# 5.6.E: Problems on Tayior's Theorem - Mathematics

Exercise (PageIndex{1})

Complete the proofs of Theorems (1,1^{prime},) and 2.

Exercise (PageIndex{2})

Verify Note 1 and Examples (b) and (left(mathrm{b}^{prime prime} ight)).

Exercise (PageIndex{3})

Taking (g(t)=(a-t)^{s}, s>0,) in ((6),) find
[
R_{n}=frac{f^{(n+1)}(q)}{n ! s}(x-p)^{s}(x-q)^{n+1-s} quad( ext { Schloemilch-Roche remainder }).
]
Obtain (left(5^{prime} ight)) and (left(5^{prime prime} ight)) from it.

Exercise (PageIndex{4})

Prove that (P_{n}) (as defined) is the only polynomial of degree (n) such that
[
f^{(k)}(p)=P_{n}^{(k)}(p), quad k=0,1, ldots, n .
]
[Hint: Differentiate (P_{n} n) times to verify that it satisfies this property.
For uniqueness, suppose this also holds for
[
P(x)=sum_{k=0}^{n} a_{k}(x-p)^{k} .
]
Differentiate (P n) times to show that
[
P^{(k)}(p)=f^{(k)}(p)=a_{k} k ! ,
]
(left. ext { so } P=P_{n} . ext { (Why?) } ight])

Exercise (PageIndex{5})

With (P_{n}) as defined, prove that if (f) is (n) times differentiable at (p,) then
[
f(x)-P_{n}(x)=oleft((x-p)^{n} ight) ext { as } x ightarrow p
]
(Taylor's theorem with Peano remainder term).
[Hint: Let (R(x)=f(x)-P_{n}(x)) and
[
delta(x)=frac{R(x)}{(x-p)^{n}} ext { with } delta(p)=0 .
]
Using the "simplified" L'Hôpital rule (Problem 3 in (§3) ) repeatedly (n) times, prove (left. ext { that } lim _{x ightarrow p} delta(x)=0 . ight])

Exercise (PageIndex{6})

Use Theorem (1^{prime}) with (p=0) to verify the following expansions, and prove that (lim _{n ightarrow infty} R_{n}=0).
(a) (sin x=x-frac{x^{3}}{3 !}+frac{x^{5}}{5 !}-cdots-frac{(-1)^{m} x^{2 m-1}}{(2 m-1) !}+frac{(-1)^{m} x^{2 m+1}}{(2 m+1) !} cos heta_{m} x)
for all (x in E^{1}) ;
(b) (cos x=1-frac{x^{2}}{2 !}+frac{x^{4}}{4 !}-cdots+frac{(-1)^{m} x^{2 m}}{(2 m) !}-frac{(-1)^{m} x^{2 m+2}}{(2 m+2) !} sin heta_{m} x) for
all (x in E^{1} .)
[Hints: Let (f(x)=sin x) and (g(x)=cos x .) Induction shows that
[
f^{(n)}(x)=sin left(x+frac{n pi}{2} ight) ext { and } g^{(n)}(x)=cos left(x+frac{n pi}{2} ight) .
]
Using formula (left(5^{prime} ight),) prove that
[
left|R_{n}(x) ight| leqleft|frac{x^{n+1}}{(n+1) !} ight| ightarrow 0 .
]
Indeed, (x^{n} / n !) is the general term of a convergent series
[
left.sum frac{x^{n}}{n !} quad ext { (see Chapter } 4, §13, ext { Example }(mathrm{d}) ight).
]
(left. ext { Thus } x^{n} / n ! ightarrow 0 ext { by Theorem } 4 ext { of the same section. } ight])

Exercise (PageIndex{7})

