The extreme value theorem for 1-variable functions

This post discusses when a real-valued 1-variable function has global extrema. As usual in rigorous mathematics we start from proper definitions of key concepts.

Proper definitions of global extrema

If a function $$f(x)$$ is defined over a domain $$D$$, then a global maximum of $$f(x)$$ is a point $$a\in D$$ such that $$f(a)\geq f(x)$$ for all $$x\in D$$.

Such a maximum is called global (or absolute), because the inequality $$f(a)\geq f(x)$$ should hold over the whole domain $$D$$, not only around the point $$x=a$$ as for a local maximum.

Similarly, a global minimum of $$f(x)$$ is a point $$a\in D$$ such that $$f(a)\leq f(x)$$ for all $$x\in D$$.

Functions without global extrema over an open interval

The simple function $$f(x)=x$$ has no global extrema over the whole real line, because the set of all possible values of $$f(x)$$ is unbounded.

If we restrict the function $$f(x)=x$$ to any open interval $$(a,b)$$, then the values of $$f(x)=x$$ are in the bounded set $$(a,b)$$, but this set does not include its boundary points $$a,b$$. Hence $$f(x)=x$$ again has no global extrema over any open interval $$(a,b)$$.

However, the function $$f(x)=x$$ on any closed interval $$[a,b]$$, has the global minimum at $$x=a$$ and the global maximum at $$x=b$$, because $$f(a)\leq f(x)\leq f(b)$$ for all $$x\in[a,b]$$.

Discontinuous functions without global extrema

The function $$f(x)=\left\{ \begin{array}{l} \tan x \mbox{ for } -\frac{\pi}{2}<x<\frac{\pi}{2},\\ 0 \mbox{ for } x=\pm\frac{\pi}{2} \end{array} \right.$$ has no global extrema over the closed interval $$[-\frac{\pi}{2},\frac{\pi}{2}]$$. Indeed, the set of all possible values of $$f(x)$$ is the unbounded real line.

A good exercise is to prove that any continuous function $$f(x)$$ over a closed interval $$[a,b]$$ is bounded, namely there are bounds $$c,d$$ such that $$c\leq f(x)\leq d$$ for all $$x\in [a,b]$$.

A proof requires a proper understanding of continuity via the epsilon-delta argument. If this theorem is examinable in your calculus course, you could let us know in the comments, which would certainly promote your university.

When does a global extremum exist?

It turns out that a bad domain and discontinuity of a function are the only reasons why global extrema do not exist in the examples above. The powerful theorem below covers the exercise on boundedness of continuous functions over closed intervals, which we stated above.

The Extreme Value Theorem.
Any continuous function over a closed interval has a global maximum and a global minimum.

• Riddle 16: is there a continuous surface with exactly 2 global
maxima, but without any global minima over the whole plane?
• How to submit: to write your full answer, simply submit a comment.
• Hint: definitions of global extrema are similar for 2-variable functions.
• Warning: give an example or justify that there is no such a surface.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.

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Next Big Thing in online education: SPOCs vs MOOCs

BBC News has reported the new term SPOC in Next Big Thing. We shall remind the short story.

Distant past: bricks-and-mortar universities.

For hundreds of years students learned in so-called bricks-and-mortar universities. The general scheme includes
(1) paying large tuition fees for
(3) enjoying social life away from parents
(4) hopefully gaining a degree at the end.

However the Internet is changing the education.

Recent past: MOOCs (Massive Open Online Courses).

Many students have probably heard the term MOOC, see more details in the Wikipedia article. The term MOOC was coined in 2008 and first 3 real MOOCs appeared in September 2011. By September 2013 there were dozens of MOOC providers in different countries. Initially MOOCs didn’t have the same factors as the conventional universities:
(1) large tuition fees
(2) real professors
(3) any social life
(4) certificates.

Several MOOCs have started to offer
(1) discussions and chats with instructors
(2) assessments with identities verified
(3) a small charge for certificates.

The main drawback of MOOCs has been the low retention rate: about 8% of all enrolled students actually pass a final test.

Next Big Thing: SPOCs (Small Private Online Courses).

It seems that term SPOC has just been coined by BBC on 24th September 2013. There is no Wikipedia article on SPOCs yet: checked on 27th September 2013. So we shall briefly describe the key points of the BBC report.
(1) Harvard says that we are already in the “post-MOOC” era.
(3) Proper guidance will be provided for the selected students.
(4) Students will be more rigorously assessed than in MOOCs.

