aarjaneiro/planetmath huggingface dataset:
proof of Rolle’s theorem

proof of Rolle’s theorem

Because

f

is

continuous

on a

compact

(closed and

bounded

)

interval

I

=

[

a

,

b

]

, it attains its

maximum

and

minimum

values. In case

f

⁢

(

a

)

=

f

⁢

(

b

)

is both the maximum and
the minimum, then there is nothing more to say, for then

f

is a

constant function

and

f

′

≡

0

on the whole interval

I

. So suppose otherwise, and

f

attains an

extremum

in the open interval

(

a

,

b

)

, and without loss of generality, let this extremum be a maximum, considering

-

f

in
lieu of

f

as necessary. We claim that at this extremum

f

⁢

(

c

)

we have

f

′

⁢

(

c

)

=

0

, with

a

<

c

<

b

.

To show this, note that

f

⁢

(

x

)

-

f

⁢

(

c

)

≤

0

for all

x

∈

I

, because

f

⁢

(

c

)

is the maximum. By definition of the

derivative

, we have that

f

′

⁢

(

c

)

=

lim

x

→

c

⁡

f

⁢

(

x

)

-

f

⁢

(

c

)

x

-

c

.

Looking at the

one-sided limits

, we note that

R

=

lim

x

→

c

+

⁡

f

⁢

(

x

)

-

f

⁢

(

c

)

x

-

c

≤

0

because the numerator in the limit is nonpositive in the interval

I

,
yet

x

-

c

>

0

, as

x

approaches

c

from the right. Similarly,

L

=

lim

x

→

c

-

⁡

f

⁢

(

x

)

-

f

⁢

(

c

)

x

-

c

≥

0

.

Since

f

is

differentiable

at

c

, the left and right limits must coincide, so

0

≤

L

=

R

≤

0

, that is to say,

f

′

⁢

(

c

)

=

0

.

Title

proof of Rolle’s

theorem

Canonical name

ProofOfRollesTheorem

Date of creation

2013-03-22 12:40:19

Last modified on

2013-03-22 12:40:19

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

5

Author

rmilson (146)

Entry type

Proof

Classification

msc 26A06

Generated on Fri Feb 9 19:45:43 2018 by

LaTeXML
proof of Taylor’s Theorem

proof of Taylor’s Theorem

Let

f

⁢

(

x

)

,

a

<

x

<

b

be a real-valued,

n

-times

differentiable
function

, and let

a

<

x

0

<

b

be a fixed base-point. We will show that
for all

x

≠

x

0

in the domain of the
function, there exists a

ξ

, strictly between

x

0

and

x

such
that

f

⁢

(

x

)

=

∑

k

=

0

n

-

1

f

(

k

)

⁢

(

x

0

)

⁢

(

x

-

x

0

)

k

k

!

+

f

(

n

)

⁢

(

ξ

)

⁢

(

x

-

x

0

)

n

n

!

.

Fix

x

≠

x

0

and let

R

be the

remainder

defined by

f

⁢

(

x

)

=

∑

k

=

0

n

-

1

f

(

k

)

⁢

(

x

0

)

⁢

(

x

-

x

0

)

k

k

!

+

R

⁢

(

x

-

x

0

)

n

n

!

.

Next, define

F

⁢

(

ξ

)

=

∑

k

=

0

n

-

1

f

(

k

)

⁢

(

ξ

)

⁢

(

x

-

ξ

)

k

k

!

+

R

⁢

(

x

-

ξ

)

n

n

!

,

a

<

ξ

<

b

.

We then have

F

′

⁢

(

ξ

)

=

f

′

⁢

(

ξ

)

+

∑

k

=

1

n

-

1

(

f

(

k

+

1

)

⁢

(

ξ

)

⁢

(

x

-

ξ

)

k

k

!

-

f

(

k

)

⁢

(

ξ

)

⁢

(

x

-

ξ

)

k

-

1

(

k

-

1

)

!

