Mathematical tools: a hammer vs a microscope

We congratulate everyone who celebrates the Knowledge Day on 1st September! This post is based on real student stories and illustrates a powerful principle about using appropriate mathematical tools.

A problem in multivariable calculus

The following problem was the first introductory question in a recent exam on multivariable calculus at a top UK university. This question was designed for students who had found the course hard. Indeed you could get a full mark within few seconds using only basic maths from school.

Q1. Find the minimum value of the function $$f(x,y)=x^2+2y^2-4x-4y$$ over all real $$x,y$$.

Derivatives may not be needed at all

Rather surprisingly, only about 2% of the 200+ class remembered the lecturer’s advice to use appropriate tools. The remaining 98% preferred the hard way finding extreme points through partial derivatives. Similarly to 1-variable functions, as we discussed in the post on global extrema, we may often find extreme values of a function without using any derivatives.

Moreover, about 50% of all students claimed that their critical point is a local minimum without justifications, hence lost almost a half of the full mark. The other 48% spent a lot of time justifying that the only critical point (where both 1st order partial derivatives vanish) is indeed a local minimum by using the Hessian of four 2nd order partial derivatives. About 10% of the class actually wrote more than 2 pages on this first question, though only a couple of lines was enough.

Here is the full solution:

$$f(x,y)=x^2+2y^2-4x-4y=(x-2)^2+2(y-1)^2-6\geq -6$$ for all real $$x,y$$.
Hence the minimum value is $$f(2,1)=-6$$.

This simple technique of completing a square illustrates the powerful principle that mastering the basics is more important than trying an advanced method without proper understanding.

Completing a square vs differentiation

Here is the simple mnemonic rule: What can you do with a quadratic polynomial? – Complete a square, of course! Most students usually try to directly write the roots of a quadratic polynomial by using the so-called quadratic formula. Actually, this formula is proved by completing a square, which is a good exercise especially if you have never done it.

From the computational point of view, the quadratic formula is equivalent to completing a square, so we essentially make the same computations in both cases. However, after completing a square, a quadratic polynomial becomes structured (similar to the simplest form $$x^2$$).

Moreover, the same idea of completing a binom $$(x+a)^n$$ works for any higher degree polynomial, but the quadratic formula doesn’t. Actually many real problems in mathematics are about putting various objects (say, functions or matrices or groups) into a normal or structured form.

So completing a square is simple and efficient like a hammer, while differentiation is powerful and delicate like a microscope. A popular student question: “Can I still use a microscope, because I like it?” Our answer: “Yes, you can, though it may look a bit unprofessional.”

• Riddle 19: find the maximum value of $$f(x)=6x^3-x^6$$ over all real $$x$$.
• How to submit: to write your full answer, submit a comment.
• Hint: find a point $$a$$ such that $$f(x)\leq f(a)$$ for any real $$x$$.
• Warning: using derivatives is possible, but is hard to justify.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.
• Update: Carlo has solved the problem, see attempts 1 and 2.

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Three easy steps to find all global extrema of a function

In this post we discuss global extrema of a real-valued 1-variable function over all reals.

Step 1: find all stationary points of a given function

By definition a stationary point of a function $$f(x)$$ is a solution of $$f'(x)=0$$. In this step many students waste their time by evaluating the 2nd derivative $$f”(x)$$ at each stationary point to determine if it is a local extremum, maximum or minimum. We shall detect global extrema simply by comparing $$f(x)$$ at all critical points, which is usually easier than $$f”(x)$$.

Step 2: find all points where the 1st derivative is undefined

This important step is often missed. For example, the derivative of $$f(x)=|x|$$ is never zero.
However, we should certainly consider the point $$x=0$$, where the derivative of $$f(x)=|x|$$ is undefined. Otherwise we miss the global minimum of $$f(x)=|x|$$ at $$x=0$$ over all reals.

We should consider all critical points and possibly more

All points where $$f'(x)=0$$ or $$f'(x)$$ is undefined are usually called critical. Sometimes we might not be sure if the derivative $$f'(x)$$ is well-defined or not.

A common mistake in sketching the graph $$y=|\cos x|$$ is to draw neighbourhoods of the points $$x=\frac{\pi}{2}+\pi n$$ for any integer $$n$$, where $$\cos x=0$$, either as smooth arcs in $$y=\cos^2 x$$ above or as vertical cusps in $$y=\sqrt{|\cos x|}$$ below.

In fact, the neighbourhoods of these points locally look like $$y=|x|$$, see the picture at the beginning of the post. Indeed, $$y=cos x$$ looks like $$y=x$$ around the points $$x=\frac{\pi}{2}+\pi n$$. So the derivative of $$f(x)=|\cos x|$$ is actually undefined at $$x=\frac{\pi}{2}+\pi n$$ for any integer $$n$$.

However we don’t need to justify this conclusion if we are interested only in extreme values of $$f(x)=|\cos x|$$. We may simply consider all “potentially critical” points, namely $$x=\frac{\pi}{2}n$$ for any integer $$n$$, where $$f'(x)=0$$ (at local extrema $$x=\pi n$$ of $$y=\cos x$$) or where $$f'(x)$$ might be undefined for $$f(x)=|\cos x|$$.

Step 3: compare the function values at all critical points

Finally we should simply compare the values of $$f(x)$$ at all (potentially) critical points and choose the points that have the largest and smallest values.