We have started our SPOCs (distance courses for maths candidates to top UK universities) in September 2010, which was 1 year before any MOOCs appeared and 3 years before Harvard started to think about SPOCs. We hope that other SPOCs (will) have the same key values as our current courses for top students preparing for MAT papers, Oxbridge interviews, STEP exams:
(1) quick and detailed feedback on regular written homework
(2) training on many problems harder than past exam questions
(3) best tips and advice from our recent successful students
(4) interactive quizzes offering gradual hints when needed.

Actually, we feel rather happy that big players like Harvard are catching up. The flexibility of small start-ups is a great advantage in tech innovations.

Small mammals are now ruling the world, while huge dinosaurs were wiped out. Here is the powerful principle: the size does not matter, but the skills do!

• Riddle 7: which black area is larger in the picture in the top left corner
at the very beginning of this post: all small squares or two big squares?
• How to submit: to write your full answer, submit a comment.
• Hint: you may assume that each small black square has side 1.
• Restriction: only the first correct answer will be rewarded.
• Prize: free 1-year access to one of our interactive web tutorials.
• Update: after Ariella’s attempt, the answer has been explained.

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How to properly state mathematical theorems: 3 key parts

At summer schools we often asked to state Pythagoras’ theorem. The most popular answer was short: $$a^2+b^2=c^2$$. Initially we were rather shocked that many students haven’t probably seen any properly stated mathematical theorems.

Now we are more experienced and will give more hints. The first light hint: could you write Pythagoras’ theorem in more details, e.g. by using words? Here is the next popular attempt (literally!): “a squared plus b squared equals c squared”.

Then we ask: what do you mean by $$a^2+b^2=c^2$$? Is it an equation you are going to solve or what? Let us give a parallel example from history.
Question: who is the Queen of England?
The numerical answer is correct, but can you expect a full mark? Similarly in mathematics, the equation $$a^2+b^2=c^2$$ says very little about Pythagoras’ theorem.

At this moment most students draw a triangle with sides a,b,c. With some extra help the triangle is marked as right-angled, not “right triangle”, which is another common confusion. So the statement now says: “A right-angled triangle with sides a,b,c satisfies $$a^2+b^2=c^2$$“. Almost correct: Pythagoras’ theorem is about any right-angled triangle, not just “a triangle”.

There is a slightly longer, but more logical statement: if a triangle with sides $$a\leq b<c$$ is right-angled, then $$a^2+b^2=c^2$$. Please notice that we have clearly separated the geometric condition (a triangle is right-angled) and the algebraic conclusion ($$a^2+b^2=c^2$$ holds).

We continue: great, this is the correct first half of Pythagoras’ theorem, could you state the second half? Even good students who have easily written the first half are often stuck and admit that they don’t know the second half. Another hint: is the triangle with sides 3,4,5 right-angled? Everyone says “yes”, so we ask why. The popular answer: by Pythagoras’ theorem.

Then we repeat: do you really claim that our statement above implies that the triangle with sides 3,4,5 is right-angled? Most students are confused, but sometimes we receive this answer: “it’s just the opposite”. Opposite sides of most coins are different and so are opposite statements in mathematics.

It is easier to reverse the longer logical statement: “if $$a,b,c>0$$ satisfy $$a^2+b^2=c^2$$, then the triangle with the sides a,b,c is right-angled.” Notice how we swapped the two key parts of the statement: if the algebraic condition is given, then the geometric conclusion holds.

Now we ask to write two statements together as full Pythagoras’ theorem. A popular attempt is to write one phrase after another: “Any right-angled triangle with sides a,b,c satisfies $$a^2+b^2=c^2$$. If $$a,b,c>0$$ satisfy $$a^2+b^2=c^2$$, then the triangle with the sides a,b,c is right-angled.” We clarify: can you write a shorter single phrase by making a logical connection between the algebraic and geometric parts?

After talking about “if… then“, “only if” or “only when“, we arrive at the following statement of full Pythagoras’ theorem: “Numbers $$a,b,c>0$$ satisfy $$a^2+b^2=c^2$$ if and only if the triangle with the sides a,b,c is right-angled.” The most important part is the keywords “if and only if” saying that the theorem works in both directions.