For any (s in E^{1}) and (n in N,) define
[
left(egin{array}{l}{s} {n}end{array} ight)=frac{s(s-1) cdots(s-n+1)}{n !} ext { with }left(egin{array}{l}{s} {0}end{array} ight)=1 .
]
Then prove the following.
(i) (lim _{n ightarrow infty} nleft(egin{array}{l}{s} {n}end{array} ight)=0) if (s>0),
(ii) (lim _{n ightarrow infty}left(egin{array}{l}{s} {n}end{array} ight)=0) if (s>-1),
(iii) For any fixed (s in E^{1}) and (x in(-1,1)).
[
lim _{n ightarrow infty}left(egin{array}{l}{s} {n}end{array} ight) n x^{n}=0 ;
]
hence
[
lim _{n ightarrow infty}left(egin{array}{c}{s} {n}end{array} ight) x^{n}=0 .
]
(left[ ext { Hints: }(mathrm{i}) ext { Let } a_{n}=left|nleft(egin{array}{l}{s} {n}end{array} ight) ight| . ext { Verify that } ight.)
[
a_{n}=|s|left|1-frac{s}{1} ight|left|1-frac{s}{2} ight| cdotsleft|1-frac{s}{n-1} ight| .
]
If (s>0,left{a_{n} ight} downarrow) for (n>s+1,) so we may put (L=lim a_{n}=lim a_{2 n} geq 0 .) (Explain!)
Prove that
[
frac{a_{2 n}}{a_{n}}]
so for large (n),
[
frac{a_{2 n}}{a_{n}}]
With (varepsilon) fixed, let (n ightarrow infty) to get (L leqleft(e^{-frac{1}{2} s}+varepsilon ight) L .) Then with (varepsilon ightarrow 0,) obtain (L e^{frac{1}{2} s} leq L).
(left. ext { As } e^{frac{1}{2} s}>1 ext { (for } s>0 ight),) this implies (L=0,) as claimed.
(ii) For (s>-1, s+1>0,) so by ((i)),
[
(n+1)left(egin{array}{c}{s+1} {n+1}end{array} ight) ightarrow 0 ; ext { i.e., }(s+1)left(egin{array}{c}{s} {n}end{array} ight) ightarrow 0 . quad( ext { Why? })
]
(iii) Use the ratio test to show that the series (sumleft(egin{array}{l}{s} {n}end{array} ight) n x^{n}) converges when (|x|<1).
( ext { Then apply Theorem } 4 ext { of Chapter } 4, §13 .])

Exercise (PageIndex{8})

Continuing Problems 6 and (7,) prove that
[
(1+x)^{s}=sum_{k=0}^{n}left(egin{array}{l}{s} {k}end{array} ight) x^{k}+R_{n}(x) ,
]
where (R_{n}(x) ightarrow 0) if either (|x|<1,) or (x=1) and (s>-1,) or (x=-1) and (s>0 .)
[Hints: (a) If (0 leq x leq 1,) use (left(5^{prime} ight)) for
[
R_{n-1}(x)=left(egin{array}{c}{s} {n}end{array} ight) x^{n}left(1+ heta_{n} x ight)^{s-n}, quad 0< heta_{n}<1 . ext { (Verify!) }
]
Deduce that (left|R_{n-1}(x) ight| leqleft|left(egin{array}{c}{s} {n}end{array} ight) x^{n} ight| ightarrow 0 .) Use Problem 7(( ext { iii) if }|x|<1 ext { or Problem } 7( ext { ii })) if (x=1).
(b) If (-1 leq x<0,) write (left(5^{prime prime} ight)) as
[
R_{n-1}(x)=left(egin{array}{c}{s} {n}end{array} ight) n x^{n}left(1+ heta_{n}^{prime} x ight) s^{-1}left(frac{1- heta_{n}^{prime}}{1+ heta_{n}^{prime} x} ight)^{n-1} . ext { (Check!) }
]
As (-1 leq x<0,) the last fraction is (leq 1 .) (Why?) Also,
[
left(1+ heta_{n}^{prime} x ight)^{s-1} leq 1 ext { if } s>1, ext { and } leq(1+x)^{s-1} ext { if } s leq 1 .
]
Thus, with (x) fixed, these expressions are bounded, while (left(egin{array}{c}{s} {n}end{array} ight) n x^{n} ightarrow 0) by Problem 7((mathrm{i})) (left. ext { or }( ext { iii }) . ext { Deduce that } R_{n-1} ightarrow 0, ext { hence } R_{n} ightarrow 0 . ight])

Exercise (PageIndex{9})

Prove that
[
ln (1+x)=sum_{k=1}^{n}(-1)^{k+1} frac{x^{k}}{k}+R_{n}(x) ,
]
where (lim _{n ightarrow infty} R_{n}(x)=0) if (-1[Hints: If (0 leq x leq 1,) use formula (left(5^{prime} ight)).
If (-1[
R_{n}(x)=frac{ln (1+x)}{(-1)^{n}}left(frac{1- heta_{n}}{1+ heta_{n} x} cdot x ight)^{n} .
]
( ext { Proceed as in Problem } 8 .])