For instance, $$f(x)=|\cos x|$$ over all real x has the largest value 1at the infinitely many global maxima $$x=\pi n$$ and the smallest value 0 at the infinitely many global minima $$x=\frac{\pi}{2}+\pi n$$ for any integer $$n$$. Here is the full justification of the extreme values: $$0\leq|\cos x|\leq 1$$ for all $$x$$.

If we are interested in extreme values of a function over a closed interval, not over the whole real line, then there is an extra step that will be discussed in one of our future posts.

The traditional riddle below is Problem 3.5 in our current MAT 2014 course, which was modified from original Question 1B in past exam MAT 2007.

• Riddle 18: Find the greatest value of $$(4\cos^4(5x-6)-3)^2$$ over all real $$x$$.
• How to submit: to write your full answer, submit a comment.
• Hint: you can write a full solution to this riddle in 1-2 lines.
• Warning: using derivatives is possible, but is hard to justify.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.
• Update: Carlo has solved the problem, see this attempt and the final solution.

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How and when school children learn the times table

This post was motivated by a long public discussion of learning the times table at UK schools.

Dropping maths at the age of 16

It seems that a public acceptance of being bad in maths may not be considered as a great shame, even by a finance minister, read BBC magazine: UK Chancellor refused to answer a “times table” question. How could that happen?

One retired maths teacher has told us that many students do not continue studying maths and English at the age of 16-18. Here is an update from BBC education: a long-term possibility might be to require all would-be teachers to study maths to 18.

This long term plan can be compared with the long existing tradition in other countries where maths and native language/literature are compulsory for all students up to the university level.

Why are UK students really lucky?

Here is the most popular comment on BBC: “I’m studying a Masters in Physics at University and STILL don’t know my times tables by heart, I got an A* at GCSE Maths and an A at A-Level.”

Does it imply that UK students may not know the times table? It is probably better not to reflect on conclusions that overseas students can make about UK qualifications and universities.

Here is our reply: you are really lucky to study in the UK, because without the times table you wouldn’t progress to a secondary school in a different country.

In many other educational systems, if children start school at the age of 6-7, then they finish learning the times table by the end of their 1st class at the age of 7-8.

Why is learning the times table so hard?

As with all other obstacles, learning mathematics becomes unnecessarily hard when pre-requisite concepts are missed. The necessary earlier step is the simpler addition table that has the sum i+j in the intersection of the i-th row and j-th column.

 + 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 10 11 3 4 5 6 7 8 9 10 11 12 4 5 6 7 8 9 10 11 12 13 5 6 7 8 9 10 11 12 13 14 6 7 8 9 10 11 12 13 14 15 7 8 9 10 11 12 13 14 15 16 8 9 10 11 12 13 14 15 16 17 9 10 11 12 13 14 15 16 17 18

Notice how the table above is beautifully symmetric like many great results in mathematics. When students learn the times table not by rote, but by actually trying to compute products in their head, they naturally start adding smaller numbers.

For instance, to find $$7\times 8$$, one can sequentially compute the following partial sums: $$8+8=16$$, $$\;16+8=24$$, $$\; 24+8=32$$, $$\; 32+8=40$$, $$\; 40+8=48$$, $$\; 48+8=56$$. These sums require the addition table above, which should be studied before the times table.

So we get $$7\times 8=56$$ after completing 6 easy additions. If we are already good at multiplying only by 2,  faster way is to do 3 multiplications: $$7\times 2=14$$, $$\; 14\times 2=28$$, $$\; 28\times 2=56$$.

Can the times table be learned only by rote?

Let us compare the times table with another basic skill of walking. Children usually start learning to walk earlier than learning the times table.

Toddlers fall (fail) many times before they become confident walkers. After a first fall a baby looks at their parent to understand how to react: if the parent looks afraid then the baby starts crying, if the parent smiles then the baby also smiles.

So almost all children successfully learn to walk, not by rote, but by actually trying to walk, despite numerous falls or even small injuries. If a kid can not walk by the age of 3, it is usually considered as a physical disability.

A criterion of success in learning

In our opinion, the same powerful principle works for learning any other basic skill. So we wish our readers to follow the excellent example of smart babies, namely more practice and consider every failure (or a fall) as one more step in a right direction.

There is even the internal criterion of success without any comparison with your peers. If people can walk, they walk easily and without making any conscious efforts. Similarly, multiplying single-digit numbers should be as effortless and enjoyable as walking or breathing.

• Riddle 17: can the square $$n^2$$ of an integer have the last digit 8?
• How to submit: to write your full answer, submit a comment.
• Hint: the required knowledge is (a small bit of) the times table.
• Warning: give an example or justify that there is no such number.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.

If you wish to receive e-mails about our new posts (for a quicker chance to answer a riddle and win a prize) or distance courses, please contact us and tick the box “keep me updated”. You can easily unsubscribe at any time by e-mailing “unsubscribe” in the subject line.

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.

If you wish to receive e-mails about our new posts (for a quicker chance to answer a riddle and win a prize) or distance courses, please contact us and tick the box “keep me updated”. You can easily unsubscribe at any time by e-mailing “unsubscribe” in the subject line.

Finding local extrema of functions: 3 mistakes and 3 tips

This post discusses common mistakes and gives tips how to easily find local extrema of real-valued 1-variable functions.