So any mathematical statement consists of 3 key parts:

• a given condition (that is given and can be used later in a proof)
• a logical connection (between the condition and the conclusion)
• a resulting conclusion (that should be deduced from the condition).

Many statements are one-sided: if A then B (if a condition A is given, then a conclusion B holds). These statements play the role of a one-side bridge between villages A and B. Namely, villagers can go from A to B, but not from B to A.

All great theorems in mathematics are two-sided: A if and only if B, which establishes a two-sided bridge between A and B, so villagers can freely travel between A and B. For instance, Pythagoras’ theorem is a two-sided bridge between geometry and algebra.

Similar and more general statements led to the development of analytic geometry where we rephrase a geometric problem in algebraic terms, solve by using algebraic methods and then translate our solution back to geometry.

You have hopefully appreciated the fascinating fact that triples of positive numbers satisfying $$a^2+b^2=c^2$$ are equivalent to geometric shapes of right-angled triangles. Simple Pythagoras’ theorem stated above can be generalised to many (even infinite) dimensions, also works for functions instead of segments and makes sense in very general so-called Hilbert spaces.

In all these cases Pythagoras’ theorem illustrates the powerful principle: 2-sided bridges are much better than 1-sided ones.

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How to solve hard problems: a key principle with examples

It would be much nicer to have a simple problem instead of a hard one. So a natural way to solve a hard mathematical problem is to make it simpler.

The key principle is to simplify.

At summer schools we often ask the simple question: how will you solve the equation $$(x-2)^2=1$$? Most students expand the brackets and then apply the quadratic formula.

The standard quadratic formula is a correct general recipe. However, a simple computer program will easily beat you. While you are solving one quadratic equation, the chip in your mobile phone can solve a billion of similar equations. So mathematics is not a cookbook of recipes, but a beautiful subject where efficiency, not brute force, is highly valued.

Efficiency often wins over brute force.

If we replace $$(x-2)^2=1$$ by a very similar equation $$(x^2-2x)^2=1$$, then expanding the brackets will produce a longer and more complicated equation of degree 4. Moreover, any computer program based on the standard quadratic formula will fail.

However, both equations $$(x-2)^2=1$$ and $$(x^2-2x)^2=1$$ have the same pattern: a square equals a number. Hence the equations can be solved in the same way. For instance, we could replace the expression under the square by a new variable.

A substitution is a common way to simplify.

In the equation $$(x-2)^2=1$$ we could set $$y=x-2$$. Then we get the simpler equation $$y^2=1$$, which has two roots $$y=\pm 1$$. Hence the original equation $$(x-2)^2=1$$ has two roots $$x=y+2=-1+2=1$$ and $$x=y+2=1+2=3$$.

Similarly, in the equation $$(x^2-2x)^2=1$$ we could set $$y=x^2-2x$$. Then we get the simpler equation $$y^2=1$$, $$y=\pm 1$$. Hence the original equation $$(x^2-2x)^2=1$$ is equivalent to the union of 2 simpler quadratic equations $$x^2-2x=-1$$ and $$x^2-2x=1$$.

How to simplify: try to keep a pattern.

Most students would probably solve the last two equations by using the standard quadratic formula again. We could actually continue the simplification and complete a simple square instead. Namely, $$x^2-2x=-1$$, $$x^2-2x+1=0$$, $$(x-1)^2=0$$, so $$x=1$$ is a double root. Similarly, $$x^2-2x=1$$, $$x^2-2x+1=2$$, $$(x-1)^2=2$$, $$x-1=\pm\sqrt{2}$$, $$x=1\pm\sqrt{2}$$.

There are many more complicated equations with the same pattern $$y^2=1$$. For instance, try to solve the equations $$(x^3-2x-1)^2=4$$,   $$(\sin x+\cos x)^2=1$$,   $$(e^{2x}-2e^x)^2=1$$.

• Riddle 3: simplify $$\sqrt{x^2}$$ for any real number x and explain your conclusion.
• How to submit: to write your full answer, simply submit a comment.
• Hint: the square root $$\sqrt{x}$$ of any positive real number $$x$$ is always positive.
• Warning: many 2nd year maths students at a top university get it wrong.
• Restriction: only the first fully correct public answer will be rewarded.
• Prize: free 1-year access to one of our interactive web tutorials.
• Update: R S has solved the riddle, read our confirmation.

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