Exercise (PageIndex{10})

Prove that if (f : E^{1} ightarrow E^{*}) is of class (mathrm{CD}^{1}) on ([a, b]) and if (-infty[
fleft(x_{0} ight)>frac{f(b)-f(a)}{b-a}left(x_{0}-a ight)+f(a) ;
]
i.e., the curve (y=f(x)) lies above the secant through ((a, f(a))) and ((b, f(b)) .)
[Hint: This formula is equivalent to
[
frac{fleft(x_{0} ight)-f(a)}{x_{0}-a}>frac{f(b)-f(a)}{b-a} ,
]
i.e., the average of (f^{prime}) on (left[a, x_{0} ight]) is strictly greater than the average of (f^{prime}) on ([a, b],) (left. ext { which follows because } f^{prime} ext { decreases on }(a, b) .( ext { Explain! }) ight])

Exercise (PageIndex{11})

Prove that if (a, b, r,) and (s) are positive reals and (r+s=1,) then
[
a^{r} b^{s} leq r a+s b .
]
(This inequality is important for the theory of so-called (L^{p}) -spaces.)
[Hints: If (a=b) , all is trivial.
Therefore, assume (a[
a=(r+s) a]
Use Problem 10 with (x_{0}=r a+s b in(a, b)) and
[
f(x)=ln x, f^{prime prime}(x)=-frac{1}{x^{2}}<0 .
]
Verify that
[
x_{0}-a=x_{0}-(r+s) a=s(b-a)
]
and (r cdot ln a=(1-s) ln a ;) hence deduce that
[
r cdot ln a+s cdot ln b-ln a=s(ln b-ln a)=s(f(b)-f(a)) .
]
After substitutions, obtain
[
left.fleft(x_{0} ight)=ln (r a+s b)>r cdot ln a+s cdot ln b=ln left(a^{r} b^{s} ight) . ight]
]

Exercise (PageIndex{12})

Use Taylor's theorem (Theorem 1') to prove the following inequalities:
(a) (sqrt[3]{1+x}<1+frac{x}{3}) if (x>-1, x eq 0).
(b) (cos x>1-frac{1}{2} x^{2}) if (x eq 0).
(c) (frac{x}{1+x^{2}}0).
(d) (x>sin x>x-frac{1}{6} x^{3}) if (x>0).

## Taylor's theorem

In calculus, Taylor's theorem gives an approximation of a k-times differentiable function around a given point by a polynomial of degree k, called the kth-order Taylor polynomial. For a smooth function, the Taylor polynomial is the truncation at the order k of the Taylor series of the function. The first-order Taylor polynomial is the linear approximation of the function, and the second-order Taylor polynomial is often referred to as the quadratic approximation. [1] There are several versions of Taylor's theorem, some giving explicit estimates of the approximation error of the function by its Taylor polynomial.

Taylor's theorem is named after the mathematician Brook Taylor, who stated a version of it in 1715, [2] although an earlier version of the result was already mentioned in 1671 by James Gregory. [3]

Taylor's theorem is taught in introductory-level calculus courses and is one of the central elementary tools in mathematical analysis. It gives simple arithmetic formulas to accurately compute values of many transcendental functions such as the exponential function and trigonometric functions. It is the starting point of the study of analytic functions, and is fundamental in various areas of mathematics, as well as in numerical analysis and mathematical physics. Taylor's theorem also generalizes to multivariate and vector valued functions.

The remainder R n + 1 ( x ) R_(x) R n + 1 ​ ( x ) as given above is an iterated integral, or a multiple integral, that one would encounter in multi-variable calculus. This may have contributed to the fact that Taylor's theorem is rarely taught this way.

and this is known as the integral form of the remainder.

By the "real" mean value theorem, this integral can be replaced by the "mean value," attained at some point ξ ∈ ( a , x ) xiin (a,x) ξ ∈ ( a , x ) , multiplied by the length x − a x-a x − a . Thus we obtain the remainder in the form of Cauchy:

This is very close to, but not quite the same as, the Roche-Schlömilch form of the remainder.

It should also be mentioned that the integral form is typically derived by successive applications of integration by parts, which avoids ever mentioning multiple integrals. However, it may be considered the same proof (up to homotopy, in some sense) because integration by parts, in essence, is saying that one could compute a certain area either by integrating over the x x x variable or over the y y y variable.

