Quartic Functions

A function is quartic if the highest order term of the variable $x$ is to the fourth power $x^4$. A quartic function can be written in expanded form

$$y=ax^4+bx^3+cx^2+dx+e$$

depressed form

$$\frac{y}{a}=t^4+pt^2+qt+s$$

or factorized form

$$y=a(x-r_1)(x-r_2)(x-r_3)(x-r_4)$$

where $x$ is the variable, $a$, $b$, $c$, $d$ and $e$ are coefficients, $t$ is the variable of the depressed form and $p$, $q$ and $s$ are its coefficients, $r_1$, $r_2$, $r_3$ and $r_4$ are the roots of the quartic.

The quartic function is plotted below where you can modify the coefficients and roots.

The local minimum and maximum can be found by taking the derivative and setting it to zero.

$$\frac{dy}{dx}=4ax^3+3bx^2+2cx+d=0$$

which is a cubic so it can be solved using the cubic formula to get the stationary points.

$$x_n=-\frac{b}{4a}+\left(\frac{-Q\pm\sqrt{Q^2+\frac{4P^3}{27}}}{2}\right)^{\frac{1}{3}}e^{\frac{2\pi i n}{3}}-\frac{P}{3\left(\frac{-Q\pm\sqrt{Q^2+\frac{4P^3}{27}}}{2}\right)^{\frac{1}{3}}e^{\frac{2\pi i n}{3}}}, \forall n\in{0,1,2}$$

where

$$P=\frac{8ac-3b^2}{16a^2}$$

$$Q=\frac{b^3-4abc+8a^2d}{32a^3}$$

The stationary points can be shown to be local minimums or maximums by substituting their location into into the second derivative and noting the sign:

• positive: local minimum,
• zero: inflection point, and
• negative: local maximum.

$$\frac{d^2y}{dx^2}=12ax^2+6bx+2c$$

The depressed quartic function is found by taking derivatives of the expanded quartic function

$$y=ax^4+bx^3+cx^2+dx+e$$

until a linear function derivative is found as we did with the cubic and could have done with the quadratic.

$$\frac{dy}{dx}=4ax^3+3bx^2+cx+d$$

$$\frac{d^2y}{dx^2}=12ax^2+6bx+c$$

$$\frac{d^3y}{dx^3}=24ax+6b$$

Letting $\frac{d^3y}{dx^3}=0$ as we are looking for a point which can be thorough of as a midpoint and solving for $x$ gives

$$x=\frac{-6b}{24a}=\frac{-b}{4a}$$

making the substitution $x=t-\frac{b}{4a}$ to shift the quartic function to this mid point gives

$$y=a\left(t-\frac{b}{4a}\right)^4+b\left(t-\frac{b}{4a}\right)^3+c\left(t-\frac{b}{4a}\right)^2+d\left(t-\frac{b}{4a}\right)+e$$

expanding, dividing both sides by $a$ and collecting like terms of $x$ gives the depressed quartic function

$$\frac{y}{a}=t^4+\frac{8ac-3b^2}{8a^2}t^2+\frac{8a^2d-4abc+b^3}{8a^3}t+\frac{256a^3e-64a^2bd+16ab^2c-3b^4}{256a^4}$$

typically written as

$$\frac{y}{a}=t^4+pt^2+qt+s$$

where

$$p=\frac{8ac-3b^2}{8a^2}$$

$$q=\frac{8a^2d-4abc+b^3}{8a^3}$$

$$s=\frac{256a^3e-64a^2bd+16ab^2c-3b^4}{256a^4}$$

The rational roots of a quartic can be found using the factorizing method for any order polynomial. Expect this method to only work on the specially crafted problems that get set as test questions in mathematics causes. You will also want to know polynomial division to get the remaining roots (assuming you find one or more root using factorizing).

