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The Taylor polynomial is just another name for the full Taylor series truncated at a finite #n#. In other words, it is a partial Taylor series (i.e. one we could write down in a reasonable amount of time).

Some common errors are:

  • Letting #x = a# within the #(x - a)^n# term.
  • Taking the derivatives at the same time as writing out the series, which is not necessarily wrong, but in my opinion, it allows more room for silly mistakes.
  • Not plugging #a# in for the #n#th derivative, or plugging in #0#.

The formula to write out the series was:

#sum_(n=0)^(oo) (f^((n))(a))/(n!)(x-a)^n#

So, we would have to take some derivatives of #f#.

#f^((0))(x) = f(x) = x/(1 + x)#

#f'(x) = x*(-(1 + x)^(-2)) + 1/(1+x)#

#= -x/(1 + x)^2 + 1/(1+x)#

#= -x/(1 + x)^2 + (1 + x)/(1+x)^2#

#= 1/(1+x)^2#

#f''(x) = -2(1 + x)^3 = -2/(1+x)^3#

#f'''(x) = 6/(1 + x)^4#

#f''''(x) = -24/(1 + x)^5#

etc.

So plugging things in gives (truncated at #n = 4#):

#=> (f^((0))(a))/(0!)(x-a)^0 + (f'(a))/(1!)(x-a)^1 + (f''(a))/(2!)(x-a)^2 + (f'''(a))/(3!)(x-a)^3 + (f''''(a))/(4!)(x-a)^4 + . . . #

#= a/(1 + a) + 1/(1+a)^2(x-a) + (-2/(1+a)^3)/(2)(x-a)^2 + (6/(1+a)^4)/(6)(x-a)^3 + (-24/(1 + a)^5)/(24)(x-a)^4 + . . . #

#= color(blue)(a/(1 + a) + 1/(1+a)^2(x-a) - 1/(1+a)^3(x-a)^2 + 1/(1+a)^4(x-a)^3 - 1/(1 + a)^5(x-a)^4 + . . . )#

Answer:

The series:

#sum_(n=1)^oo ((n!)^2)/((kn)!)#

is convergent for #k>=2#

Explanation:

The ratio test states that a necessary condition for a series #sum_(n=1)^oo a_n# to converge is that:

#L= lim_(n->oo) abs (a_(n+1)/a_n) <= 1#

if # L < 1# the condition is also sufficient.

In our case:

#abs (a_(n+1)/a_n) = ((((n+1)!)^2)/((k(n+1))!))/(((n!)^2)/((kn)!)) = (((n+1)!)^2)/((n!)^2) ((kn)!)/ ((kn+k)!) = (n+1)^2/((kn+k) (kn+k-1)...(kn+1))#

Now we have that:

  1. For #k= 1#

#lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) (n+1)^2/(n+1) = lim_(n->oo) (n+1) = +oo#

and the series is divergent.

  1. For #k= 2#

#lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) (n+1)^2/((n+2)(n+1)) = 1#

and the test is inconclusive, so we have to look at the series in more detail:

#sum_(n=1)^oo ((n!)^2)/((2n)!) = (n(n-1)(n-2)...2*1)/ (2n(2n-1)...(n+1)#

We can note that the numerator has #n# factors from #1# to #n# and the denominator has #n# factors from #n+1# to #2n#, so ordering them appropriately we have:

# (n(n-1)(n-2)...2)/ (2n(2n-1)...(n+1)) = prod_(q=1)^n q/(n+q)#

Now as #q <= n#,

#q/(n+q) <= q/(q+q) = 1/2#

So we have:

# (n(n-1)(n-2)...2)/ (2n(2n-1)...(n+1)) <= (1/2)^n#

And as:

#sum_(n=1)^oo (1/2)^n = 1#

is convergent, then also our series is convergent by direct comparison.

  1. For #k >= 3#

#lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) (n+1)^2/((n+1)(n+2)...(n+k)) = 0#

and the series is convergent.

Answer:

Maximum area is #2.244 " unit"^2# (3dp)

Explanation:

I assume that you man bounded by the x-axis also, otherwise the largest rectangle would be unbounded and therefore infinite.