1st mistake: any local extremum = a stationary point?

This myth has been dethroned in riddle 14. Actually, the function $$f(x)=|x|$$ has a local (even global) minimum at $$x=0$$, because $$|x|\geq 0$$ for all $$x$$. However, the derivative $$f'(x)=\left\{ \begin{array}{l} 1 \mbox{ for } x>0,\\ -1 \mbox{ for } x<0 \end{array} \right.$$ is undefined at $$x=0$$.

On the other hand, the function $$f(x)=x^3$$ from the previous post has a stationary point at $$x=0$$, which is not a local extremum. So the concepts of a stationary point and a local extremum are independent in the sense that one doesn’t follow from another.

1st tip: find all points where the derivative is undefined

The examples above imply that we should study all point where $$f'(x)$$ is undefined to find local extrema of $$f(x)$$. If the derivative $$f'(x)$$ is a fraction, we may start from
finding all points where the denominator of $$f'(x)$$ vanishes.

For example, the function $$f(x)=\sqrt{x}$$ is well-defined for all $$x\geq 0$$, but the derivative $$f'(x)=\frac{1}{2\sqrt{x}}$$ is not defined at $$x=0$$, where the tangent line to $$y=\sqrt{x}$$ is vertical.

2nd mistake: sufficient or necessary conditions of extrema

Recall that a stationary point of a function $$f(x)$$ is by definition a solution of the equation $$f'(x)=0$$. Here is a theorem stating when a stationary point $$x=a$$ is a local extremum.

Sufficient conditions of a local extremum. For a stationary point $$x=a$$ of a function $$f(x)$$ with a well-defined derivative $$f'(x)$$ for all $$x$$, if $$f”(a)<0$$ then $$x=a$$ is a local maximum, if $$f”(a)>0$$ then $$x=a$$ is a local minimum.

The conditions above are sufficient, but are not necessary as the example of $$f(x)=|x|$$ at $$x=0$$ shows. If $$f”(a)=0$$, then more analysis is needed. This is a typical question of singularity theory, which is richer for more than 1 variable.

2nd tip: check your rule for basic shapes of $$\pm x^2$$.

Many students often forget which inequality $$f”(a)<0$$ or $$f”(a)>0$$ corresponds to a local maximum or a local minimum. The simple trick is to remember the basic shapes of $$x^2$$ and $$-x^2$$. Namely, the positive parabola $$f(x)=x^2$$ has $$f”(0)=2>0$$ and a local minimum at $$x=0$$. Similarly, the negative parabola $$f(x)=-x^2$$ has $$f”(0)=-2<0$$ and a local maximum at $$x=0$$.

3rd mistake: does $$f”(a)=0$$ mean that $$x=a$$ is a point of inflection?

We have seen dozens of student scripts wrongly claiming that “$$f”(0)=0 \Rightarrow x=0$$ is a point of inflection”. Here is a proper geometric definition of a point of inflection of $$f(x)$$: if the graph $$y=f(x)$$ goes from one side of its the tangent line at $$x=a$$ to another side in a small neighbourhood of $$x=a$$, then $$x=a$$ is a point of inflection.

Analytically, if $$L(x)=f(a)+f'(a)(x-a)$$ is the tangent line to $$y=f(x)$$, then the difference $$f(x)-L(x)$$ changes its sign at a point of inflection $$x=a$$. For instance, at a stationary point $$x=a$$ the tangent line $$L(x)=f(a)$$ is horizontal and the same difference $$f(x)-f(a)$$ keeps its sign around $$x=a$$. Hence a point of inflection can not be a local extremum of $$f(x)$$.

For example, $$x=0$$ is a point of inflection of $$f(x)=x^3$$, because $$y=x^3$$ intersects the tangent line $$y=0$$ and goes from the lower half-plane to the upper half-plane.

3rd tip: use well-known inequalities for justifying extrema

The function $$f(x)=x^4$$ has a local minimum, but not a point of inflection at $$x=0$$. The sufficient conditions from the theorem above are not satisfied as $$f'(0)=0=f”(0)$$. However, the justification is even easier: $$x=0$$ is a global minimum of $$f(x)=x^4$$ over all real $$x$$ as $$x^4\geq 0$$.

• Riddle 15: state conditions when $$x=a$$ is a point of inflection of $$f(x)$$.
• How to submit: to write your full answer, simply submit a comment.
• Hint: give sufficient conditions in derivatives of $$f(x)$$ without proof.
• Warning: not all conditions are equalities, $$f'(a)=0$$ isn’t enough.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.

If you wish to receive e-mails about our new posts (for a quicker chance to answer a riddle and win a prize) or distance courses, please contact us and tick the box “keep me updated”. You can easily unsubscribe at any time by e-mailing “unsubscribe” in the subject line.

Traffic jams and 3 common mistakes on turning points

This post discusses the commonly confused concept of a turning point.

1st mistake: stationary = turning?

Let $$f(x)$$ be the position of a moving car in the $$x$$-axis (on a highway if you wish). Then the derivative $$f'(x)$$ is the speed (or the length of the velocity vector) of the car.

The solutions of $$f'(x)=0$$ are the points where the speed is 0, so the car is stationary. Hence the solutions of $$f'(x)=0$$ are called stationary points of the function $$f(x)$$.