I recommend using newtons method to solve quartic functions in any hand calculations as it requires the least amount of input into a calculator (minimizing your chances of making a mistake and the question is solved quickly). The method can find all roots but your guess needs to be close to a particular solution to converge to that solution. With a random guess, you are likely to find at least one root. (If there are no real roots and your guess is real it will jump around not converging to a solution.) Finding all roots using this method is possible but it may be difficult to find a starting guess which will converge on all the different roots. You can use polynomial division to address this issue. Newton’s method is

$$x_\text{new}=x-\frac{y}{\left(\frac{dy}{dx}\right)}$$

so for the quartic function, the roots can be found using Newton’s method with an appropriate starting point and replacing $x$ on the right-hand side with ans on the calculator

$$x_\text{new}=x-\frac{ax^4+bx^3+cx^2+dx+e}{4ax^3+3bx^2+2cx+d}$$

This method also works if the solution is a complex number (assuming your calculator supports complex numbers), you just need to give it a complex initial guess.

What should the initial guess be?

Any number should work provided it does not cause serious numerical issues such as a guess resulting in the maximum value the calculator can represent being exceeded. But guesses further from a root will require more iterations. This is not really an issue when using a calculator as you may need to press the equals sign seven times rather than five but if you were doing this without a calculator or solving a billion of these on a computer saving a few iterations will make a big difference in the computation time. To choose a reasonable guess we could find what the maximum and minimum root could plausibly be. As this works for any order polynomial we will find it for the general polynomial and assuming $z$ is a root and $z\ge1$

$$0=a_nz^n+a_{n-1}z^{n-1}+…+a_1z+a_0$$

subtracting the leading term from both sides

$$-a_nz^n=a_{n-1}z^{n-1}+…+a_1x+a_0$$

$$-a_nz^n=\sum_{k=0}^{n-1}a_kz^k$$

applying an absolute value function to both sides

$$|a_n||z^n|=\left|\sum_{k=0}^{n-1}a_kz^k\right|$$

taking the absolute maximum leading coefficient out of the sum and taking the absolute value of the root results in the inequality

$$|a_n||z^n|\le\max(|a_k|)\sum_{k=0}^{n-1}|z^k|$$

multiplying both sides by $|z|-1$

$$(|z|-1)|a_n||z^n|\le\max(|a_k|)(|z|-1)(|z^0|+|z^1|+…+|z^{n-1}|)$$

expanding the right hand side

$$(|z|-1)|a_n||z^n|\le\max(|a_k|)(|z|+…+|z|^{n}-|1|-|z|-…-|z^{n-1}|)$$

combining like terms cancels out all but two terms

$$(|z|-1)|a_n||z^n|\le\max(|a_k|)(|z^n|-1)$$

the left hand side is still less than the right if we do not subtract the one on the right hand side

$$(|z|-1)|a_n||z^n|<\max(|a_k|)|z^n|$$

solving for $z$ gives a bound for the maximum and minimum value a root can take

$$(|z|-1)|a_n|<\max(|a_k|)$$

$$|z|<1+\max\left(\left|\frac{a_k}{a_n}\right|\right) \forall k\in\{0,..,n-1\}$$

If an initial guess of the root of the quartic is chosen between

$$-1-\max\left(\left\{\left|\frac{b}{a}\right|,\left|\frac{c}{a}\right|,\left|\frac{d}{a}\right|,\left|\frac{e}{a}\right|\right\}\right)<x<1+\max\left(\left\{\left|\frac{b}{a}\right|,\left|\frac{c}{a}\right|,\left|\frac{d}{a}\right|,\left|\frac{e}{a}\right|\right\}\right)$$

it will be in the range of a potential root. There are other ways of calculating the bounds the roots are between, potentially with tighter bounds but this method has a minimal amount of calculation.

The first step to obtain a formula to find the roots of the quartic function

$$y=ax^4+bx^3+cx^2+dx+e$$

is to find its depressed form as found in a previous section.