This is a diagram depicting the problem:

enter image source here

Where #P(alpha,beta)# is the point in Quadrant 1 where the rectangle intersects the curve #y=2cosx#, and #P'(-alpha,beta)# is the corresponding point in quadrant 2.

Let us set up the following variables:

# { (alpha, x"-coordinate of point "P), (beta, y"-coordinate of point "P),(A, "Total Area of inscribed rectangle") :} #

Our aim is to find #A#, as a function of a single variable and to maximize the total area, #A# (It won't matter which variable we do this with as we will get the same result). ie we want a critical point of #A# wrt the variable.

As #P# lies on the curve #y=2cosx#, we have:

# beta = 2cos alpha \ \ \ \ \ ..... [1]#

And the total Area is that of a rectagle of widt #2alpha# and height #beta#, so:

# A = 2 alpha beta #
# \ \ \ = 2 alpha (2cos alpha) \ \ \ \ \# (from [1] )
# \ \ \ = 4 alpha cos alpha #

We now have the Area, #A#, as a function of a single variable #alpha#, so differentiating wrt #alpha# (using the product rule) we get:

# (dA)/(d alpha) = (4alpha)(-sin alpha) + (4)(cos alpha) #
# \ \ \ \ \ \ = 4(cos alpha - alpha sin alpha )#

At a critical point we have #(dA)/(d alpha) =0 => #

# 4(cos alpha - alpha sin alpha ) = 0 #
# :. cos alpha - alpha sin alpha = 0 #

In order to solve this equation we use Newton-Rhapson which gives #alpha =0.860333589 ...#

With this value of #alpha# we have:

# beta = 1.30436924 ... #
# A = 2.24438535 ... #

We can visually verify that this corresponds to a maximum by looking at the graph of #y=A(alpha)#:

graph{4xcosx [-4, 4, -5.5, 5.5]}

So maximum area is #2.244 " unit"^2# (3dp)

Answer:

At points #(1,1)# and #-1.-1)#

Explanation:

Slope of a tangent at a point on the curve #x^2y^2+xy=2# will be given by the value of #(dy)/(dx)# at that point. So let us find its derivative, which is given by,

#2x xxy^2+x^2xx2yxx(dy)/(dx)+1xxy+x xx(dy)/(dx)=0#

i.e. #(dy)/(dx)[2x^2y+x]=-2xy^2-y# and

#(dy)/(dx)=-(2xy^2+y)/(2x^2y+x)#

= #-(y(2xy+1))/(x(2xy+1))=-y/x#

and as we have to identify, where slope of tangent is #-1#, we have solution #y=x#

and as #x^2y^2+xy=2#, this is at

#x^4+x^2-2=0#

or #(x^2-1)(x^2+2)=0#

or #x^2-1=0#

i.e. #x=+-1# and as #y=x#

we have slope of tangent as #-1# at #(1,1)# and #(-1,-1)#
graph{(x^2y^2+xy-2)(x-y)=0 [-10, 10, -5, 5]}

Answer:

#sum_(n=1)^oo 1/n^2 = pi^2/6#

Explanation:

Finding the sum of this series is the "Basel Problem", first posed in 1644 by Pietro Mengoli and solved by Leonhard Euler in 1734.

My favourite way of looking at it is to consider the function:

#sin(x)/x#

Note that:

#lim_(x->0) sin(x)/x = 1#

and when #n# is any non-zero integer:

#sin(npi)/(npi) = 0/(npi) = 0#

Consider the function:

#f(x) = prod_(n in ZZ, n != 0) (1-x/(npi))#

#color(white)(f(x)) = prod_(n = 1)^oo (1-x/(npi))(1+x/(npi))#

#color(white)(f(x)) = prod_(n = 1)^oo (1-x^2/(n^2pi^2))#

#color(white)(f(x)) = 1-1/(pi^2)(sum_(n=1)^oo 1/n^2)x^2+O(x^4)#

From its definition, we find:

#{(f(0) = 1), (f(npi) = 0 " for " n in ZZ " with " n != 0) :}#

We can tell that #f(x)# is well defined for any real value of #x# since given any #x# then for all #n > abs(x)/pi#, we have #0 <= (1-x^2/(n^2pi^2)) < 1#. So the product definition converges and so does the Maclaurin expansion.