)

-

R

⁢

(

x

-

ξ

)

n

-

1

(

n

-

1

)

!

=

f

(

n

)

⁢

(

ξ

)

⁢

(

x

-

ξ

)

n

-

1

(

n

-

1

)

!

-

R

⁢

(

x

-

ξ

)

n

-

1

(

n

-

1

)

!

=

(

x

-

ξ

)

n

-

1

(

n

-

1

)

!

⁢

(

f

(

n

)

⁢

(

ξ

)

-

R

)

,

because the sum telescopes.
Since,

F

⁢

(

ξ

)

is a differentiable function, and since

F

⁢

(

x

0

)

=

F

⁢

(

x

)

=

f

⁢

(

x

)

, Rolle’s

Theorem

imples that there exists a

ξ

lying
strictly between

x

0

and

x

such that

F

′

⁢

(

ξ

)

=

0

. It follows that

R

=

f

(

n

)

⁢

(

ξ

)

, as was to be shown.

Title

proof of Taylor’s Theorem

Canonical

name

ProofOfTaylorsTheorem

Date of creation

2013-03-22 12:33:59

Last modified on

2013-03-22 12:33:59

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

8

Author

rmilson (146)

Entry type

Proof

Classification

msc 26A06

Generated on Fri Feb 9 19:45:46 2018 by

LaTeXML
quotient rule

quotient rule

The

quotient rule

says that the

derivative

of the quotient

f

/

g

of two

differentiable functions

f

and

g

exists at all values of

x

as long as

g

⁢

(

x

)

≠

0

and is given by the formula

d

d

⁢

x

⁢

[

f

⁢

(

x

)

g

⁢

(

x

)

]

=

g

⁢

(

x

)

⁢

f

′

⁢

(

x

)

-

f

⁢

(

x

)

⁢

g

′

⁢

(

x

)

[

g

⁢

(

x

)

]

2

The Quotient Rule and the other

differentiation

formulas allow us to compute the derivative of any

rational function

.

Title

quotient rule

Canonical name

QuotientRule

Date of creation

2013-03-22 12:38:51

Last modified on

2013-03-22 12:38:51

Owner

Luci (289)

Last modified by

Luci (289)

Numerical id

13

Author

Luci (289)

Entry type

Theorem

Classification

msc 26A06

Generated on Fri Feb 9 19:45:47 2018 by

LaTeXML
ratio test

ratio test

Let

(

a

n

)

be a

sequence

of

real numbers

. Then the series

∑

n

=

1

∞

a

n

converges absolutely

if

lim sup

n

→

∞

⁡

|

a

n

+

1

a

n

|

<

1

and

diverges

if

lim inf

n

→

∞

⁡

|

a

n

+

1

a

n

|

>

1

.

Title

ratio test

Canonical name

RatioTest

Date of creation

2013-03-22 12:24:14

Last modified on

2013-03-22 12:24:14

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

12

Author

Koro (127)

Entry type

Theorem

Classification

msc 26A06

Classification

msc 40A05

Synonym

quotient criterion

Generated on Fri Feb 9 19:45:49 2018 by

LaTeXML
Rolle’s theorem

Rolle’s theorem

Rolle’s

theorem

.

If

f

is a

continuous function

on

[

a

,

b

]

, such that

f

⁢

(

a

)

=

f

⁢

(

b

)

and

differentiable

on

(

a

,

b

)

then there exists a point

c

∈

(

a

,

b

)

such that

f

′

⁢

(

c

)

=

0

.