$$\frac{y}{a}=t^4+pt^2+qt+s$$

where

$$p=\frac{8ac-3b^2}{8a^2}$$

$$q=\frac{8a^2d-4abc+b^3}{8a^3}$$

$$s=\frac{256a^3e-64a^2bd+16ab^2c-3b^4}{256a^4}$$

The incite required to find the roots of the quartic function is that a quartic function is a multiplication of two quadratic functions. If we let the depressed quartic function equal the multiplication of two quadratics

$$0=t^4+pt^2+qt+s=(t^2+mt+n)(t^2+ut+v)$$

we need to solve for $m$, $n$, $u$, and $v$ to obtain the roots of the quartic function. No leading coefficient on the quadratic functions is needed as the depressed quartic function does not have a leading coefficient. The roots of the depressed quartic function are the roots of the two quadratic functions, using the quadratic formula they are

$$t=\begin{cases}\frac{-m\pm\sqrt{m^2-4n}}{2}\\\frac{-u\pm\sqrt{u^2-4v}}{2}\end{cases}$$

However, we still need to find $m$, $n$, $u$, and $v$ to do this we begin by expanding the right-hand side of

$$t^4+pt^2+qt+s=(t^2+mt+n)(t^2+ut+v)$$

$$t^4+pt^2+qt+s=t^4+ut^3+vt^2+mt^3+mut^2+mvt+nt^2+nut+nv$$

$$t^4+pt^2+qt+s=t^4+(m+u)t^3+(mu+n+v)t^2+(mv+nu)t+nv$$

which gives us four equations to find the four unknowns $m$, $n$, $u$ and $v$ as different powers of $t$ can not be added, they are

$$0=m+u$$

$$p=mu+n+v$$

$$q=mv+nu$$

$$s=nv$$

These equations will be solved by rearranging each one to make one letter the subject then substituting the result into the next one, beginning with the first equation

$$m=-u$$

substituting the result $m$ into the next equation

$$p=mu+n+v$$

$$p=-u^2+n+v$$

solving for $v$

$$v=p+u^2-n$$

substituting $m$ and $v$ into the next equation

$$q=mv+nu$$

$$q=-u(p+u^2-n)+nu$$

solving for $n$

$$q=-pu-u^3+nu+nu$$

$$q=-pu-u^3+2nu$$

$$n=\frac{u^3+pu+q}{2u}$$

substituting $n$ into the equation for $v$ found earlier

$$v=p+u^2-n$$

$$v=p+u^2-\frac{q+up+u^3}{2u}$$

$$v=\frac{u^3+pu-q}{2u}$$

and substituting $n$ and $v$ into the final equation

$$s=nv$$

$$s=\frac{u^3+pu+q}{2u}\frac{u^3+pu-q}{2u}$$

which can be rearranged to obtain a cubic equation with the substitution $u^2=U$

$$4u^2s=u^6+pu^4-qu^3+pu^4+p^2u^2-pqu+qu^3+pqu-q^2$$

$$0=u^6+2pu^4+(p^2-4s)u^2-q^2$$

$$0=U^3+2pU^2+(p^2-4s)U-q^2$$

This cubic is a resolvent cubic of the depressed quartic. Using one version of the cubic formula the solution is

$$U=\frac{-2p}{3}+V-\frac{P}{3V}$$

where

$$V=\left(\frac{-Q\pm\sqrt{Q^2+\frac{4P^3}{27}}}{2}\right)^{\frac{1}{3}}\exp\left(\frac{2\pi in}{3}\right)$$

$$P=-\frac{p^2+12s}{3}=\frac{3bd-12ae-c^2}{3a^2}$$

$$Q=\frac{72ps-2p^3-27q^2}{27}=\frac{72ace-27ad^2-27b^2e+9bcd-2c^3}{27a^3}$$

and $u=\pm\sqrt{U}$ where either solution can be chosen as it only switches the order of the cases. Any solution of $U$ which does not equal zero can be used. Substituting these results back into our formula for the roots of the depressed quartic using the results obtained to find $u$: $m=-u$, $n=\frac{u^2}{2}+\frac{p}{2}+\frac{q}{2u}$ and $v=\frac{u^2}{2}+\frac{p}{2}-\frac{q}{2u}$

$$t=\begin{cases}\frac{-m\pm\sqrt{m^2-4n}}{2}\\\frac{-u\pm\sqrt{u^2-4v}}{2}\end{cases}$$

$$t=\begin{cases}\frac{u\pm\sqrt{-u^2-2p-\frac{2q}{u}}}{2}\\\frac{-u\pm\sqrt{-u^2-2p+\frac{2q}{u}}}{2}\end{cases}$$