If a car comes to a stationary point (a stop), it doesn’t mean that the car will make a U-turn. You have certainly been in a traffic jam, where a car stops for a while and then starts moving again in the same direction. Despite this overwhelming real-life practice, stationary points are often (and wrongly) called turning points even at university.

2nd mistake: stationary points = extrema?

Another common mistake is to confuse stationary points with local extrema. We have seen hundreds of scripts claiming something like “$$f'(a)=0$$, hence $$x=a$$ is a minimum (or a maximum)”. The simple counter-example is $$f(x)=x^3$$. Indeed, $$f'(x)=3x^2$$, so $$x=0$$ is a stationary point. However, $$x=0$$ is neither a local minimum nor a local maximum.

A stationary point and a local extremum are different concepts. A stationary point is analytically defined as a solution of $$f'(x)=0$$. A local extremum is geometrically defined below.

A proper definition of a local maximum

A point $$x=a$$ is called a (strict) local maximum of $$f(x)$$ if $$f(x)<f(a)$$ for all points $$x\neq a$$ sufficiently close to the point $$a$$. Notice that the inequality $$f(x)<f(a)$$ can’t hold at $$x=a$$ and is not required over the whole domain of $$f(x)$$.

For example, $$f(x)=\cos x$$ has a local maximum at $$x=0$$, because $$\cos x<1$$ for all $$x\neq 0$$ over $$-2\pi<x<2\pi$$.

A proper definition of a local minimum

Similarly, a point $$x=a$$ is called a (strict) local minimum of $$f(x)$$ if $$f(x)>f(a)$$ for all points $$x\neq a$$ sufficiently close to the point $$a$$.

The word strict refers to the strict inequality $$f(x)>f(a)$$. Non-strict local extrema allow the condition $$f(x)\geq f(a)$$ for all $$x$$ sufficiently close to the point $$a$$. For example, $$x=0$$ can be considered as a non-strict local minimum of $$f(x)=\left\{ \begin{array}{l} x \mbox{ for } x\geq 0,\\ 0 \mbox{ for } x\leq 0. \end{array}\right.$$

Local minima and maxima can be called local extrema. The word extremum means either a minimum or a maximum. A turning point of a function $$f(x)$$ is the same concept as a strict local extremum. Indeed, the graph $$y=f(x)$$ “turns” (or makes a U-turn) at any strict local extremum of $$f(x)$$. However, experts usually say a local extremum, not a “turning point”.

3rd mistake: extreme point = extreme value?

The final common mistake is to confuse extrema with extreme values by writing, for instance, “$$f(x)=x^2+1$$ has the minimum $$f(0)=1$$“. The point $$x=0$$ is indeed a local minimum of $$f(x)=x^2+1$$. However, the value $$f(0)=1$$ is called a local minimum value, not a local minimum point. So a point and the value of a function at this point are very different.

• Riddle 14: for a function $$f(x)$$, should any local extremum be a stationary point?
• How to submit: to write your full answer, simply submit a comment to this post.
• Hint: the derivative $$f'(a)$$ should be well-defined at any stationary point $$x=a$$.
• Warning: justify that any extremum has $$f'(a)=0$$ or give a counter-example.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.
• Update: IS has solved the riddle giving the counter-example f(x)=|x|.

If you wish to receive e-mails about our new posts (for a quicker chance to answer a riddle and win a prize) or distance courses, please contact us and tick the box “keep me updated”. You can easily unsubscribe at any time by e-mailing “unsubscribe” in the subject line.

Cultural differences: failure vs life-changing experience

Happy New Persian Year Nowruz if you are celebrating on March 21!

In our previous post on common mistakes with square roots we have dethroned the popular myth that $$\sqrt{x^2}=x$$. Below we shall tell our readers a personal story how we gained a life-changing experience due to the basic fact that $$\sqrt{x^2}=|x|$$. Students have told us that the absolute value $$|x|$$ is taught in the UK only at A levels (age 17-18). Moreover, the last year STEP examiners’ report again highlighted this common mistake about $$\sqrt{x^2}$$. So we have shown the graph $$y=|x|$$ in the picture above just in case.

Entrance exams in Physical-Mathematical Schools

We started to learn maths at overseas schools that specialise in mathematics and physics. All such schools are state-funded, but have rigorous entrance exams, usually at the age of 12 or 13. Our traditional riddle below contains Question 2 (of 8) from a recent entrance exam.

So we were in a class of 30 best maths students selected in a 1-million city. Everyone was a maths champion in their previous school. However our new Physical-Mathematical School was rather different. Grades were not given for past successes or for social connections, but only for hard independent work.

Answering in front of a class

Solving unexpected problems at a blackboard in front of a class was a routine exercise every lesson unless we had a written paper for independent solving. Once the teacher called one of the prettiest girls, who might have been named as a class queen in the UK. At that time she actually was in the top 20% of the class due to her background knowledge from previous years.

While the girl was trying to solve a new problem at the board, our teacher realised that the girl confuses some basic things. So she directly asked: “What is the square root of $$x^2$$?” The girl replied: “x”. The teacher: “That’s a fail, sit down.”

There was complete silence, nobody laughed. Indeed, despite we were the best in our previous schools, I’m sure that everyone from our class had been in a similar sutation before, probably with more advanced stuff than $$\sqrt{x^2}$$, however we understood all feelings anyway. The key lesson in this story is not about $$\sqrt{x^2}$$ at all, but about the exemplary reaction to such a failure.