## 5.6.E: Problems on Tayior's Theorem - Mathematics

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## Question 1

### Solution to question 1

a)
f(0) = 1 and f(2π) = 1 therefore f(0) = f(2π)
f is continuous on [0 , 2π]
Function f is differentiable in (0 , 2π)
Function f satisfies all conditions of Rolle's theorem
b)
function g has a V-shaped graph with vertex at x = 2 and is therefore not differentiable at x = 2.
Function g does not satisfy all conditions of Rolle's theorem
c)
Function h is undefined at x = 0.
Function h does not satisfy all conditions of Rolle's theorem.
d)
k(x) = |sin(x)| , for x in [0 , 2π]
The graph of function k is shown below and it shows that function k is not differentiable at x = π.
Function k does not satisfy all conditions of Rolle's theorem.

Figure 4. Graph of k(x) = |sin(x)| , for x in [0 , 2π]

## Brook Taylor

Brook Taylor's father was John Taylor and his mother was Olivia Tempest. John Taylor was the son of Natheniel Taylor who was recorder of Colchester and a member representing Bedfordshire in Oliver Cromwell's Assembly, while Olivia Tempest was the daughter of Sir John Tempest. Brook was, therefore, born into a family which was on the fringes of the nobility and certainly they were fairly wealthy.

Taylor was brought up in a household where his father ruled as a strict disciplinarian, yet he was a man of culture with interests in painting and music. Although John Taylor had some negative influences on his son, he also had some positive ones, particularly giving his son a love of music and painting. Brook Taylor grew up not only to be an accomplished musician and painter, but he applied his mathematical skills to both these areas later in his life.

As Taylor's family were well off they could afford to have private tutors for their son and in fact this home education was all that Brook enjoyed before entering St John's College Cambridge on 3 April 1703 . By this time he had a good grounding in classics and mathematics. At Cambridge Taylor became highly involved with mathematics. He graduated with an LL.B. in 1709 but by this time he had already written his first important mathematics paper ( in 1708) although it would not be published until 1714 . We know something of the details of Taylor thoughts on various mathematical problems from letters he exchanged with Machin and Keill beginning in his undergraduate years.

In 1712 Taylor was elected to the Royal Society. This was on the 3 April, and clearly it was an election based more on the expertise which Machin, Keill and others knew that Taylor had, rather than on his published results. For example Taylor wrote to Machin in 1712 providing a solution to a problem concerning Kepler's second law of planetary motion. Also in 1712 Taylor was appointed to the committee set up to adjudicate on whether the claim of Newton or of Leibniz to have invented the calculus was correct.

The paper we referred to above as being written in 1708 was published in the Philosophical Transactions of the Royal Society in 1714 . The paper gives a solution to the problem of the centre of oscillation of a body, and it resulted in a priority dispute with Johann Bernoulli. We shall say a little more below about disputes between Taylor and Johann Bernoulli. Returning to the paper, it is a mechanics paper which rests heavily on Newton's approach to the differential calculus.

The year 1714 also marks the year in which Taylor was elected Secretary to the Royal Society. It was a position which Taylor held from 14 January of that year until 21 October 1718 when he resigned, partly for health reasons, partly due to his lack of interest in the rather demanding position. The period during which Taylor was Secretary to the Royal Society does mark what must be considered his most mathematically productive time. Two books which appeared in 1715 , Methodus incrementorum directa et inversa and Linear Perspective are extremely important in the history of mathematics. The first of these books contains what is now known as the Taylor series, though it would only be known as this in 1785 . Second editions would appear in 1717 and 1719 respectively. We discuss the content of these works in some detail below.

Taylor made several visits to France. These were made partly for health reasons and partly to visit the friends he had made there. He met Pierre Rémond de Montmort and corresponded with him on various mathematical topics after his return. In particular they discussed infinite series and probability. Taylor also corresponded with de Moivre on probability and at times there was a three-way discussion going on between these mathematicians.

Between 1712 and 1724 Taylor published thirteen articles on topics as diverse as describing experiments in capillary action, magnetism and thermometers. He gave an account of an experiment to discover the law of magnetic attraction (1715) and an improved method for approximating the roots of an equation by giving a new method for computing logarithms (1717) . His life, however, suffered a series of personal tragedies beginning around 1721 . In that year he married Miss Brydges from Wallington in Surrey. Although she was from a good family, it was not a family with money and Taylor's father strongly objected to the marriage. The result was that relations between Taylor and his father broke down and there was no contact between father and son until 1723 . It was in that year that Taylor's wife died in childbirth. The child, which would have been their first, also died.