Title

Rolle’s theorem

Canonical name

RollesTheorem

Date of creation

2013-03-22 11:51:26

Last modified on

2013-03-22 11:51:26

Owner

drini (3)

Last modified by

drini (3)

Numerical id

14

Author

drini (3)

Entry type

Theorem

Classification

msc 26A06

Related topic

IntermediateValueTheorem

Related topic

MeanValueTheorem

Related topic

ZeroesOfDerivativeOfComplexPolynomial

Generated on Fri Feb 9 19:45:55 2018 by

LaTeXML
S

S

L

Title

S

Canonical name

S

Date of creation

2013-03-22 19:00:41

Last modified on

2013-03-22 19:00:41

Owner

Altair (23757)

Last modified by

Altair (23757)

Numerical id

6

Author

Altair (23757)

Entry type

Topic

Classification

msc 26A06

Classification

msc 12D10

Classification

msc 26-01

Related topic

C

Generated on Fri Feb 9 19:45:57 2018 by

LaTeXML
sinc is L^2

sinc is

L

2

Our objective will be to prove the

integral

∫

ℝ

f

2

⁢

(

x

)

⁢

𝑑

x

exists in the Lebesgue sense when

f

⁢

(

x

)

=

sinc

⁡

(

x

)

.

The integrand is an

even function

and so we can restrict our proof to the set

ℝ

+

.

Since

f

is a

continuous function

, so will

f

2

be and thus for every

a

>

0

,

f

∈

L

2

⁢

(

[

0

,

a

]

)

.

Thus, if we prove

f

∈

L

2

(

[

π

,

∞

[

)

, the result will be proved.

Consider the

intervals

I

k

=

[

k

⁢

π

,

(

k

+

1

)

⁢

π

]

and

U

k

=

⋃

i

=

1

k

I

k

=

[

π

,

(

k

+

1

)

⁢

π

]

.

and the succession of

functions

f

n

⁢

(

x

)

=

f

2

⁢

(

x

)

⁢

χ

U

n

⁢

(

x

)

, where

χ

U

n

is the

characteristic function

of the set

U

n

.

Each

f

n

is a continuous function of compact support and will thus be

integrable

in

ℝ

+

. Furthermore

f

n

⁢

(

x

)

↗

f

2

⁢

(

x

)

(

pointwise

) in this set.

In each

I

k

,

0

≤

f

2

⁢

(

x

)

≤

sin

2

⁡

(

x

)

(

k

⁢

π

)

2

, for

k

>

0

.

So:

∫

x

≥

π

f

n

⁢

(

x

)

⁢

𝑑

x

=

∑

k

=

1

n

∫

k

⁢

π

(

k

+

1

)

⁢

π

sin

(

x

)

2

x

2

⁢

𝑑

x

≤

∑

k

=

1

n

∫

k

⁢

π

(

k

+

1

)

⁢

π

sin

(

x

)

2

(

k

⁢

π

)

2

=

∑

k

=

1

n

1

2

⁢

k

2

⁢

π

1

1

we have used the well known result

∫

0

π

sin

2

⁡

(

x

)

⁢

𝑑

x

=

π

2

So:

lim

n

→

∞

⁡

∫

x

≥

π

f

n

⁢

(

x

)

⁢

𝑑

x

≤

lim

n

→

∞

⁡

∑

k

=

1

n

1

2

⁢

k

2

⁢

π

and since the series on the right side

converges

2

2

asymptotic

behaviour as

k

-

2

and

f

n

↗

f

2

we can use the

monotone convergence theorem

to state that

f

2

∈

L

(

[

π

,

∞

[

)

.

So we get the result that

sinc

∈

L

2

⁢

(

ℝ

)

Title

sinc is

L

2

Canonical name

SincIsL2

Date of creation

2013-03-22 15:44:44

Last modified on

2013-03-22 15:44:44

Owner

cvalente (11260)

Last modified by

cvalente (11260)

Numerical id

9

Author

cvalente (11260)

Entry type

Result

Classification

msc 26A06

Generated on Fri Feb 9 19:46:08 2018 by

LaTeXML
sinc is not L^1

sinc is not

L

1

The main results used in the proof will be that

f

∈

L

1

⁢

(

A

)

⇔

|

f

|

∈

L

1

⁢

(

A

)

and the

dominated convergence theorem

.