The girl hold the nerves and quietly sat down and was actually ok after the lesson. The whole class was really impressed by her level of self-control at the age of 13. I’m not sure if it was a life-long lesson for the girl, but definitely for me since I can vividly recall all the details more than 20 years after this lesson. Later when I was in many tough situations I remembered how to properly withstand a punch with dignity and never give up!

There is no success without failures

Without our failures at school, we wouldn’t become professional mathematicians, but this is another story for our future post. The key arguments are outlined in the article The science of success by NewScientist. You may also enjoy the recent success story of the new billionaires behind WhatsApp from BBC. Here is a couple of the very inspiring quotes: “Got denied by Twitter HQ. That’s ok. Would have been a long commute.” “Facebook turned me down. It was a great opportunity to connect with some fantastic people. Looking forward to life’s next adventure.”

It is impossible to imagine in the UK anything similar to the school story above. Students are usually not called to solve unexpected problems in front of a class, even at university. Many vice-chancellors are proud of a 2% dropout rate at their universities. At the opposite extreme, we can mention Bear Grylls, who was one of 4 (among 180) candidates selected for SAS.

Weeping because of Pythagoras’ theorem

When we were running summer schools for the gifted and talented in the UK, we knew that students should not be asked, at least if they prefer to hide. Once we were discussing Pythagoras’ theorem with year 9-11 students (of age 14-16). Most students knew at least the first half of full Pythagoras’ theorem even if without a proof. However, one girl who was hiding quite well unfortunately decided that Pythagoras’ theorem is beyond her ability and started to cry.

This accident has taught us that we should teach Pythagoras’ theorem in  a much more careful way. That is why we have designed our free tutorial on Pythagoras’ theorem including its full statement, proofs in both directions and applications to diaphontine equations. Actually, our riddle on a proof of converse Pythagoras’ theorem is still open for more than 6 months. Hence those students who have enough motivation to self-register and master our free tutorial, can still solve this Pythagoras’ riddle and gain access to other tutorials on advanced topics.

Your personal choice: crying or learning

You could make your own conclusions from the cultural differences described above. We shall highlight only our own personal choice. We are human and also make mistakes, which is the human nature. Even robots fail and actually more frequently than humans. However, every failure makes us only stronger, because we learn not to repeat the same mistake again. And the harder we fail, the stronger we become. So we call all our failures the important life-changing experience. Here is the powerful principle: the harder the training, the easier the exam!

Our riddle is from an entrance exam to a Physical-Mathematical School at the age of 12-13.

• Riddle 13: factorise the polynomial $$6m^4-3m^3-4m^2+1$$.
• How to submit: to write your full answer, submit a comment.
• Hint: any factor of a polynomial gives a root and vice versa.
• Warning: justify that you found all factors with real coefficients.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.
• Update: Chen has excellently solved the riddle at first attempt.

If you wish to receive e-mails about our new posts (for a quicker chance to answer a riddle and win a prize) or distance courses, please contact us and tick the box “keep me updated”. You can easily unsubscribe at any time by e-mailing “unsubscribe” in the subject line.

Mistakes with square roots: easy, common, potential

We congratulate all readers of our blog on Chinese New Year!

In this post we discuss common mistakes with square roots of real numbers. As usual in mathematics, we start from the definition: the square root $$\sqrt{x}$$ is well-defined only for a non-negative real number $$x\geq 0$$ and equals a unique number $$y$$ such that $$x=y^2$$.

The graph of the square root function $$y=\sqrt{x}$$ is shown in the picture above. The function is always increasing and has the vertical tangent at the origin. If you understand derivatives, you may check that $$y'(x)\to+\infty$$ as $$0<x\to 0$$. We discuss only real numbers here. The square root of a complex number is another (interesting and more advanced) story.

Easy mistake: simplifying $$\sqrt{x^2}$$

Rather surprisingly, it is a very popular myth that $$\sqrt{x^2}=x$$. At least, this is the first answer we got at many summer schools and also from 2nd year maths undergraduates at a top UK university. Our next question was to check if the identity holds for $$x=-1$$. Then most students realised that $$\sqrt{x^2}=x$$ fails for $$x=-1$$.

Some students were still in doubt and claimed that $$\sqrt{1}$$ is $$1$$ or $$-1$$. We knew that all students have calculators at hand and asked them to compute $$\sqrt{1}$$ with their calculators. The expected correct answer $$\sqrt{1}=1$$ left no chance for the myth that $$\sqrt{x^2}=x$$ can hold for all real $$x$$.

Indeed, $$\sqrt{x^2}=|x|=\left\{\begin{array}{c} x \mbox{ for } x\geq 0,\\ -x \mbox{ for } x<0. \end{array} \right.$$ At one of the summer schools we were told that the absolute value $$|x|$$ is learned in the UK only at A-levels (at the age of 16-18). So there is little chance to get the correct identity $$\sqrt{x^2}=|x|$$. Later is better than never, so just in case we have sketched the graph $$y=|x|$$ on the left.

Common mistake: squaring both sides

In the riddle from the post about success at Oxbridge interviews we asked to find all real $$x$$ such that $$x\geq\sqrt{3x-2}$$. Unfortunately, public attempts 1 and 2 (and more private attempts by e-mail) started from squaring both sides: $$x^2\geq 3x-2$$.

