Methods To Solve Problems

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)*(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[((x^n*x)/(n+1))*(n/x^n)]

(write this out so you understand)

Cancelling out gives: Abs: [x*(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. If x=1 then the series becomes 1/n which is divergent (you should memorize this) by the integral test. BUT if x=-1 then series becomes an alternating harmonic series (1+-1/2+1/3+-1/4…) which you should memorize as convergent (the alternating harmonic series is convergent because an+1∞ 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…. which is 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(0)=1

f’’(x)=2(1-x)^-3 f(0)=2 etc… The series then becomes: 1+x+x2+ x3+ x4+ x5+ …xn 3.b) Since you know that series is x^n then you can use the ratio test to find its convergence. I won’t go over the minor steps of the ratio test because I already explained them in an earlier question. Using the ratio test you’ll end up with Abs(xnx/xn )<1 which simplifies to Abs(x)<1; therefore -1<x<1. BUT we must remember to go back and plug in for x=1 and x=1. We then find that x=-1 oscillates between -1 and 1 and therefore diverges (like the sine curve) and when x=1 the series is exponential and so diverges as well. Therefore the final solution (excuse the pun) is -1<x<1. 3.c. ) Use the same idea as question 3a. Just write the Maclaurin series: f(x)=-(x-1)-2

f(0)=-1

f’(x)=2(x-1)-3

f(0)=-2

f’’(x)=-6(x-1)-4

f(0)=-(3!)

etc… The Maclaurin is then written: (-1)-(2x)-(3x^2)-(4x^3)-…-nx^(n-1) 3.d. Same idea here as 3b. Just use the ratio test and basic algebra to simplify: In this case you’ll end up 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 the series tends to negative infinity. If x=-1 in the original equation the series becomes a prime example of a convergent alternating series (you may apply the alternating series test to be doubly sure which is an+1∞ an =0).

4. 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). 4. 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.

5. 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*3!)+…+(x^n)/[(2^n)*n!] The general term is: (x^n)/[(2^n)*n!]

5. 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*3!))-1]/x This becomes: [0.5x+(0.25x^2)/2+(x^3)/(8*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.