The roots of the general quartic

$$y=ax^4+bx^3+cx^2+dx+e$$

can be found by the substitution $t=x+\frac{b}{4a}$ to reverse the linear transform used to find the depressed quartic function. We will also let $S=\frac{u}{2}$ to present the formula how it is typically written

$$x=-\frac{b}{4a}+\begin{cases}S\pm\frac{1}{2}\sqrt{-4S^2-2p-\frac{q}{S}}\\-S\pm\frac{1}{2}\sqrt{-4S^2-2p+\frac{q}{S}}\end{cases}$$

where

$$p=\frac{8ac-3b^2}{8a^2}$$

$$q=\frac{8a^2d-4abc+b^3}{8a^3}$$

and $S$ is half the square root of the solution to the resolvent cubic we found earlier

$$S=\frac{1}{2}\sqrt{U}=\frac{1}{2}\sqrt{\frac{-2p}{3}+V-\frac{P}{3V}}$$

where

$$V=\left(\frac{-Q\pm\sqrt{Q^2+\frac{4P^3}{27}}}{2}\right)^{\frac{1}{3}}\exp\left(\frac{2\pi in}{3}\right)$$

$$P=\frac{3(p^2-4s)-4p^2}{3}=\frac{3bd-12ae-c^2}{3a^2}$$

$$Q=\frac{16p^3-18p(p^2-4s)-27q^2}{27}=\frac{72ace-27ad^2-27b^2e+9bcd-2c^3}{27a^3}$$

For $S$ any cube root can be chosen and therefore any $n\in\mathbb{Z}$ can be used. Also, any cube root formula can be used to solve the resolvent cubic. If the real roots are to be found using only real numbers the trigonometric or hyperbolic cubic formulas are required and a positive resolvent cubic root needs to be chosen. If $q=0$ the cubic is biquadratic can be solved using the quadratic formula.

You can try to solve the quartic formula yourself. Be aware of the comments I made for the quadratic and cubic if you choose to solve them as well. You will need to input the equations relating $m$, $n$, $u$ and $v$ to $p$, $q$ and $s$ as separate equations. It’s probably not a good idea to substitute in the result of the resolvent cubic before you have finished as you will get an intractable mess after it is substituted.

The nature of the roots of a quartic function depends on its discriminant. As the discriminant was not obtained as part of the solution process as it was with the quadratic and cubic we will need to calculate it from its definition. Using the method to calculate the discriminant for any order polynomial the discriminant is given by

$$\Delta=\frac{(-1)^\frac{n^2-n}{2}\text{res}\left(y,\frac{dy}{dx}\right)}{a_n}$$

Substituting in the quartic function properties and evaluating

$$\Delta=\frac{\text{det}\left(\begin{bmatrix}a&b&c&d&e&0&0\\0&a&b&c&d&e&0\\0&0&a&b&c&d&e\\4a&3b&2c&d&0&0&0\\0&4a&3b&2c&d&0&0\\0&0&4a&3b&2c&d&0\\0&0&0&4a&3b&2c&d\end{bmatrix}\right)}{a}$$

$$\Delta=256a^3e^3 – 192a^2bde^2 – 128a^2c^2e^2 + 144a^2cd^2e – 27a ^2d^4 + 144ab^2ce^2 – 6ab^2d^2e\\- 80abc^2de + 18abcd^3 + 16ac^4 e – 4ac^3d^2 – 27b^4e^2 + 18b^3cde – 4b^3d^3 – 4b^2c^3e + b^2c^2d^2$$

This discriminant can also be shown to equal the discriminant of the resolvent cubic multiplied by $a^6$

$$\Delta=-\left(4P^3+27Q^2\right)a^6$$

where

$$P=\frac{3bd-12ae-c^2}{3a^2}$$

$$Q=\frac{72ace-27ad^2-27b^2e+9bcd-2c^3}{27a^3}$$

Using the root definition of the discriminant for a quartic function

$$\Delta=a^{6}\prod^4_{\substack{i=1,j=1\\i<j}}(r_i-r_j)^2=a^6(r_1-r_2)^2(r_1-r_3)^2(r_1-r_4)^2(r_2-r_3)^2(r_2-r_4)^2(r_3-r_4)^2$$

where $r$ is a root of the quartic the following is observed:

• If any root is repeated the discriminant is zero as a term of $r_i-r_j$ will equal zero.
• If all roots are real numbers the discriminant is positive as all real numbers squared are positive.
• If all roots are complex numbers the discriminant is also positive because the difference in roots will either be a complex number without a real part in two cases and two negatives multiplied to give a positive or form a complex contingent with one of the other four cases which when multiplied give a positive number.
• If two roots are real distinct numbers and the other two complex numbers, the discriminant is negative because one of the differences will be a real number, positive after squaring, another complex without a real part which when squared is negative and the remainder form a complex contingent with one of the other four cases, positive after multiplying. This results in one bracket being negative so the discriminant is negative.

To further classify the roots we need to look at the case when the depressed quartic function has two double roots. This will happen when $q=0$ and $(p^2-4s)=0$. We need to calculate $V$ for this case by first substituting $q=0$ and $(p^2-4s)=0$ into $P$ and $Q$

$$P=\frac{-4p^2}{3}$$

$$Q=\frac{16p^3}{27}$$

$$V=\left(\frac{-\frac{16p^3}{27}\pm\sqrt{\left(\frac{16p^3}{27}\right)^2+\frac{4\left(\frac{-4p^2}{3}\right)^3}{27}}}{2}\right)^{\frac{1}{3}}$$

$$V=\left(-\frac{8p^3}{27}\right)^{\frac{1}{3}}$$

$$V=-\frac{2p}{3}$$

Now we calculate $S$ for this case by substituting in the values for $V$ and $P$

$$S=\frac{1}{2}\sqrt{\frac{-2p}{3}-\frac{2p}{3}-\frac{2p}{3}}=\frac{1}{2}\sqrt{-2p}$$

we now use this result to find $V$ whenever $S=\frac{1}{2}\sqrt{-2p}$

$$\frac{1}{2}\sqrt{-2p}=\frac{1}{2}\sqrt{\frac{-2p}{3}+V-\frac{P}{3V}}$$

$$-2p=\frac{-2p}{3}+V-\frac{P}{3V}$$

$$-6pV=-2pV+3V^2-P$$

$$0=3V^2+4pV-P$$

$$V=\frac{-2p\pm2\sqrt{4p^2+3P}}{3}$$

and the discriminant of this quadratic for $V$ is

$$D=4p^2+3P$$

substituting in $p$ and $P$, expanding, and multiplying both sides by $\frac{16a^4}{3}$

$$D=4\left(\frac{3bd-12ae-c^2}{3a^2}\right)^2+3\frac{3bd-12ae-c^2}{3a^2}$$

$$\frac{16a^4D}{3}=-64a^3e+16a^2bd+16a^2c^2-16ab^2c+3b^4$$

gives the discriminant of the quadratic in another form without a denominator. This quadratic discriminant (either form) in combination with other constants allows us to fully classify the roots as follows:

• when $\Delta>0$ and
  • $p<0$ and $D>0$ there are four distinct real number roots.
    
  • $p>0$ or $D<0$ there are two pairs of complex conjugate roots.
    
• when $\Delta<0$ there are two distinct real number roots and two complex conjugate roots.
  
• when $\Delta=0$ there are repeated roots as follows:
  • when $p<0$ and $D>0$ and $P\ne0$ there are 2 distinct real number roots and the third is a repeated root real number root.
    
  • when $D<0$ or $p>0\cap(D\ne0\cup q\ne0)$ there are two complex conjugate roots and the third is a repeated root real number root.
    
  • when $P=0$ and $D\ne0$ there is a triple real number root and another real number root.
    
  • when $D=0$ and
    • $p<0$ there are two real number double roots.
      
    • $p>0$ and $q=0$ there are two complex conjugate double roots.
      
    • $P=0$ there are four repeated real roots equal to $-\frac{b}{4a}$.
      