The resulting inequality is not equivalent to the original one, because $$x=-1$$ satisfies $$x^2\geq 3x-2$$, but not $$x\geq\sqrt{3x-2}$$. So the popular myth that $$\sqrt{x^2}=x$$ has led to serious consequences that squaring both sides is a safe operation, but it is not!

When a given equation or inequality contains functions that are not well-defined for all real numbers, experts start from writing the domain when the problem makes sense. Best students always remember that a good style to finish a solution is to substitute answers back into an original equation.

However, this final check should not be considered as a part of a solution, but only as an opportunity to develop self-criticism, a key mathematical skill. Indeed, if we have infinitely many answers (as in the case of an inequality), we can not substitute back all our answers or such a substitution could be too hard. So if a final check fails, our solution fails, but the final check is ok.

For the inequality $$x\geq\sqrt{3x-2}$$, we first write that square root makes sense only when $$3x-2\geq 0,\; x\geq\frac{2}{3}$$. In this domain squaring both sides produces the equivalent inequality $$x^2\geq 3x-2,\; x^2-3x+2\geq 0,\; (x-1)(x-2)\geq 0$$.

The last inequality has the solutions $$x\leq 1$$ and $$x\geq 2$$. If we remember the domain $$x\geq\frac{2}{3}$$, the final answer is $$\frac{2}{3}\leq x\leq 1$$ and  $$x\geq 2$$. Just in case, we may check that the boundary values $$\frac{2}{3},1,2$$ satisfy the original inequality $$x\geq\sqrt{3x-2}$$.

Potential confusion: Q6(ii) in STEP I exam 2007

Here is the exact quote: “Given that $$x^3-y^3=(x-y)^4$$ and that $$x-y=d\neq 0$$, show that $$3xy=d^3-d^2$$. Hence show that $$2x=d\pm d\sqrt{\frac{4d-1}{3}}$$.”

Even without trying to solve the problem, any professional mathematician could spot within a few seconds that the expression $$\frac{4d-1}{3}$$ under the square root can be negative. Indeed, the only given restriction $$d\neq 0$$ allows $$d=-2$$ when we get the complex roots $$x=-1\pm i\sqrt{3}$$.

It is a good exercise in complex numbers to check that the pairs $$(x,y)=(-1+i\sqrt{3},1+i\sqrt{3})$$ and $$(x,y)=(-1-i\sqrt{3},1-i\sqrt{3})$$ actually satisfy the given equation $$x^3-y^3=(x-y)^4$$, so there is no mistake in the problem. However, complex numbers are not in the STEP I syllabus and we suspect that the case of complex roots was probably missed in the original problem.

The STEP examiners’ solution contains the identity $$\pm\sqrt{9d^2+12(d^3-d^2)}=\pm d\sqrt{12d-3}$$, which is luckily correct for all real d, but only if we use complex numbers with the sign $$\pm$$ in both sides. For real numbers, as was probably expected, we should write $$\sqrt{|9d^2+12(d^3-d^2)|}=|d|\sqrt{|12d-3|}$$, simply because the left hand side is not negative and so should the expressions under the square roots.

Unfortunately, there are other STEP problems when the sign $$\pm$$ doesn’t help. So one of the homework problem from our STEP I course has a parametric equation that can’t be solved by careless squaring both sides. The average mark for this homework is less than 16/20.

Just in case the picture above shows the curve $$x^3-y^3=(x-y)^4$$ excluding the line $$y=x$$ of easy solutions. The straight line $$x-y=d=\frac{1}{4}$$ meets the curve at the red point where $$x=\frac{1}{8}$$, hence $$y=-\frac{1}{8}$$. If $$d<\frac{1}{4}$$, the straight line $$x-y=d$$ doesn’t intersect the curve (in the real domain) and has only complex intersection points $$(x,y)$$, which are invisible on the usual plane.

• Riddle 12: find all real $$x,y$$ when the inequality $$\frac{x+y}{2}\geq\sqrt{xy}$$ holds.
• How to submit: to write your full answer, submit a comment.
• Warning: justify that you found all (not only some) real solutions.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.

If you wish to receive e-mails about our new posts (for a quicker chance to answer a riddle and win a prize) or distance courses, please contact us and tick the box “keep me updated”. You can easily unsubscribe at any time by e-mailing “unsubscribe” in the subject line.

Mistakes in trigonometry: easy, common, unexpected

We congratulate all Oxbridge candidates who have received their offers in January 2014!

In this post we discuss common mistakes in solving trigonometric equations.

Easy mistake: basic values often forgotten

If you remember basic values of trigonometric functions such as $$\sin\frac{\pi}{6}$$, $$\cos\frac{\pi}{4}$$, $$\tan\frac{\pi}{3}$$ without using a calculator, then at a real exam you will save time for more interesting problems.

A good exercise is to check the following basic values (for any integer n): $$\sin(n\pi)=0$$, $$\sin(\frac{\pi}{2}+n\pi)=(-1)^n=\cos(n\pi)$$, $$\tan(\frac{\pi}{4}+n\frac{\pi}{2})=(-1)^n=\cot(\frac{\pi}{4}+n\frac{\pi}{2})$$. You may check these identities for n=0, n=1 (possibly for n=2, n=3) and then use periodicity.

A good style to finish your solution to a trigonometric equation is to substitute your answers back into the original equation. If something is wrong, please don’t write “something is wrong”, but try to correct your computations. In our distance courses we give more specific tips to locate a computational error in a long chain of equations.

