1

My teacher said that

If we have$$ f(x)=x^4 $$ Then there will be 4 same root $0$ satisfying the equation .

He said that it is because $$f'(x)=4x^3$$ $$f''(x)=12x^2$$ $$f'''(x)=24x^1$$ All are zero at $0$

Another example of such type of question is $$e^{x-1} -x = 0$$

I want to know why this happen

Aakash Kumar
  • 3,480
  • 2
    You want to know why what happen? That a function can have the same roots as its derivatives? Or why we call it a multiple root? Or something else? – Arthur Jul 13 '16 at 16:12
  • Well if a and b are distinct roots of a polynomial p(x) that means p(x) = (x - a)(x - b)q(x). For $f(x) = x^4$ then $x^4 = (x - 0)(x-0)(x-0)(x-0)$. Thus 0 is a multiple (quadruple) root. Similar for $(x -2)(x-2)(x-2) = x^3 - 6x^2 + 12x - 8$ has a multiple (triple) root of 3. – fleablood Jul 13 '16 at 16:13
  • Yes , this is what I wants – Aakash Kumar Jul 13 '16 at 16:13
  • @AakashKumar Do you know the definition of a multiple root? – Wojowu Jul 13 '16 at 16:17

5 Answers5

4

The equation $e^{x-1}-x=0$ has a double root at $x=1$ because:

for $x=1$ we have $y=e^{x-1}-x=\frac{e^x}{e}-x=\frac{e}{e}-1=0$ and also: $$ y'=\frac{e^x}{e}-1=\frac{e}{e}-1=0 $$

We say that a function $f(x)$ has a double root at $x=a$ if it can be factorized as $f(x)=(x-a)^2h(x)$ and, in this case we have $$ f'(x)=2(x-a)h(x)+(x-a)^2h'(x) $$ so that also $f'(a)=0$

On the other side, you can use Taylor series to show that if $f(a)=f'(a)=0$ than $f(x)$ can be factorized as $f(x)=(x-a)^2h(x)$.

Emilio Novati
  • 62,675
  • I want to know why this happens ? – Aakash Kumar Jul 13 '16 at 16:13
  • It happens because the math works. Do you want to know why we call it a multiple root? Do you want to know why if $p(x)$ is a polynomial it follows that $p(x)= (x - a)^n q(x)$ iff $p'(a) = 0; p''(a) = (0);... p'^{n-1}(a)=0$? What exactly are you asking.? – fleablood Jul 13 '16 at 16:21
  • I've added to my answer for a double root. You can extend to a multiple root . – Emilio Novati Jul 13 '16 at 16:26
  • @EmilioNovati How would you necessarily factor $e^{x-1} - x$ as the form $f(x) = (x-a)^2 h(x)$ though? It doesn't seem clear to me how to do it. – Aarony Jamesys Nov 12 '19 at 05:26
1

Read this: Multiplicity of a root of a polynomial (especially the answer by Bernard) to find some information about multiple roots.

For $e^{x-1}-x$: what are you asking about? It indeed vanishes at $x=1$ and the same happens to the derivative, but "multiple root" refers to polynomials.

Community
  • 1
tong_nor
  • 3,994
1

Every polynomial of degree 4 can be decomposed as $c(x-a_1)(x-a_2)(x-a_3)(x-a_4)$, where the $a_i$ are the roots of the polynomial.

It may happen that not all $a_i$ are distinct from each other, in which case one says one has a multiple root.

In particular this is the case for $x^4=(x-0)(x-0)(x-0)(x-0)$ which has $0$ as a 4-fold root.

user39082
  • 2,124
1

I think:

For a polynomial, $p(x) = a_nx^n + ..... + a_1x + a_0$ we say that $b$ is a root if $p(b) = 0$. This happens if and only if $p(x) = (x -b)(c_{n-1}x^{n-1} + .... + c_1x + c_0)$ where $(c_{n-1}x^{n-1} + .... + c_1x + c_0)$ is another polynomial.

Example $2,-2, 3$ are all roots to $p(x)= x^3 -3x^2 - 4x + 12$ and $p(x) = (x-2)(x+2)(x-3)$.

We say $b$ is a "multiple" root if $p(b) = 0$ and $p(b) = (x-b)(x-b)(c_{n-2}x^{n-2} + .... + c_1x + c_0)$ because $(x -b)$ does just divide into $p(x)$ once; it divides into it multiple times.

Now. Here is a neat thing about multiple roots of polynomials:

If $b$ is an "k multiple root of $p(x)$ that means $p(x) = (x - b)^kq(x)$ for some other polynomial $q(x)$. So $p'(x) = k(x - b)^{k-1}q(x) + (x - b)^kq'(x)$ and so $p'(b) = k(b-b)^{k-1}q(b) + (b -b)^kq'(b) = 0$. So not only is $p(b) = 0$ but $p'(b) = 0$.

We can repeat this down to the n-1 derivative.

Okay, that was because $p(x)$ is a polynomial.

$f(x) = e^{x-1} - x$ is not a polynomial. We can not factor $f(x)$. (Not for any significant purpose at any rate.) But $f(1) = e^{1-1} - 1 = 1-1 = 0$ so $1$ is a root of $f(x)$ because $f(1) = 0$.

Furthermore $f'(x) = e^{x-1} - 1$ and $f'(1) = e^{1-1} -1 = 0$. So both $f(1) = f'(1)=0$ so $1$ is a root of both $f(x)$ and $f'(x)$. Just as a double root for a polynomial was.

Although I haven't ever heard anyone call $1$ a "double root" of $f(x)$ but maybe some instructors do refer to "double roots" meaning $f(x) = f'(x) = 0$.

fleablood
  • 124,253
0

So, if I can understand you correctly, you want to know how we can have a function $f$ and a number $a$ so that $f(a) = f'(a) = 0$. The reason is exactly what it says in the equation: In addition to the funciton value being $0$, the derivative is also $0$, which means that the graph of the function aligns itself with the $x$-axis. So the graph not only crosses the $x$ -axis, but is actually tangent to it.

Arthur
  • 199,419