After the tragedy of losing his wife and child, Taylor returned to live with his father and relations between the two were repaired. Two years later, in 1725 , Taylor married again to Sabetta Sawbridge from Olantigh in Kent. This marriage had the approval of Taylor's father who died four years later on 4 April 1729 . Taylor inherited his father's estate of Bifons but further tragedy was to strike when his second wife Sabetta died in childbirth in the following year. On this occasion the child, a daughter Elizabeth, did survive.

Taylor added to mathematics a new branch now called the "calculus of finite differences", invented integration by parts, and discovered the celebrated series known as Taylor's expansion. These ideas appear in his book Methodus incrementorum directa et inversa of 1715 referred to above. In fact the first mention by Taylor of a version of what is today called Taylor's Theorem appears in a letter which he wrote to Machin on 26 July 1712 . In this letter Taylor explains carefully where he got the idea from.

It was, wrote Taylor, due to a comment that Machin made in Child's Coffeehouse when he had commented on using "Sir Isaac Newton's series" to solve Kepler's problem, and also using "Dr Halley's method of extracting roots" of polynomial equations. There are, in fact, two versions of Taylor's Theorem given in the 1715 paper which to a modern reader look equivalent but which, the author of [ 8 ] argues convincingly, were differently motivated. Taylor initially derived the version which occurs as Proposition 11 as a generalisation of Halley's method of approximating roots of the Kepler equation, but soon discovered that it was a consequence of the Bernoulli series. This is the version which was inspired by the Coffeehouse conversation described above. The second version occurs as Corollary 2 to Proposition 7 and was thought of as a method of expanding solutions of fluxional equations in infinite series.

We must not give the impression that this result was one which Taylor was the first to discover. James Gregory, Newton, Leibniz, Johann Bernoulli and de Moivre had all discovered variants of Taylor's Theorem. Gregory, for example, knew that

and his methods are discussed in [ 13 ] . The differences in Newton's ideas of Taylor series and those of Gregory are discussed in [ 15 ] . All of these mathematicians had made their discoveries independently, and Taylor's work was also independent of that of the others. The importance of Taylor's Theorem remained unrecognised until 1772 when Lagrange proclaimed it the basic principle of the differential calculus. The term "Taylor's series" seems to have used for the first time by Lhuilier in 1786 .

There are other important ideas which are contained in the Methodus incrementorum directa et inversa of 1715 which were not recognised as important at the time. These include singular solutions to differential equations, a change of variables formula, and a way of relating the derivative of a function to the derivative of the inverse function. Also contained is a discussion on vibrating strings, an interest which almost certainly come from Taylor's early love of music.

Taylor, in his studies of vibrating strings was not attempting to establish equations of motion, but was considering the oscillation of a flexible string in terms of the isochrony of the pendulum. He tried to find the shape of the vibrating string and the length of the isochronous pendulum rather than to find its equations of motion. Further discussion of these ideas is given in [ 14 ] .

Taylor also devised the basic principles of perspective in Linear Perspective (1715) . The second edition has a different title, being called New principles of linear perspective. The work gives first general treatment of vanishing points. Taylor had a highly mathematical approach to the subject and made no concessions to artists who should have found the ideas of fundamental importance to them. At times it is very difficult for even a mathematician to understand Taylor's results. The phrase "linear perspective" was invented by Taylor in this work and he defined the vanishing point of a line, not parallel to the plane of the picture, as the point where a line through the eye parallel to the given line intersects the plane of the picture. He also defined the vanishing line to a given plane, not parallel to the plane of the picture, as the intersection of the plane through the eye parallel to the given plane. He did not invent the terms vanishing point and vanishing line, but he was one of the first to stress their importance. The main theorem in Taylor's theory of linear perspective is that the projection of a straight line not parallel to the plane of the picture passes through its intersection and its vanishing point.

There is also the interesting inverse problem which is to find the position of the eye in order to see the picture from the viewpoint that the artist intended. Taylor was not the first to discuss this inverse problem but he did make innovative contributions to the theory of such perspective problems. One could certainly consider this work as laying the foundations for the theory of descriptive and projective geometry.