Common mistake: not all solutions are found

We have seen hundreds of scripts that contain “$$\sin x=0\Rightarrow x=0$$“. This logical implication fails, because x=π also satisfies the equation $$\sin x=0$$. Here is the full correct solution “$$\sin x=0, x=n\pi$$ for any integer n”.

We wouldn’t write the arrow $$\Rightarrow$$, because this logical implication doesn’t say that all found values are correct. For instance, the implication “$$\sin x=0 \Rightarrow x=n\frac{\pi}{2}$$ for any integer n” is also logically correct. So this confusing notation will be a topic for one of our future posts.

Since trigonometric functions are periodic, a trigonometric equation usually has infinitely many solutions. To make problems simpler, the authors of MAT (entrance exams to Oxford and Imperial College London) often specify a short range of expected solutions, usually from 0 to 2π.

Unexpected mistake: Q1 in STEP III exam 2007

Here is the exact quote: “The four real numbers θ1, θ2, θ3 and θ4 lie in the range $$0\leq \theta_i< 2\pi$$ and satisfy the equation $$p\cos(2\theta)+\cos(\theta-\alpha)+p=0$$, where p and $$\alpha$$ are independent of θ. Show that θ1234=nπ for some integer n.”

Let us consider the simple value of the parameter p=0. Then the given equation becomes $$\cos(\theta-\alpha)=0$$ and has only two real solutions in the range $$0\leq\theta<2\pi$$. For instance, if $$\alpha=\frac{\pi}{6}$$, then the two solutions of $$\cos(\theta-\frac{\pi}{6})=0$$ in the range $$0\leq\theta<2\pi$$ are
$$\theta=\frac{2\pi}{3}$$ and $$\theta=\frac{5\pi}{3}$$.

There is no way to assign these two values $$\frac{2\pi}{3}$$ and $$\frac{5\pi}{3}$$ to 4 variables whose sum θ1234 is a multiple of π. Indeed, if the value $$\frac{2\pi}{3}$$ is taken k (of 4) times, then the value $$\frac{5\pi}{3}$$ is taken 4-k times. Then $$k\frac{2\pi}{3}+(4-k)\frac{5\pi}{3}=n\pi$$, hence $$2k+5(4-k)=3n$$ and $$20=3(k+n)$$, which is impossible for integer k,n.

A brute-force alternative is to consider all 5 cases for possible values of the 4 angles $$0\leq\theta_1\leq\theta_2\leq\theta_3\leq\theta_4<2\pi$$ as follows.

(1) If $$\theta_1=\theta_2=\theta_3=\theta_4=\frac{2\pi}{3}$$, then $$\theta_1+\theta_2+\theta_3+\theta_4=\frac{8\pi}{3}$$.

(2) If $$\theta_1=\theta_2=\theta_3=\frac{2\pi}{3}$$ and $$\theta_4=\frac{5\pi}{3}$$, then $$\theta_1+\theta_2+\theta_3+\theta_4=\frac{11\pi}{3}$$.

(3) If $$\theta_1=\theta_2=\frac{2\pi}{3}$$ and $$\theta_3=\theta_4=\frac{5\pi}{3}$$, then $$\theta_1+\theta_2+\theta_3+\theta_4=\frac{14\pi}{3}$$.

(4) If $$\theta_1=\frac{2\pi}{3}$$ and $$\theta_2=\theta_3=\theta_4=\frac{5\pi}{3}$$, then $$\theta_1+\theta_2+\theta_3+\theta_4=\frac{17\pi}{3}$$.

(5) If $$\theta_1=\theta_2=\theta_3=\theta_4=\frac{5\pi}{3}$$, then $$\theta_1+\theta_2+\theta_3+\theta_4=\frac{20\pi}{3}$$.

In all 5 cases (1)-(5) above the sum $$\theta_1+\theta_2+\theta_3+\theta_4$$ is not a multiple of π, which disproves the statement from Q1 STEP III exam 2007. The STEP examiners’ solution to this problem is unfortunately too short and doesn’t consider exceptional cases when a denominator can be zero.

Rather surprisingly, neither the problem nor the solution were corrected in the official STEP resources for more than 6.5 years (checked on 24th January 2014). Fortunately, we have modified this problem and carefully considered all exceptional cases in our STEP III course.

• Riddle 11: solve the equation $$\sin x=a$$ for any real parameter a.
• How to submit: to write your full answer, submit a comment.
• Hint: there is a short formula for x depending on an integer n.
• Warning: a full answer should cover all possible values of a.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.
• Update: K Wright has solved the problem, see attempts 1 and 2.

If you wish to receive e-mails about our new posts (for a quicker chance to answer a riddle and win a prize) or distance courses, please contact us and tick the box “keep me updated”. You can easily unsubscribe at any time by e-mailing “unsubscribe” in the subject line.

The unlucky deduction: mathematics vs mathology

We wish a Merry Christmas and a Happy New Year 2014 to all mathematics enthusiasts!

The unlucky deduction in teacher training

This post is motivated by few news stories. The first one is a personal story from our friend who is completing a teaching qualification and had some practice in a UK private school.

On one day the future maths teacher was discussing many examples with a class how to find the length of a round arc using the angle at the centre. For instance, if r is the radius of the circle, the full angle 2π corresponds to the full circumcircle of length 2πr, where . The angle π gives only the half-circumcircle of length πr and so on.