Proof by contradiction

:

Let

f

⁢

(

x

)

=

|

sinc

⁡

(

x

)

|

and suppose it’s

Lebesgue integrable

in

ℝ

+

.

Consider the

intervals

I

k

=

[

k

⁢

π

,

(

k

+

1

)

⁢

π

]

and

U

k

=

⋃

i

=

0

k

I

k

=

[

0

,

(

k

+

1

)

⁢

π

]

.

and the succession of

functions

f

n

⁢

(

x

)

=

f

⁢

(

x

)

⁢

χ

U

n

⁢

(

x

)

, where

χ

U

n

is the

characteristic function

of the set

U

n

.

Each

f

n

is a

continuous function

of compact support and will thus be

integrable

in

ℝ

+

. Furthermore

f

n

⁢

(

x

)

↗

f

⁢

(

x

)

(

pointwise

)

in each

I

k

,

f

⁢

(

x

)

≥

|

sin

⁡

(

x

)

|

(

k

+

1

)

⁢

π

.

So

∫

ℝ

+

f

n

=

∑

k

=

0

n

∫

k

⁢

π

(

k

+

1

)

⁢

π

|

sin

⁡

(

x

)

|

x

⁢

𝑑

x

≥

∑

k

=

0

n

∫

k

⁢

π

(

k

+

1

)

⁢

π

|

s

⁢

i

⁢

n

⁢

(

x

)

|

(

k

+

1

)

⁢

π

=

∑

k

=

0

n

2

(

k

+

1

)

⁢

π

.

Suppose

f

is integrable in

ℝ

+

. Then by the dominated convergence theorem

lim

n

→

∞

⁡

∫

ℝ

+

f

n

=

∫

ℝ

+

f

.

But

lim

n

→

∞

⁡

∫

ℝ

+

f

n

≥

lim

n

→

∞

⁡

∑

k

=

0

n

2

(

k

+

1

)

⁢

π

=

+

∞

and we get the

contradiction

∫

ℝ

+

f

≥

+

∞

.

So

f

cannot be integrable in

ℝ

+

.
This implies that

f

cannot be integrable in

ℝ

and since a function is integrable in a set iff its

absolute value

is

sinc

⁡

(

x

)

∉

L

1

⁢

(

ℝ

)

Title

sinc is not

L

1

Canonical name

SincIsNotL1

Date of creation

2013-03-22 15:44:32

Last modified on

2013-03-22 15:44:32

Owner

cvalente (11260)

Last modified by

cvalente (11260)

Numerical id

14

Author

cvalente (11260)

Entry type

Result

Classification

msc 26A06

Generated on Fri Feb 9 19:46:10 2018 by

LaTeXML
sum of series depends on order

sum of series depends on order

According to the Leibniz’ test (

http://planetmath.org/LeibnizEstimateForAlternatingSeries

),
the

alternating series

1

-

1

2

+

1

3

-

1

4

+

1

5

-

1

6

+

1

7

-

1

8

+

1

9

-

1

10

+

1

11

-

1

12

+

-

…

is

convergent

and has a

positive

sum (

=

ln

⁡

2

; see the

natural logarithm

(

http://planetmath.org/NaturalLogarithm2

)).  Denote it by

S

.  We can by

1

2

getting the two series

S

=

(

1

-

1

2

)

+

(

1

3

-

1

4

)

+

(

1

5

-

1

6

)

+

(

1

7

-

1

8

)

+

(

1

9

-

1

10

)

+

…

,

1

2

S

=

1

2

-

1

4

+

1

6

-

1

8

+

1

10

-

+

…

.

Then we add these two series termwise getting the sum

1

⁤

1

2

⁢

S

=

1

+

1

3

-

2

4

+

1

5

+

1

7

-

2

8

+

1

9

+

1

11

-

2

12

+

…

.