Taylor challenged the "non-English mathematicians" to integrate a certain differential. One has to see this challenge as part of the argument between the Newtonians and the Leibnitzians. Conte in [ 7 ] discusses the answers given by Johann Bernoulli and Giulio Fagnano to Taylor's challenge. We mentioned above the arguments between Johann Bernoulli and Taylor. Taylor, although he did not win all the arguments, could certainly dispute with Johann Bernoulli on fairly equal terms. Jones describes these arguments in [ 1 ] :-

Always look for new problems especially after successfully solving one - you may get much more than you expected to start with.

#### Pairs of statements in which one is a clear generalization of another whereas in fact the two are equivalent.

The Intermediate Value Theorem - The Location Principle (Bolzano Theorem) Rolle's Theorem - The Mean Value Theorem Existence of a tangent parallel to a chord - existence of a tangent parallel to the x-axis. Binomial theorem for (1 + x) n and (x + y) n The Maclaurin and Taylor series. Two properties of Greatest Common Divisor Pythagoras' Theorem and the Cosine Rule Pythagoras' Theorem and its particular case of an isosceles right triangle Pythagoras' Theorem and Larry Hoehn's generalization Combining pieces of 2 and N squares into a single square Measurement of inscribed and (more generally) secant angles Probability of the union of disjoint events and any pair of events Butterfly theorem in orthodiagonal and arbitrary quadrilaterals Pascal's theorem in ellipse and in circle A triangle is embeddable into a rectangle twice its area and so is any convex polygon Brahmagupta's Theorem and Heron's Theorem

There are cases where a more general statement highlights the features of the original problem not otherwise obvious and by doing so spells a solution that works in both cases.

#### Pairs of statements in which one is a clear generalization of another and the more general statement is not more difficult to prove.

Bottema's Theorem and McWorter's generalization Butterfly theorem Butterflies in a Pencil of Conics Carnot's theorem and Wallace's theorem. Concyclic Circumcenters: A Dynamic View. Concyclic Circumcenters: A Sequel. Find a plane through a point outside of an octahedron such that the plane bisects the volume of the octahedron - same statement but replace octahedron with a solid with a center of symmetry A Fine Feature of the Stern-Brocot tree Four Construction Problems Lucas' Theorem and its variant Matrix Groups Napoleon's Theorem and one theorem about similar triangles Four Pegs That Form a Square On the Difference of Areas One Trigonometric Formula and Its Consequences Pythagorean Theorem and the Parallelogram Law Pythagorean Theorem - General Pythagorean Theorem Asymmetric Propeller and Several of Its Generalizations A construction problem that combines several apparently unrelated ones Two-Parameter Solutions to Three Almost Fermat Equations Three circles with centers on their pairwise radical axes Square Roots and Triangle Inequality Miguel Ochoa's van Schooten is a Slanted Viviani Scalar Product Optimization Barycenter of cevian trangle generalizes Marian Dinca's criterion.

#### Problems that allow a meaningful generalization.

In a plane, given 3n points. Is it possible to draw n triangles with vertices at these points so that no two of them have points in common? Lines in a triangle intersecting at a common point. Ceva's theorem. 5109094x171709440000 = 21!, find x.
A number is written with 81 ones. Prove it's a multiple of 81. Prove that the number 1010101. 0101, 81 ones and 80 alternating zeros, is a multiple of 81.
Given a 1x1 square. Is it possible to put into it not intersecting circles so that the sum of their radii will be 1996? Criteria of divisibility by 9 and 11. The game of Fifteen and Puzzles on Graphs Weierstrass Product Inequality Fermat's Little Theorem and Euler's Theorem

#### Problems that allow more than one generalization.

Pythagorean Theorem Napoleon's Theorem Fermat Point and Generalizations A problem in extension fields Butterfly theorem A System of Equations Begging for Generalization

Thus we see that generalization is quite useful and often enjoyable. It's a great vehicle for discovering new facts. However, if unchecked, generalization may lead to erroneous results. I'd call such situations

## 5.6.E: Problems on Tayior's Theorem - Mathematics

Meetings:
M-F 9:30-10:20am at 380-380Y (M-Th Lecture, F TA section)

Instructor:
Jonathan Luk (jluk AT stanford DOT edu)
Office: 382-Z
Office hours: TuTh 3-4:30pm or by appointment

Teaching Assistant:
Alexander Dunlap (ajdunl2 AT stanford DOT edu)
Office: 380-J
Office hours: M 5:15-7:15pm at 380-J, W 5:35-7:35pm at 380-J, F 10:30am-12:20pm at Sapp 105 or by appointment

Textbook:
An Introduction to Multivariable Mathematics by Leon Simon
(If you are not on campus, you will first need to configure your browser.)