The intersections between a quartic function

$$y=ax^4+bx^3+cx^2+dx+e$$

and another quartic, cubic, quadratic, linear or constant function

$$y=jx^4+kx^3+lx^2+mx+n$$

where the leading coefficients can be set to zero to obtain the lower order polynomials, can be found by substituting $y$ from one equation into the other

$$ax^4+bx^3+cx^2+dx+e=jx^4+kx^3+lx^2+mx+n$$

rearranging to obtain another quartic equal to zero

$$(a-j)x^4+(b-k)x^3+(c-l)x^2+(d-m)x+e-n=0$$

which can be solved for $x$ using any of the methods on this page. Once the values of $x$ which solve the above equation have been found they can be substituted into either $y=ax^4+bx^3+cx^2+dx+e$ or $y=jx^4+kx^3+lx^2+mx+n$ to obtain the $y$ coordinates of the intersection location.

To find the equation of a quartic from data points five coordinates are required lets choose $(w,f)$, $(g,h)$, $(j,k)$, $(m,n)$ and $(u,v)$. Substitute these points into $x$ and $y$ of the quartic equation $y=ax^4+bx^3+cx^2+dx+e$ to obtain the set of equations:

$$f=aw^4+bw^3+cw^2+dw+e$$

$$h=ag^4+bg^3+cg^2+dg+e$$

$$k=aj^4+bj^3+cj^2+dj+e$$

$$n=am^4+bm^3+cm^2+dc+e$$

$$v=au^4+bu^3+cu^2+du+e$$

To solve for $a$, $b$, $c$ $d$ and $e$ we could solve for one variable then substitute it into the next equation solve for the next variable and so on (and probably get it wrong by forgetting to copy something in the mess this will create) or assemble the equation set into matrices.

$$\begin{bmatrix}f\\h\\k\\n\\v\end{bmatrix}=\begin{bmatrix}w^4&w^3&w^2&w&1\\g^4&g^3&g^2&g&1\\j^4&j^3&j^2&j&1\\m^4&m^3&m^2&m&1\\u^4&u^3&u^2&u&1\end{bmatrix}\begin{bmatrix}a\\b\\c\\d\\e\end{bmatrix}$$

left multiplying both sides by the inverse coefficient matrix gives the coefficients of the quartic passing through the four points

$$\begin{bmatrix}a\\b\\c\\d\\e\end{bmatrix}=\begin{bmatrix}w^4&w^3&w^2&w&1\\g^4&g^3&g^2&g&1\\j^4&j^3&j^2&j&1\\m^4&m^3&m^2&m&1\\u^4&u^3&u^2&u&1\end{bmatrix}^{-1}\begin{bmatrix}f\\h\\k\\n\\v\end{bmatrix}$$

If you solve this by hand, use Gaussian elimination, and it will be very tedious. I recommend using octave online and replace the letters with numbers first or your result will not fit on the screen.

In the following examples, we will be finding the coefficients of a quartic function using different conditions at the endpoints. Rather than just solving for the stated conditions we will solve for the deflection of beams under uniform load, such as their own weight.

The deflection of a beam is given by

$$\frac{d}{dx^2}\left(EI\frac{d^2y}{dx^2}\right)=w$$

where $E$ is Young’s modulus the stiffness of the material and $I$ is the second moment of area, a property of the shape of the beam. We will take both of these values as constants. When the load $w$ is applied along the entire length of the beam and is constant, the deflected shape of the beam $y$ is given by a quartic function. This function is typically solved by integrating the load four times applying the conditions after each integral. Each integral of the load has a physical meaning and a name. The first integral is called the shear force, the second is called the bending moment, the third is the slope and the fourth is the deflected shape of the beam. As the focus here is on finding the coefficients of a quartic function not on finding the deflected shape of a beam with a generic loading we will set up our equations from the quartic function rather than integrating. We will consider the following cases:

• A simply supported beam, a beam that rests between two supports. Many bridges are supported this way.
  
• A cantilever, a beam where one end is prevented from moving or rotating and the other end is free to move. A balcony is an example.
  
• A propped cantilever, a beam where one end is prevented from moving or rotating and the other end is supported but not clamped so it can still rotate.
  
