Taylor series and MacLaurin series?

Question

If anyone can guide me on the following problems it would be great. Thanks!

1)A function f has a MacLaurin given by: (x^5)/2! + (x^6)/3!+...+[x^(n+4)]/ (n+1)!+..
what is the expression for f(x)?

2) The graph of function represented by MacLaurin 1 - x + (x^2)/2!-(x^3)/3!+...+[(( -1)^n)*x^n]/n!+... intersects the graph of y=2x^3 at x=?

3) When using e^x = 1+ x+(x^2)/2 to estimate sqrt(e), the Lagrange remainder is no greater than _____?

4) if the sum of [b(subscript n)]*x^n from n=0 to infitity is a tyalor seires the converges to f(x) for all real x, what is f'(1)?

5) What is the interval of convergence of the Taylor Series
f(x)= the sumer of [[(-1)^(k+1)]*(2x-1)^k]/k from (k=1) to infinity = (2x-1) - [(2x-1)^2]/2 + [(2x-1)^3]/3...on its interval of convergence

Answer 1

1) Note that the difference between the exponent and the factorial is 3: x^5 and 2!, x^6 and 3!, etc. So factor out an x^3:

x^3 [ x^2/2 + x^6/6 + ... ].

The part inside the brackets is the Maclaurin series for e^x, except for the first two terms, which are 1+x. Therefore, you have

x^3(e^x-1-x).

2) The function represented by the series is e^(-x), so you can use that to try to find the intersection. e^(-x)>0 for all x, but 2x^3 is only positive for positive x. Also, e^(-x)<1 for all positive x, but 2x^3 < 1 only for x < cbrt(1/2), so that gives you an idea of where the interesction is.

3) The Lagrange remainder for the nth-degree Taylor polynomial of f(x), centered at x=x0, is f^(n+1)(x*) / (n+1)! * (x-x0)^(n+1), where x* is between x and x0.

In this case, x0=0, f(x)=e^x, x=1/2, and n=2, so you have

e^(x*)/3!*(1/2)^3 = e^(x*)/48. But 0 < x* < 1/2, and e^x is increasing, so the largest the remainder can be is sqrt(e)/48.

4) Differentiate the series term-by-term and you get

sum n=1^infinity n[b_n]*x^(n-1).

Plug in x=1:

sum n=1^infinity n[b_n].

5) Ratio test gives you lim k->inf |2x-1||k/(k+1)| < 1 only if |2x-1|<1, so use that to obtain the interval.

Answer 2

a) first by potential of definition, the superb mclaurin sequence enhance for the functionality f(x) = sin(x) is ? ? ( -a million )^n * [ x^(2n+a million) / (2n+a million)! ] ..............yet how can we get this? because of the fact here: n=0 first, i'll define some words which includes n = the form of derivatives & form of factorials f^(n)(x) = is first unique and the derived equations. f^(n)(0) = plugging 0 contained in the unique equation and derived equations. an = is coefficient numbers of the variables. it truly is going to likely be AS table: n ---- f^(n)(x) ------f^n(0)-------an 0------sin(3x)---------0-----------(0)..... = 0 (0! is comparable to one ) a million-----3cos(3x)-------3-----------3/a million! = 3 2----(-9sin(3x)-------0-----------0/2! = 0 3---(-27cos(3x))---(-27)--------(-27/3..... 4---(81sin(3x)) ------0------------ 0/4! = 0.5---(243cos(3x))-----243--------(243/5..... now sin(3x) = 0 + 3*x^a million + 0 - [ 27*x^3/3! ] + 0 + [ 243*x^5/5! ] then it truly is going to likely be on condition that its alternating, and its increment for the two: the exponent of the variable and to the denominator (with factorials) are unusual huge form, it truly is going to likely be as 2n+a million yet while its increment even numbers, it truly is going to likely be 2n so the sequence is ? ? ( -a million )^n * [ 3^(2n+a million) * x^(2n+a million) / (2n+a million)! ] = 3*x^a million - [ 27*x^3/3! ] + [ 243*x^5/5! ] - .... n = 0 on condition that we've x * sin(3x), it truly is going to likely be because of the fact here: ? ? x * ( -a million )^n * [ 3^(2n+a million) * x^(2n+a million) / (2n+a million)! ] n = 0 ? ? ( -a million )^n * [ 3^(2n+a million) * x^(2n+2) / (2n+a million)! ] n = 0