#intermediate value theorem

The Intermediate Value Theorem guarantees that for every real number "y" between two outputs of a real continuous function f(x) there exists at least one value "c" between the corresponding inputs such that y = f(x) when x = c.

This is part two of a translation of an article that appeared in the February 2007 issue of Pythagoras.

The completeness of R, the set of real numbers, is the key to many theorems from Mathematical Analysis. You can use it to prove that many equations have a solution.

The Theorem

Can we distill a general theorem from yesterday’s post about the equation cos(x)=x? Yes we can and that theorem was formulated and proven in 1817 by the Czech mathematician Bernard Bolzano in an article with the wonderful title Rein analytischer Beweis des Lehrsatzes, daß zwischen je zwey Werthen, die ein entgegengesetztes Resultat gewähren, wenigstens eine reelle Wurzel der Gleichung liege. Here is the theorem, in Bolzano’s own words:

Lehrsatz. Wenn sich zwey Functionen von x, f(x) und φ(x), entweder für alle Werthe von x, oder doch für alle, die zwischen α und β liegen, nach dem Gesetze der Stetigkeit ändern, wenn ferner f(α)<φ(α), und f(β)>φ(β) ist: so gibt es jedesmahl einen gewissen zwischen α und β liegenden Werth von x, f&uumlr welchen f(x)=φ(x) wird.

(Very) freely translated: when two functions are continuous on an interval [α,β] and if in addition f(α)<φ(α) and f(β)>φ(β) then the equation f(x)=φ(x) has at least one solution between α and β.

There is one thing still undefined: what does it mean for a function to be continuous? Bolzano was probably the first to give a proper definition of that.

Nach einer richtigen Erklärung versteht man unter der Redensart, das eine Function f(x) für alle Werthe von x die inner- oder außerhalb gewisser Grenzen liegen, nach dem Gesetze der Stetigkeit sich ändre, nur so viel, daß, wenn x irgend ein solcher Werth ist, der Unterschied f(x+ω)-f(x) kleiner als jede gegebene Größe gemacht werden könne, wenn man ω so klein als man nur immer will, annehemen kan.

Translated freely: one says that a function f(x) varies for all values of x according to the laws of continuity if for each value of x the difference f(x+ω)-f(x) can be made smaller than any given magnitude by taking ω small enough.

In a more strict formulation, given by Weierstraß: for any given positive number ε there is an other positive number δ such that whenever |y-x|<δ one has |f(y)-f(x)|<ε.

Yesterday’s example dealt with f(x)=x, φ(x);=cos(x), α=0, and β=1. Both functions are indeed continuous. In both cases we can always take δ=ε, for we showed that always |cos(y)-cos(x)|≤|y-x|, so that if |y-x|<δ then |cos(y)-cos(x)|<ε.

The proof

The proof of the theorem is much like yesterday’s proof. We let A={y: α≤y≤β and f(y)if |y-x|then both |f(y)-f(x)| and |φ(y)-φ(x)| are smaller than ε (apply the definition to f and to φ and take the smaller of the two δs that you get).

Because x is the least upper bound of A there is y&in;A such that x-δ

Next take z such that x0.

This shows that -2εall positive ε and that means that f(x)=φ(x).

Squaring is continuous

Proving that a function is continuous is not always easy, as we saw with the cosine function: we needed a trigonometric formula and an estimate for |sin(y)|. Let us see how to show that the function g, given by g(x)=x2, is continuous. Fix x and a positive ε. How do we find a suitable positive δ?

Well, we first see what we can do with |y2-x2|, how we can relate it to |y-x|. We can factor it: |y2-x2|=|y+x|×|y-x|. That gives us |y-x| but there is also the variable factor |y+x|. Since we may take δ as small as necessary (as long as it is positive) we specify at the outset that it must not be larger than 1. Why? Because then |y-x|<δ will imply that |y-x|<1 and hence that |y+x| is less than the maximum of 2|x+1| and 2|x-1|, call that maximum M. This ensures that |y2-x2|≤M|y-x| when |y-x|<δ. Now we know how to specify δ given ε: take the minimum of 1 and ε/M;. Because then: if |y-x|<δ then |y2-x2|≤M|y-x| (because |y-x|<1) and M|y-x|

Conclusion: g(x)=x2 varies according to the laws of continuity.

Now let ψ(x)=2 (constant function), and α=0 and β=2. Then g(0)<ψ(0) and g(2)>&psi(2). So there is a solution of the equation g(x)=ψ(x) between 0 and 2. So Bolzano’s theorem shows the existence of √2 in an analytic, rather than geometric, way.

Why `Intermediate Value Theorem’?

Nowadays you will mostly see the the following theorem in many books:

Intermediate Value Theorem. If f is a continuous function on an interval [a,b] then f takes on all values between f(a) and f(b) on that interval.

It does not matter whether f(a)>f(b) or f(a)

This version follows from Bolzano’s version by comparing f with the constant function with value c.

The converse is also true: work with g=f-φ. Then g(α)<0 and g(β)>0, so there is an x between α and β such that g(x)=0, or f(x)=φ(x).

the Intermediate Value Theorem

Sometimes you just need to check the Intermediate Value Theorem is still there. It's been nine years since I started undergrad and fell in love with maths, and it's due to maths, friends and music that I have stayed alive this long.