Hence, this last series exactly the same as the original, but its sum is fifty percent

greater

.  This is possible because the original series is not absolutely convergent:  the series which is formed of the

absolute values

of its is the divergent

harmonic series

.

P. S.  – For justification of the used manipulations of the series, see the entry.

Title

sum of series depends on order

Canonical name

SumOfSeriesDependsOnOrder

Date of creation

2013-03-22 14:50:59

Last modified on

2013-03-22 14:50:59

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

16

Author

pahio (2872)

Entry type

Example

Classification

msc 26A06

Classification

msc 40A05

Related topic

AbsoluteConvergence

Related topic

OrderOfFactorsInInfiniteProduct

Related topic

AlternatingHarmonicSeries

Related topic

ConditionallyConvergentSeries

Related topic

ConvergingAlternatingSeriesNotSatisfyingAllLeibnizConditions

Related topic

FiniteChangesInConvergentSeries

Related topic

FiniteChangesInConvergentSeries2

Generated on Fri Feb 9 19:46:13 2018 by

LaTeXML
addition and subtraction formulas for sine and cosine

addition and subtraction formulas for sine and cosine

The

rotation matrix

(

cos

⁡

θ

-

sin

⁡

θ

sin

⁡

θ

cos

⁡

θ

)

will be used to obtain the

addition formulas for sine and cosine

.

Recall that a vector in

ℝ

2

can be rotated

θ

radians in the counterclockwise

direction

by multiplying on the left by the rotation matrix

(

cos

⁡

θ

-

sin

⁡

θ

sin

⁡

θ

cos

⁡

θ

)

. Because rotating by

α

+

β

radians is the same as rotating by

β

radians followed by rotating by

α

radians, we obtain:

(

cos

⁡

(

α

+

β

)

-

sin

⁡

(

α

+

β

)

sin

⁡

(

α

+

β

)

cos

⁡

(

α

+

β

)

)

=

(

cos

⁡

α

-

sin

⁡

α

sin

⁡

α

cos

⁡

α

)

⁢

(

cos

⁡

β

-

sin

⁡

β

sin

⁡

β

cos

⁡

β

)

=

(

cos

⁡

α

⁢

cos

⁡

β

-

sin

⁡

α

⁢

sin

⁡

β

-

cos

⁡

α

⁢

sin

⁡

β

-

sin

⁡

α

⁢

cos

⁡

β

sin

⁡

α

⁢

cos

⁡

β

+

cos

⁡

α

⁢

sin

⁡

β

-

sin

⁡

α

⁢

sin

⁡

β

+

cos

⁡

α

⁢

cos

⁡

β

)

Hence,

sin

⁡

(

α

+

β

)

=

sin

⁡

α

⁢

cos

⁡

β

+

cos

⁡

α

⁢

sin

⁡

β

and

cos

⁡

(

α

+

β

)

=

cos

⁡

α

⁢

cos

⁡

β

-

sin

⁡

α

⁢

sin

⁡

β

.

Note that sine is an

even function

and that cosine is an

odd function

, i.e. (

http://planetmath.org/Ie

)

sin

⁡

(

-

x

)

=

-

sin

⁡

x

and

cos

⁡

(

-

x

)

=

-

cos

⁡

x

. These facts enable us to obtain the subtraction formulas for sine and cosine.

sin

⁡

(

α

-

β

)

=

sin

⁡

(

α

+

(

-

β

)

)

=

sin

⁡

(

α

)

⁢

cos

⁡

(

-

β

)

+

cos

⁡

(

α

)

⁢

sin

⁡

(

-

β

)

=

sin

⁡

(

α

)

⁢

cos

⁡

(

β

)

-

cos

⁡

(

α

)

⁢

sin

⁡

(

β

)

cos

⁡

(

α

-

β

)

=

cos

⁡

(

α

+

(

-

β

)

)