Syllabus can be found here.

Schedule can be found here.

Midterms:
Midterm 1 Solutions (first quartile 15, median19.5, third quartile 25.7 mean 19.4)
Midterm 2 (Part 1) Midterm 2 (Part 2) Solutions (first quartile 23.125, median 26.5, third quartile 28.875 mean 25.6)

Final distribution (out of 40):
(first quartile 25.56, median 30, third quartile 34.19 mean 28.98)

## Taylor's Theorem

Taylor's theorem (without the remainder term) was devised by Taylor in 1712 and published in 1715, although Gregory had actually obtained this result nearly 40 years earlier. In fact, Gregory wrote to John Collins, secretary of the Royal Society, on February 15, 1671, to tell him of the result. The actual notes in which Gregory seems to have discovered the theorem exist on the back of a letter Gregory had received on 30 January, 1671, from an Edinburgh bookseller, which is preserved in the library of the University of St. Andrews (P. Clive, pers. comm., Sep. 8, 2005).

However, it was not until almost a century after Taylor's publication that Lagrange and Cauchy derived approximations of the remainder term after a finite number of terms (Moritz 1937). These forms are now called the Lagrange remainder and Cauchy remainder.

Most modern proofs are based on Cox (1851), which is more elementary than that of Cauchy and Lagrange (Moritz 1937), and which Pringsheim (1900) referred to as "leaving hardly anything to wish for in terms of simplicity and strength" (Moritz 1937).

Cox, H. Cambridge and Dublin Math. J. 6, 80, 1851.

Dehn, M. and Hellinger, D. "Certain Mathematical Achievements of James Gregory." Amer. Math. Monthly 50, 149-163, 1943.

Jeffreys, H. and Jeffreys, B. S. "Taylor's Theorem." §1.133 in Methods of Mathematical Physics, 3rd ed. Cambridge, England: Cambridge University Press, pp. 50-51, 1988.

Malet, A. Studies on James Gregorie (1638-1675). Ph.D. thesis. Princeton, NJ: Princeton University, 1989.

Malet, A. "James Gregorie on Tangents and the 'Taylor' Rule for Series Expansions." Archive for History of Exact Science 46, 97-137, 1993-1994.

Moritz, R. E. "A Note on Taylor's Theorem." Amer. Math. Monthly 44, 31-33, 1937.

Pringsheim, A. "Zur Geschichte des Taylorschen Lehrsatzes." Bibliotheca Math. 1, 433-479, 1900.

## Taylor Series Approximation

In some cases, in engineering or real world technical problems, we are not interested to find the exact solution of a problem. If we have a good enough approximation, we can consider that we’ve found the solution of the problem.

For example, if we want to compute the trigonometric function f(x)=sin(x) with a hand held calculator, we have two options:

• use the actual trigonometric function sin(x), if the calculator has the function embedded (available if it’s a scientific calculator)
• use a polynomial as an approximation of the sin(x) function and compute the result with any calculator, or even by hand

In general, any mathematical function f(x), with some constraints, can be approximated by a polynomial P(x):

### Weierstrass approximation theorem

First, let’s put down what the theorem sounds like. After, we’ll try to explain it a bit.

Theorem : For a given function f(x), which is defined and continuous on the interval [a, b], there is always a polynomial P(x), also defined on the interval [a, b], with the property:

[left | f(x)-P(x) ight | < varepsilon]

for any x ∈ [a, b]. and a given ε > 0.

Image: Polynomial approximation of a function f(x)

The theorem says that for any function f(x), which is continuous and defined between the points a and b, there is always a polynomial P(x), which can approximate the function f(x) with a small error ε, in the same interval [a, b].

### Taylor’s polynomials

Example: Let’s approximate the function f(x)=sin(x) with a polynomial of order 3, around the point x0 = 0. Using the determined polynomial, approximate the value of sin(0.1).

Explanation : “around the point x0” means that f (n) (x0) = P (n) (x0), which means that the evaluation of the function and its derivatives in the point x0 is equal to the evaluation of the polynomial and its derivatives.

Step 1. Write the polynomial of order 3.