We should mention that our friend has completed 5 years of proper undergraduate maths studies in another country, where students do not expect a lot of help from their lecturers and would consider such basic examples as a strange spoon feeding at the age of 13-14.

On the next day the hopeful trainee decided to revise and introduced the formula βr for the length of a round arc with a radius r and an angle β (in radians). Then the pupils were asked to refresh the yesterday computations by substituting simple angles β to get the required length of the round arc. At this moment the observer who was a qualified teacher immediately stopped the lesson and explained that in this country the maths teachers are not allowed to use general statements for deducing more specific results.

Deductive reasoning in mathematics.

Making a logical conclusion in a specific case from a general theorem is called deductive reasoning, or simply deduction (not subtraction). Mathematical induction is (in a sense) the opposite approach when a general claim is proved via easier base and inductive step.

If the deduction is forbidden, it seems very logical that most British students unfortunately struggle with a full statement of Pythagoras’ theorem, let alone its proof in both directions. That is why we designed our free web tutorial on Pythagoras’ theorem and applications.

The Britain is world-wide known for its famous Sherlock Holmes whose crime stories are actually based on deductive reasoning. Reading a few novels by Arthur Conan Doyle may help the authors of the UK school curriculum improve the career chances of UK students.

Deduction was accepted in Ancient Greece.

We are not professional novelists, but could tell a short mathematical story how deductive reasoning was formalised in Pythagoras’ school more than two thousands years ago in Ancient Greece. The word “geometry” is combined from the Greek words “geo” (earth) and “metron” (measurement). If you start to measure a square piece of land (as ancient Greeks certainly tried), the first obstacle is to get the ideal right-angle 0.5π (90 degrees).

The students who are familiar with full Pythagoras’ theorem could quickly give the practical recipe: measure the lengths of sides a,b,c of a triangle, if a2+b2=c2 then the angle opposite to side c is right (90 degrees). So converse Pythagoras’ theorem (not the usual direct theorem) was practical, because measuring lengths by a standard stick was easier than angles.

Using the basic deductive reasoning, we can conclude from direct Pythagoras’ theorem that the right-angled triangle with two smaller sides 1 has the longer side $$\sqrt{2}$$ (the square root of 2). This expression looks obvious for us now, but ancient Greeks operated only with rational numbers. Indeed, using fractions of a stick or a rope, they could get any rational number, but they couldn’t really measure $$\sqrt{2}$$. This “mysterious” number led to a philosophical crisis.

Does Ofsted hide the deductive reasoning from pupils to avoid such a crisis in British schools for health and safety reasons? The government will not worry, because most UK students take a calculator and use an approximate value, say 1.414213562373. However, in real life exact computations are much more valuable. Victims wouldn’t be happy after a fatal accident caused by the fact that 1.4142135623732<2, for instance in the space industry.

Mathematics vs mathology: one example.

Ancient Greece should be famous not only for widely known Greek mythology, but also for initiating rigorous logic, for proving first formal theorems and for making justified conclusions by deductive reasoning. We haven’t made a typo in the heading , but propose the term mathology for all improper numerical practices contrary to rigorous mathematics. So the comparison of mathematics vs mathology is very similar to astronomy (real science) vs astrology (rubbish pseudo-science). Mathology ignores the deductive reasoning.

A more specific example is the following question that is popular even in entrance tests at UK grammar schools. A few (usually four) objects are drawn and students are asked to select the odd one. This “problem” is mathological, because without extra restrictions any of four objects can be considered as the odd one by a properly justified mathematical argument.

These mathological “problems” are easily marked in a multiple-choice form and may not be suitable for bright individuals  who can “think outside the box”. As a result, parents pay special tutors who show their children how to guess standard patterns and pass mathological tests. A proper mathematical problem would be to find four different reasons when each of the given objects can be considered as the odd one, see our riddle below.

Congratulations on the PISA results!

According to the facts, the UK gained good international ranks, read BBC Education: PISA test (Programme for International Student Assessment): rank 26 in maths, rank 23 in reading, rank 21 in science. It remains to hope that the next results can be only better in 2016.

We wish a long life to all maths heros!

Finally, we highlight the real British Hero, read BBC Technology: royal pardon for codebreaker Alan Turing. Alan Turing is now considered as the father of modern computers, but was convicted for homosexuality in 1952, punished by being chemically castrated, lost his security clearence for code-breaking work and finally committed suicide in 1954.

In our opinion, by breaking the German code of the Enigma machine, Alan Turing contributed much more to the victory over nazis in the World War II than most generals and politicians. In fact, Churchill was the UK prime minister in 1940-1945 and then again in 1951-1955 and could had known about Alan Turing’s impact on the world history.

Here is the highest rated comment on the BBC news article (222 votes): “Somehow this seems the wrong way round – her majesty’s government should be asking for a pardon over Turing’s treatment, not granting one sixty years too late.” It is unfortunate, but not surprising, that the computer industry flourished after Alan Turing in another more tolerant valley, not around the Thames river. So we wish all contemporary mathematics enthusiasts to live long enough
and get rewarded for their contribution to the world knowledge (and peace).

• Riddle 10: find 4 different reasons when each of 4 geometric objects
(cycle, disk, square, sphere) in the picture above can be the odd one.