Note that #f(x)# matches #sin(x)/x# as #x->0# and for #x = npi#.

Hence by the Weierstrass Factorisation Theorem, they are equal functions.

Now the Maclaurin expansion for #sin(x)/x# is:

#1/(1!)-x^2/(3!)+x^4/(5!)-x^6/(7!)+... = 1-x^2/6+O(x^4)#

Equating the coefficient of #x^2# for this expansion and the one we found for #f(x)#, we find:

#1/6 = 1/pi^2 sum_(n=1)^oo 1/n^2#

Multiplying both sides by #pi^2# and transposing, we find:

#sum_(n=1)^oo 1/n^2 = pi^2/6#

Answer:

Please see below. Warning: long answer due to explanatory analysis.

Explanation:

The explanation has two sections. There is a preliminary analysis to find the values used in the proof, then there is a presentation of the proof itself.

Finding the proof

By definition,

#lim_(xrarrcolor(green)(a))color(red)(f(x)) = color(blue)(L)# if and only if

for every #epsilon > 0#, there is a #delta > 0# such that:
for all #x#, #" "# if #0 < abs(x-color(green)(a)) < delta#, then #abs(color(red)(f(x))-color(blue)(L)) < epsilon#.

We have been asked to show that

#lim_(xrarrcolor(green)(-1.5))color(red)((9-4x^2)/(3+2x) = color(blue)(6)#

So we want to make #abs(underbrace(color(red)((9-4x^2)/(3+2x) ))_(color(red)(f(x)) )-underbrace(color(blue)(6))_color(blue)(L))# less than some given #epsilon# and we control (through our control of #delta#) the size of #abs(x-underbrace(color(green)((-1.5))))_color(green)(a))#

We want: #abs((9-4x^2)/(3+2x) )-6) < epsilon#

Look at the thing we want to make small. Rewrite this, looking for the thing we control.

#abs((9-4x^2)/(3+2x) - 6) = abs(((3-2x)(3+2x))/(3+2x) - 6)#

# = abs(((3-2x) - 6)#

#=abs(-2x-3)#

Recall that we control the size of #abs(x-(-1.5)) = abs(x-(-3/2))#

I see #abs(x+3# which is the same as #abs(x-(-3))# so let's 'factor out #-2#.

# = abs(-2)abs(x+3/2)#

# = 2 abs(x-(-3/2))#

In order to make this less than #epsilon#, it suffices to make #abs(x-(-1.5))# less than #epsi/2#

Proving our L is correct -- Writing the proof

Claim: #lim_(xrarr-1.5)((9-4x^2)/(3+2x) ) = 6#

Proof:

Given #epsilon > 0#, choose #delta = epsilon/2#. (Note that #delta# is positive.)

Now if < delta# then

#abs((9-4x^2)/(3+2x) -6) = abs(((3-2x)(3+2x))/(3+2x) - 6)#

# = abs(((3-2x) - 6)#

#=abs(-2x-3)#

# = abs(-2)abs(x+3/2)#

# = 2 abs(x-(-3/2))#

# < 2 delta#

# = 2 (epsi/2)#

# = epsilon#

We have shown that for any positive #epsilon#, there is a positive #delta# such that for all #x#, if #0 < abs(x-(-1.5)) < delta#, then #abs((9-4x^2)/(3+2x) -6) < epsilon#.

So, by the definition of limit, we have #lim_(xrarr-1.5)((9-4x^2)/(3+2x) ) = 6#.

Note

This is an exampole of a limit in which the strict inequality #0 < abs(x-delta)# is very important. If we allowed #0 = abs(x-(-1.5))#, then a choice of #x = -1.5# would result in an undefined expression. #abs((9-4x^2)/(3+2x) -6) #.

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