# Finding the number of solutions to a given function/equation

1. Oct 24, 2015

### Ocata

Hi,

Suppose I have a function, f(x). Is there a way to find how many solutions exist at a given x? For instance, suppose there is an equation,
f(x) = y = x^{2} - 1 = 0.

This is a parabola shifted down one unit and we can easily solve to find x = +1 and -1. And thus we know there are two solutions.

But is there a way to first arrive at the quantity 2, before (or instead of) actually solving the equation for x?

Thank you.

2. Oct 24, 2015

### aikismos

The answer is yes. (See https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra)

"The theorem is also stated as follows: every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n roots."

Essentially, think of it this way. For a quadratic, there are two solutions. For a cubic, there are three. For a quartic there are four. But there's a catch. Some of the solutions might be complex numbers which mean they aren't real numbers. They'll have imaginary portions.

Let's take an example. For quadratic equations (degree of two, that is they have a leading term $ax^2$) we have the quadratic formula and it has a discriminant.

$$D = b^2 - 4ac$$

In your equation, the coefficients are (a, b, c) = (1, 0 , -1), so we can plug it in and discriminate among the roots.

$$D = 0^2 - 4 \cdot 1 \cdot (-1) \rightarrow D = 4$$

If the discriminant is positive, there are two real roots. ($D > 0 \rightarrow 2$ real roots)
If the discriminant is zero, there is one real root. ( $D = 0 \rightarrow 1$ real root)
If the discriminant is negative, there are no real roots. ( $D > 0 \rightarrow 0$ real roots)

So, you should be able to get two roots, and in fact, it's trivial to show that $f(-1) = 0, f(1) = 0$ in your example. Two real roots at (-1, 0) and (1, 0).

For higher order polynomials, it gets more complicated. The best way for a beginner to see how many real roots exist is to graph with an application like payware Geometer sketchpad or freeware Geogebra . Wherever you have the curve cross the x-axis, you have a real root. To wit:

Last edited: Oct 24, 2015
3. Oct 24, 2015

### Ocata

Hi aikismos, thank you for responding. May I ask, what if I have a function:

f(t) = $\frac{-75}{5pi}cos(\frac{pi*t}{10})-t+5$

I graphed it on the program you suggested:

zoomed in:

Is it possible to determine the quantity of roots this function will have before graphing or solving for t?

4. Oct 24, 2015

### jbriggs444

There is the obvious way.

Take the first derivative of this function. Note that this derivative is periodic and solve for its two principle zeroes. The other zeroes can be found trivially. These are the places where the original function can have a local minimum or maximum. The original function must be either monotone increasing or monotone decreasing in the intervals between each pair of consecutive zeroes of its first derivative.

Evaluate the original function at the points where the first derivative has zeroes near -12, 2, 8 and 22. It is alternatively positive and negative at each of these points. So it must have a zero in each of the three intervals thus defined. It is strictly monotone in each of the intervals. So there can only be one zero in each interval.

There is a clear pattern in the values of the original function at its local minima and maxima. It is strictly positive at all such extreme points to the left of -12 and strictly negative at all extreme points to the right. So the three zeroes deduced above are the only ones that exist.

5. Oct 27, 2015

### Ocata

Hi jbriggs444,

Step one: Solve for the two principle zeros:

f(t) = $\frac{-75}{5pi}cos(\frac{pi*t}{10})-t+5$

f'(t) = $\frac{d}{dt} \frac{-75}{5pi}cos(\frac{pi*t}{10})-t+5$

= $\frac{7.5}{5}sin(\frac{pi*t}{10}) -1$ = 0

$[ sin(\frac{pi*t}{10}) = \frac{5}{7.5} ]$ = $[sin^{-1}(\frac{5}{7.5}) = \frac{pi}{10}t ]$ = $[\frac{10}{pi}sin(\frac{5}{7.5})$ = t_{1} = 2.322795272

and because of sin(x) = sin(pi - x) ===> $[\frac{pi}{10}t = pi ]$ => $[t = pi(frac{10}{pi})]$ = > $[t_{1} = 10]$

so t_{2} = 10 - 2.322795272 = 7.677204728

The two principle zeros are: t_{1} = 2.322795272 and t_{2} = 7.677204728

The original function: f(t) = $\frac{-75}{5pi}cos(\frac{pi*t}{10})-t+5$

The derivative function = f'(t) = $\frac{7.5}{5}sin(\frac{pi*t}{10}) -1$

so you are saying where ever f'(t) = 0, that will be where f(t) has a maximum or minimum, correct?

So since the f'(t) is periodic, then the same value will repeat every full period or cycle?

$[\frac{pi}{10}t = 2(pi)]$ => [t = $[2(pi)\frac{10}{pi}]$ => [t = 20]

So adding 20 to each of the principle values will produce the other t values where f(t) = 0.

t_{1} + 20 = 2.322795272 + 20 = 22.322795272 = t_{3}

t_{2} + 20 = 7.677204728 + 20 = 27.677204728 = t_{4}

t_{1} - 20 = 2.322795272 - 20 = -17.67720473 = t_{5}

t_{2} - 20 = 7.677204728 - 20 = -12.32279527 = t_{6}

Renaming all the values to reflect position, from least to greatest, on the t axis, we have:

$t_{1} = -17.677$, $t_{2} = -12.322$, $t_{3} = 2.322$, $t_{4} = 7.677$, $t_{5} = 22.322$, $t_{6} = 27.677$

f(t_{1}) = + 19.1184

f(t_{2}) = +20.8816

f(t_{3}) = -0.881608

f(t_{4}) =+0.881608

f(t_{5}) = -20.8816

f(t_{6}) = -19.1184

The corresponding graph would be:

And we don't know the exact t values where f(t) = 0. We only know that the t values where f(t) = 0 will fall somewhere between the minimums and maximums with alternating signs.

That is, f(t) = 0 when t is somewhere between -12 and +2.3, then somewhere between +2.3 and 7.6, and somewhere between 7.6 and 22.

How was it determined that these are the only points where f(t) = 0? How can it be determined that there are no x-intercepts to the left of -17 or to the right of +27?

Thank you

Last edited: Oct 27, 2015
6. Oct 28, 2015

### jbriggs444

In the range between t2 and t5, the original function is strictly monotone increasing or strictly monotone decreasing in each sub-interval. It can have at most one zero in each interval. This can be proven by using the mean value theorem.

The mean value theorem states that if a [differentiable] function f is defined over a closed interval [a,b] then there is at least one point x within the open interval (a,b) where $f'(x) = \frac{f(b)-f(a)}{b-a}$.

Suppose that there were two zeroes of our original function f within one of the three sub-intervals. Then by the mean value theorem, there would need to be a point between those two zeroes where f'(x) is equal to zero. But we've already identified all of the zeroes of f'. There are no extra zeroes to be found in any of these three sub-intervals. Contradiction. So the three zeroes we know about in this range are the only ones there are.

Now consider an arbitrary interval from tn to tn+1 for n less than or equal to 1. f(tn) and f(tn+1) are both clearly positive. f is either strictly monotone increasing or strictly monotone decreasing over each such interval. All of the values taken on by f over the interval must be between its values at the endpoints. So its value must be strictly positive throughout the interval. It cannot have a zero. f is strictly positive everywhere to the left of t2

The same argument applies to the right of t5. f will be strictly negative everywhere to the right of t5