Step 2. Calculate the 3 rd order derivatives of P(x). We need them in order to find out the values of the coefficients a0, a1, a2 and a3.

Step 3. Calculate P (n) (x0).

Step 4. Calculate the 3 rd order derivatives of f(x).

Step 5. Calculate f (n) (x0).

Step 7. Write the final form of the approximation polynomial

Step 8. Approximate the value of sin(0.1) using the polynomial.

[f(0.1)=sin(0.1)cong P(0.1)=0.1-frac<1> <6>cdot 0.1^<3>approx 0.0998333]

Using Scilab we can compute sin(0.1) just to compare with the approximation result:

As you can see, the approximation with the polynomial P(x) is quite accurate, the result being equal up to the 7 th decimal.

### Accuracy of the approximation

Since the approximation is done around the point x0, evaluating the polynomial in x0 gives pretty good results. But what happens if we move away from x0, how accurate the approximation will still be?

To evaluate the polynomial for several points around x0, we are going to use a Scilab script.

What the Scilab script does:

• defines the polynomial P(x) as a function
• defines the range of x points from -π/2 up to π/2, using π/10 increments
• calculates the relative error e(x) between sin(x) and P(x)
• plots the values of sin(x) and P(x)
• plots the relative error for each x point

The relative error e(x), in percentage, is calculated as:

[e(x) = left | 1-frac ight | cdot 100]

Running the Scilab script will plot the following:

Image: Taylor approximation of sin(x) – Scilab plot

As you can see, the polynomial approximation of sin(x) is quite accurate around x0 but starts to diverge as we move toward the margins of the interval. For the extreme points, -π/2 and π/2, the relative error is around 7.5%.

In order to increase the accuracy of the polynomial approximation we need to increase the order of the polynomial. The higher the degree of the polynomial the better the approximation for a wider interval.

Image: Taylor approximation of sin(x) for P(x) of degree 1, 3, 5, 7, 9, 11 and 12

### Taylor’s theorem

The polynomial P(x) used in the example above is a specific case of a Taylor series for function approximation.

Theorem : Any function f(x) can be written as:

with P(x) being Taylor’s polynomial and R(x) being Taylor’s remainder:

P(x) can be written is a short form, as:

• f ∈ C n ([a, b]) – which means that f(x) is continuous and derivable on an interval [a, b]
• f (n+1) ∈ [a, b] – which means that all n+1 derivatives of f(x) exist in [a, b]
• x0 ∈ [a, b] – which means that x0 is a point between a and b
• ξ(x) ∈ (a, b) – which means that there is a point ξ between a and b which is greater than a and smaller than b

Example. For the function f(x), defined as:

1. find the Taylor polynomial of order 3 around x0 = 0.
2. approximate the value of √1.1
3. find the maximum error for the approximation

Step 1. Calculate the 3 rd order derivatives of f(x).

Step 2. Calculate f (n) (x0).

Step 3 (solves a). Calculate Taylor’s polynomial P3(x)

Step 4 (solves b). Approximate the value of √1.1

1.1 = 1 + 0.1, which means that x=0.1.

To verify our result let’s use Scilab function sqrt() :

Step 5. Calculate the 4 th derivative of f(x).

Step 6. Calculate the 4 th derivative of f(ξ).

Step 7 (solves c). Calculate R3(x) for a given ξ.

ξ must be chosen in such a way that the 4 th derivative of f(ξ) has maximum value. In our case this makes ξ = 0.

This means that the error of the polynomial approximation will not exceed 3.91·10 -6 .

### Examples with Scilab programming

Let’s use Scilab to calculate the Taylor series approximations for a couple of functions. To visualise the impact of the order of the approximation polynomial, we’ll use Scilab plot() function. For the functions f(x) and P(x) given below, we’ll plot the exact solution and Taylor approximation using a Scilab script.

Example 1. Exponential function f(x)=e x .

The Scilab script will define a custom Scilab function for P(x).

The output of the Scilab script is plotted below, together with an animation of the same approximation.

Image: Taylor approximation of exp(x)

Image: Taylor approximation of exp(x) – animation

Example 2. Natural logarithm function f(x)=ln(1+x).

The Scilab script will define a custom Scilab function for P(x).

The output of the Scilab script is plotted below.

Image: Taylor approximation of ln(1+x)

In order to try out other functions and their Taylor series approximation, redefine the P(x,n) function from the Scilab script.