="" lim="" n-="">∞ of (n+1/n) is equal to 1. So therefore you have Abs(x2)<1; therefore -1<x<1. placing="" x="1" in="" the="" original="" equation,="" we="" see="" that="" series="" tends="" to="" negative="" infinity.="" if="" equation="" becomes="" a="" prime="" example="" of="" convergent="" alternating="" (you="" may="" apply="" test="" be="" doubly="" sure="" which="" is="" an+1<a="" for="" all="" n="" and="" lim="" n-="">∞ an =0).</x<1.></max></x<1.></p>
<ol>
<li>a) I not very sure but I would suppose that the question asks to rewrite it in terms of partial fractions (like what we do for integration).
So (x-5)/(x2-x-2) simplifies to (x-5)/[(x-2)(x+1)]. To write this in partial form we do the following:
(x-5)/[(x-2)(x+1)]=A/(x-2) +B/(x+1)
Cross multiplying we get:
x-5=A(x-1)+B(x-2).
When x=1, B=4
When x=2,A=-3.
We can then write the original function as: -3/(x-2)+4/(x-1).</li>
<li><p>b. We just write the power series for the above answer:
The power series for -3/(x-2) is : -1.5-(3x/4)-(3x^2/8)-(3x^3/16)?-3x^(n-1)/(2^n)
The power series for 4/(x-1) is: -4-(4x)-(4x^2)-(4x^3)?-4x^(n-1).
4.c. Sorry I?m not sure what to do from here. Maybe you have to add both series and then find the ratio test but then you?ll have to work through an algebraic mess. I think that I misunderstood the previous question and didn?t do ?partial functions? properly. Sorry about that. </p></li>
<li><p>Here you just find the Maclaurin series.
f(x)= e^(0.5x) f(0)= 1
f?(x)=0.5 e^(0.5x) f(0)=0.5
f??(x)= 0.25e^(0.5x) f(0)=0.25
f???(x)= 1/8e^(0.5x) f(0)=1/8
The first 4 non-zero terms are: 1, 0.5, 0.25, and 1/8
The Maclaurin series is: 1+0.5x+(0.25x^2)/2+(x^3)/(8<em>3!)+?+(x^n)/[(2^n)</em>n!]<br>
The general term is: (x^n)/[(2^n)*n!] </p></li>
<li><p>b. Just replace the Maclaurin series from above into the equation.
((e^(x/2))-1)/ x =>[(1+0.5x+(0.25x^2)/2+(x^3)/(8<em>3!))-1]/x
This becomes:
[0.5x+(0.25x^2)/2+(x^3)/(8</em>3!)]x
The first three terms therefore are: 0.5+(0.25x)/2+(x^2)/(8*3!)
The general term is found by inspection and is: [(0.5^n)(x^(n-1))]/n!
As a final tip I suggest you buy and learn how to use a Ti-89 which has some not very well known functions which can the Taylor series of any function, and it can also do partial fraction integration +many more things. A builder is only as good as his tools and the ti-89 is the best. Just my 2cents.</p></li>
</ol>
<p>Geez all that text looks hideous. Here is the link to the Word 2003 file which is much easier on the eyes:</p>
<p><a href="http://www.esnips.com/nsdoc/93a4544b-169d-40ec-9829-588365d3c7e0%5B/url%5D">http://www.esnips.com/nsdoc/93a4544b-169d-40ec-9829-588365d3c7e0</a></p>
<p>By the way I did a stupid mistake on question 1. There is a much easier way to solve:
Abs [x*(n/n+2)]<1 </p>
<p>Basically you know when n tends to infinity that n/(n+2) becomes 1.Now you simply replace n/n+2 with 1 in the equation and it simplifies to:</p>
<p>Abs(x*1)<1.</p>
<p>Its is much much faster than pointlessly doing the ratio test again
Everything else is perfect.</p>
<p>Also my email address is <a href="mailto:atomic1221@gmail.com">atomic1221@gmail.com</a><a href="which%20is%20where%20I'd%20like%20you%20to%20please%20send%20the%20exam%20papers">/email</a>. Thanks a million</p>
<p>Ok 1 u can solve so much easier
NO Tedious tests
1) The top term has to grow more slowly than bottom for it to converge.
So it must be between -1 and 1.
Using the alternating series tests you know that by -1 its less than or equal to and by the p series test by 1 its less than.</p>
<p>yea but on a free response question you'll recieve maybe 1 or 2 points if you don't use a "tedious test" and IF the examiner is feeling generous. In multiple choice doing mathwiz's method could be time saving (its actually a good idea, i never thought about it like that).</p>
<p>But on free response questions, trust me, do the tests.</p>
<p>thank you so much, atomicbomb22.
check your PM.</p>