=

cos

⁡

(

α

)

⁢

cos

⁡

(

-

β

)

-

sin

⁡

(

α

)

⁢

sin

⁡

(

-

β

)

=

cos

⁡

(

α

)

⁢

cos

⁡

(

β

)

+

sin

⁡

(

α

)

⁢

sin

⁡

(

β

)

Title

addition and subtraction formulas for sine and cosine

Canonical name

AdditionAndSubtractionFormulasForSineAndCosine

Date of creation

2013-03-22 16:59:01

Last modified on

2013-03-22 16:59:01

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

16

Author

Wkbj79 (1863)

Entry type

Derivation

Classification

msc 26A09

Classification

msc 15-00

Classification

msc 33B10

Synonym

addition and subtraction formulae for sine and cosine

Synonym

addition formulas for sine and cosine

Synonym

addition formulae for sine and cosine

Synonym

subtraction formulas for sine and cosine

Synonym

subtraction formulae for sine and cosine

Synonym

addition formula for sine

Synonym

subtraction

Related topic

AdditionFormula

Related topic

DefinitionsInTrigonometry

Related topic

DoubleAngleIdentity

Related topic

MeanCurvatureAtSurfacePoint

Related topic

DAlembertAndDBernoulliSolutionsOfWaveEquation

Related topic

AdditionFormulas

Generated on Fri Feb 9 19:46:24 2018 by

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angle multiplication and division formulae for tangent

angle multiplication and division formulae for tangent

From the

angle addition formula

for the

tangent

, we may derive formulae
for tangents of multiples of angles:

tan

⁡

(

2

⁢

x

)

=

2

⁢

tan

⁡

x

1

-

tan

2

⁡

x

tan

⁡

(

3

⁢

x

)

=

3

⁢

tan

⁡

x

-

tan

3

⁡

x

1

-

3

⁢

tan

2

⁡

x

tan

⁡

(

4

⁢

x

)

=

4

⁢

tan

⁡

x

-

3

⁢

tan

3

⁡

x

1

-

6

⁢

tan

2

⁡

x

+

tan

4

⁡

x

These formulae may be derived from a

recursion

. Write

tan

⁡

x

=

w

and
write

tan

⁡

(

n

⁢

x

)

=

u

n

/

v

n

where the

u

’s and the

v

’s are

polynomials

in

w

. Then we have the initial values

u

1

=

w

and

v

1

=

1

and the recursions

u

n

+

1

=

u

n

+

w

⁢

v

n

v

n

+

1

=

v

n

-

w

⁢

u

n

,

which follow from the

addition formula

. Moreover, if we know the tangent
of an angle and are interested in finding the tangent of a multiple of
that angle, we may use our recursions directly without first having to derive
the multiple angle formulae. From these recursions, one may show that the

u

’s will only involve odd powers of

w

and the

v

’s will only involve
even powers of

w

.

Proceeding in the opposite direction, one may consider bisecting an angle.
Solving for

tan

⁡

x

in the duplication formula above, one arrives at the
following

half-angle formula

:

tan

⁡

(

x

2

)

=

1

+

1

tan

2

⁡

x

-

1

tan

⁡

x

Expressing the tangent in terms of sines and cosines and simplifying, one
finds the following

equivalent

formulae:

tan

⁡

(

x

2

)

=

1

-

cos

⁡

x

sin

⁡

x

=

sin

⁡

x

1

+

cos

⁡

x

=

±

1

-

cos

⁡

x

1

+

cos

⁡

x

Title

angle multiplication and division formulae for tangent

Canonical name

AngleMultiplicationAndDivisionFormulaeForTangent

Date of creation

2013-03-22 17:00:15

Last modified on

2013-03-22 17:00:15

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

8

Author

rspuzio (6075)

Entry type

Result

Classification

msc 26A09

Generated on Fri Feb 9 19:46:27 2018 by

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ceiling

ceiling

The

ceiling

of a

real number

is the smallest integer greater than or equal to the number. The ceiling of

x

is usually denoted by

⌈

x

⌉

.

