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Featured 1 month ago

The **Taylor polynomial** is just another name for the full Taylor series truncated at a finite

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^((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

#=> (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 + . . . )#

Featured 2 months ago

The series:

is convergent for

The ratio test states that a necessary condition for a series

if

In our case:

Now we have that:

- For
#k= 1#

and the series is divergent.

- For
#k= 2#

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

We can note that the numerator has

Now as

So we have:

And as:

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

- For
#k >= 3#

and the series is convergent.

Featured 2 months ago

Maximum area is

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:

Where

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

As

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

And the total Area is that of a rectagle of widt

# A = 2 alpha beta #

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

# \ \ \ = 4 alpha cos alpha #

We now have the Area,

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

# \ \ \ \ \ \ = 4(cos alpha - alpha sin alpha )#

At a critical point we have

# 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

With this value of

# beta = 1.30436924 ... #

# A = 2.24438535 ... #

We can visually verify that this corresponds to a maximum by looking at the graph of

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

So maximum area is

Featured 1 month ago

At points

Slope of a tangent at a point on the curve

i.e.

=

and as we have to identify, where slope of tangent is

and as

or

or

i.e.

we have slope of tangent as

graph{(x^2y^2+xy-2)(x-y)=0 [-10, 10, -5, 5]}

Featured 3 weeks ago

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

#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

Note that

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

Now the Maclaurin expansion for

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

Equating the coefficient of

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

Multiplying both sides by

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

Featured 1 week ago

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

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,

**if and only if**

for every

for all

We have been asked to show that

So we want to make

We want:

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

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

#=abs(-2x-3)#

Recall that we control the size of

I see

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

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

In order to make this less than

**Proving our L is correct -- Writing the proof**

Claim:

Proof:

Given

Now if < delta# then

# = 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

So, by the definition of limit, we have

**Note**

This is an exampole of a limit in which the strict inequality

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