This is part one of a translation of an article that appeared in the February 2007 issue of Pythagoras.

The completeness of R, the set of real numbers, is the key to many theorems from Mathematical Analysis. You can use it to prove that many equations have a solution. Here we illustrate what it says by way of an example. Next time we discuss the general statement.

In the previous five posts we used the completeness of R, the number line, to prove that all n-th roots exist, that 2x exists for all x, and that we can make logarithms.

The number line is complete. Every non-empty set of real numbers that has an upper bound has a least upper bound.

In a few more words: if A is a non-empty subset of R for which there is an x such that a≤x for all a∈A (x is an upper bound) then there is an upper bound a* for A that is smaller than all other upper bounds.

Solving equations

We will use this property of the number line to prove that the equation cos(x)=x has a solution. When you look at the graphs of y=cos(x) and y=x then you will `see' that there is a solution, but a picture does not qualify as a proof.

At x=0 we see that x lies below cos(x) (for x=0 and cos(0)=1) and at x=1 we have cos(x) below x (for cos(1)<1). You would expect that the graphs intersect between 0 and 1 and that there should be a number α for which cos(α)=α.

To turn that expectation in certainty we can do two things. We can try to solve the equation is some way or another but when you try this you will find it hard to get a handle on the equation and that you are getting nowhere. The other thing we can do is prove that there must be a solution and give a way to get good approximations of α.

Finding a candidate α

Consider the set A={x:x<cos(x)}. This set is not empty, because 0∈A. The set also has an upper bound: 1∈A and if x>1 then certainly x>cos(x) (because cos(x)≤1).

This gives us a candidate for α: the least upper bound of A. For, just below α there are x-es with x<cos(x) and just above α there are x-es with x≥cos(x). This should imply that α=cos(α).

Proving that cos(α)=α

To prove that α=cos(α) we need a trigonometric formula, to wit

Using this we can deduce that cos(x) is close to cos(α) whenever x is close to α. The absolute value of the difference is equal to

And because the absolute value of the sine function is at most 1 we know that this is not larger than

Furthermore, as the picture below illustrates we always have |sin(y)|≤|y|

We find that always

With this we can show that cos(α)≥α and cos(α)≤α and hence that cos(α)=α.

Take a natural number n and pick x∈A such that x>α-2-n. We rewrite the difference cos(α)-α as the sum of three differences: cos(α)-cos(x), cos(x)-x, and x-α. Because |cos(α)-cos(x)|≤|x-α| we find that the sum of the three differences is larger than -2×2-n: cos(x)-x is positive and both cos(α)-cos(x) and x-α are larger than -2-n.

It follows that cos(α)-α is larger than -2×2-n for all n. But this then means that cos(α)-α is at least as large as the least upper bound of the numbers -2×2-n, which is 0. We find that cos(α)-α≥0.

Exercise. Show in the same way that cos(α)-α≤0.

An algorithm

We can now describe an algorithm to find approximations to α. What we do is cut the interval [0,1] in two halves, [0,½] and [½,1], and determine which half contains α. Then cut that half in halves, see where α must lie, and so on, until we are satisfied with the accuracy to which we have approximated α. A measure for that accuracy is the length of the interval that we find at each step. At step 0 (when we start) that length is 1-0=1, at step 1 it is ½ at step 2 we have ¼, …, at step i we have length (½)i.

It is easy to write a program for this: fix your desired accuracy ε, and set a0=0 and b0=1. Now repeat the following recipe.

set mi=½(ai-1+bi-1) and calculate cos(mi)

if cos(mi)=mi then PRINT "the solution is mi" and STOP

if cos(mi)>mi then set ai=mi and bi=bi-1

if cos(mi)<mi then set ai=ai-1 and bi=mi

if bi-ai<ε then STOP and PRINT "½(ai+bi) is the desired approximation"

the first `if' is there to see if we are lucky and have stumbled on the exact solution; the next two `if's ensure that always cos(ai)>ai and cos(bi)<bi and hence that α lies between ai and bi. The final `if' tests to see if we have reached the desired accuracy.

Exercise. How many iterations are needed when ε=1/1000? When ε=1/1000000?

The algorithm described above is known, for obvious reasons, as the bisection method. It will always give good approximations but it is not very efficiënt.

Introduction against intermediate meaning theorem for derivatives:Intermediate fair-trade theorem says that ' A round-the-clock operational purpose on a closed and bounded interval attains every value between any two taken for granted points in the range. ' Look at the following example,Repression f: ]0, 5] `->` `RR` happen to be defined nearby f(x) = `1\(x-3)` + 2 for x `!=` 3.We have f(0) = `5\3` and f(5) = `5\2`. Now we have `5\3` `Here f is not continuous. So consonance of f is crazy important to apply inteermediate value theorem.Except once you learn ' intermediate value theorem for derivates' anon other self will obtain astonished to see that though we are not satisfied of continuity of a function, the article satisfies the front man value property. This happens so as to some special functions. Those are differentiable functions and we will be present talking anywise the intermediate high order demesne for their derivatives.<\p>