Some examples:

⌈

6.2

⌉

=

7

,

⌈

0.4

⌉

=

1

,

⌈

7

⌉

=

7

,

⌈

-

5.1

⌉

=

-

5

,

⌈

π

⌉

=

4

,

⌈

-

4

⌉

=

-

4

.

Note that this function is not the

integer part

(

[

x

]

), since

⌈

3.5

⌉

=

4

and

[

3.5

]

=

3

.

The notation for floor and ceiling was introduced by Iverson in 1962

[

1

]

.

References

1

N. Higham, Handbook of writing for the mathematical sciences, Society for Industrial and Applied Mathematics, 1998.

Title

ceiling

Canonical

name

Ceiling

Date of creation

2013-03-22 11:48:21

Last modified on

2013-03-22 11:48:21

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

17

Author

yark (2760)

Entry type

Definition

Classification

msc 26A09

Classification

msc 11-00

Synonym

ceiling function

Synonym

smallest integer function

Synonym

smallest integer greater than or equal to

Related topic

BeattysTheorem

Related topic

Floor

Generated on Fri Feb 9 19:46:33 2018 by

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concavity of sine function

concavity of sine function

Theorem 1

.

The

sine function

is

concave

on the

interval

[

0

,

π

]

.

Proof.

Suppose that

x

and

y

lie in the interval

[

0

,

π

/

2

]

.
Then

sin

⁡

x

,

sin

⁡

y

,

cos

⁡

x

, and

cos

⁡

y

are all
non-negative. Subtracting the identities

sin

2

⁡

x

+

cos

2

⁡

x

=

1

and

sin

2

⁡

y

+

cos

2

⁡

y

=

1

from each other, we conclude that

sin

2

⁡

x

-

sin

2

⁡

y

=

cos

2

⁡

y

-

cos

2

⁡

x

.

This implies that

sin

2

⁡

x

-

sin

2

⁡

y

≥

0

if and only if

cos

2

⁡

y

-

cos

2

⁡

x

≥

0

, which is

equivalent

to stating
that

sin

2

⁡

x

≥

sin

2

⁡

y

if and only if

cos

2

⁡

x

≤

cos

2

⁡

y

. Taking

square roots

, we conclude that

sin

⁡

x

≤

sin

⁡

y

if and only if

cos

⁡

x

≥

cos

⁡

y

.

Hence, we have

(

sin

⁡

x

-

sin

⁡

y

)

⁢

(

cos

⁡

x

-

cos

⁡

y

)

≤

0

.

Multiply out both sides and move terms to conclude

sin

⁡

x

⁢

cos

⁡

x

+

sin

⁡

y

⁢

cos

⁡

y

≤

sin

⁡

x

⁢

cos

⁡

y

+

sin

⁡

y

⁢

cos

⁡

x

.

Applying the angle addition and

double-angle identities

for the
sine function, this becomes

1

2

⁢

(

sin

⁡

(

2

⁢

x

)

+

sin

⁡

(

2

⁢

y

)

)

≤

sin

⁡

(

x

+

y

)

.

This is equivalent to stating that, for all

u

,

v

∈

[

0

,

π

]

,

1

2

⁢

(

sin

⁡

u

+

sin

⁡

v

)

≤

sin

⁡

(

u

+

v

2

)

,

which implies that

sin

is concave in the interval

[

0

,

π

]

.
∎

Title

concavity of sine function

Canonical name

ConcavityOfSineFunction

Date of creation

2013-03-22 17:00:26

Last modified on

2013-03-22 17:00:26

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

8

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 26A09

Classification

msc 15-00

Generated on Fri Feb 9 19:46:35 2018 by

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convexity of tangent function

convexity of tangent function

We will show that the

tangent function

is convex on the

interval

[

0

,

π

/

2

)

.
To do this, we will use the

addition formula

for the

tangent

and the fact that
a

continuous

real function

f

is convex (

http://planetmath.org/ConvexFunction

) if and only if

f

⁢

(

(

x

+

y

)

/

2

)

≤

(

f

⁢

(

x

)

+

f

⁢

(

y

)

)

/

2

.

We start with the observation that, if

0

≤

x

<

1

and

0

≤

y

<

1

, then by the

arithmetic-geometric mean inequality

(

http://planetmath.org/ArithmeticGeometricMeansInequality

),

-

2

⁢

x

⁢

y

≥

-

x

2

-

y

2

1

-

2

⁢

x

⁢

y

+

x

2

⁢

y

2

≥

1

-

x

2

-

y

2

+

x

2

⁢

y

2

(

1

-

x

⁢

y

)

2

≥

(

1

-

x

2

)

⁢

(

1

-

y

2

)

,

so

(

1

-

x

⁢

y

)

2

(

1

-

x

2

)

⁢

(

1

-

y

2

)

≥

1

.

Let

u

and

v

be two numbers in the interval

[

0

,

π

/

4

)

. Set

x

=

tan

⁡

u

and

y

=

tan

⁡

v

.
Then

0

≤

x

<

1

and

0

≤

y

<

1

.

By the addition formula, we have

tan

⁡

(

2

⁢

u

)

=

2

⁢

x

1

-

x

2

tan

⁡

(

u

+

v

)

=

x

+

y

1

-

x

⁢

y

tan

⁡

(

2

⁢

v

)

=

2

⁢

y

1

-

y

2

.

Hence,

1

2

⁢

(

tan

⁡

(

2

⁢

u

)

+

tan

⁡

(

2

⁢

v

)

)

=

x

+

y

-

x

2

⁢

y

-

x

⁢

y

2

(

1

-

x

2

)

⁢

(

1

-

y

2

)

=

(

x

+

y

)

⁢

(

1

-

x

⁢

y

)

(

1

-

x

2

)

⁢

(

1

-

y

2

)

=

x

+

y

1

-

x

⁢

y

⁢

(

1

-

x

⁢

y

)

2

(

1

-

x

2

)

⁢

(

1

-

y

2

)

≥

x

+

y

1

-

x

⁢

y

=

tan

⁡

(

u

+

v

)

,

so the tangent function is convex.

Title

convexity of tangent function

Canonical

name

ConvexityOfTangentFunction

Date of creation

2013-03-22 17:00:12

Last modified on

2013-03-22 17:00:12

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

14

Author

rspuzio (6075)

Entry type

Result

Classification

msc 26A09

Generated on Fri Feb 9 19:46:46 2018 by

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cosine at multiples of straight angle

cosine at multiples of straight angle

Everybody remembers the cosine values

cos

⁡

0

=

1

,

cos

⁡

(

±

π

)

=

-

1

,

cos

⁡

(

±

2

⁢

π

)

=

1

,

cos

⁡

(

±

3

⁢

π

)

=

-

1

,

…

The thing can be concisely expressed as the

cos

⁡

n

⁢

π

=

(

-

1

)

n

for each integer

n

.  The values of sine at the same angles are simply 0.

Title

cosine at multiples of straight angle

Canonical

name

CosineAtMultiplesOfStraightAngle

Date of creation

2013-03-22 15:45:40

Last modified on

2013-03-22 15:45:40

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Definition

Classification

msc 26A09

Related topic

multiple

Related topic

ComplexSineAndCosine

Related topic

HigherOrderDerivativesOfSineAndCosine

Related topic

FourierSineAndCosineSeries

Related topic

GeneralAssociativity

Related topic

ValueOfRiemannZetaFunctionAtS4

Related topic

ValueOfDirichletEtaFunctionAtS2

Generated on Fri Feb 9 19:46:48 2018 by

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