• A beam fixed at both ends. Rotation and displacement are prevented at both ends.
  

Simply Supported Beam

To solve for the coefficients of

$$EIy=ax^4+bx^3+cx^2+dx+e$$

where $EI$ can be taken out of the integral as it is a constant, for a simply supported beam with length $L$ and uniform load $w$ the constraints are:

• Both ends can not move vertically $$y(0)=0=e$$ $$y(L)=0=aL^4+bL^3+cL^2+dL+e$$
• There is no resistance to rotation at both ends (the bending moment or second derivative is zero at the ends) $$\frac{d^2y(0)}{dx^2}=0=2c$$ $$\frac{d^2y(L)}{dx^2}=0=12aL^2+6bL+2c$$
• The difference in shear force (third derivative) at the ends equals the applied load $$\frac{d^3y(L)}{dx^3}-\frac{d^3y(0)}{dx^3}=wL=24aL+6b-6b$$

Assembling these constraint equations into a matrix and rearranging for the coefficients gives

$$\begin{bmatrix}a\\b\\c\\d\\e\end{bmatrix}=\begin{bmatrix}0&0&0&0&1\\L^4&L^3&L^2&L&1\\0&0&2&0&0\\12L^2&6L&2&0&0\\24L&0&0&0&0\end{bmatrix}^{-1}\begin{bmatrix}0\\0\\0\\0\\wL\end{bmatrix}$$

Therefore the equation for the displacement of the simply supported beam is

$$y=\frac{w}{24EI}x^4-\frac{wL}{12EI}x^3+\frac{wL^3}{24EI}x$$

Cantilever

To solve for the coefficients of

$$EIy=ax^4+bx^3+cx^2+dx+e$$

for a cantilever with length $L$ and uniform load $w$ the constraints are:

• One end can not move vertically $$y(0)=0=e$$
• One end can not rotate $$\frac{dy(0)}{dx}=0=d$$
• There is no resistance to rotation at the free end (the bending moment or second derivative is zero at one end) $$\frac{d^2y(L)}{dx^2}=0=12aL^2+6bL+2c$$
• The difference in shear force (third derivative) at the ends equals the applied load $$\frac{d^3y(L)}{dx^3}-\frac{d^3y(0)}{dx^3}=wL=24aL+6b-6b$$ and at the free end the shear force equals zero $$\frac{d^3y(0)}{dx^3}=0=24aL+6b$$

Assembling these constraint equations into a matrix and rearranging for the coefficients gives

$$\begin{bmatrix}a\\b\\c\\d\\e\end{bmatrix}=\begin{bmatrix}0&0&0&0&1\\0&0&0&1&0\\12L^2&6L&2&0&0\\24L&0&0&0&0\\24L&6&0&0&0\end{bmatrix}^{-1}\begin{bmatrix}0\\0\\0\\wL\\0\end{bmatrix}$$

Therefore the equation for the displacement of the cantilever is

$$y=\frac{w}{24EI}x^4-\frac{wL}{6EI}x^3+\frac{wL^2}{4EI}x^2$$

Propped Cantilever

To solve for the coefficients of

$$EIy=ax^4+bx^3+cx^2+dx+e$$

for a propped cantilever with length $L$ and uniform load $w$ the constraints are:

• Both ends can not move vertically $$y(0)=0=e$$ and $$y(L)=0=aL^4+bL^3+cL^2+dL+e$$
• One end can not rotate $$\frac{dy(0)}{dx}=0=d$$
• There is no resistance to rotation at the other end (the bending moment or second derivative is zero at one end) $$\frac{d^2y(L)}{dx^2}=0=12aL^2+6bL+2c$$
• The difference in shear force (third derivative) at the ends equals the applied load $$\frac{d^3y(L)}{dx^3}-\frac{d^3y(0)}{dx^3}=wL=24aL+6b-6b$$

Assembling these constraint equations into a matrix and rearranging for the coefficients gives

$$\begin{bmatrix}a\\b\\c\\d\\e\end{bmatrix}=\begin{bmatrix}0&0&0&0&1\\L^4&L^3&L^2&L&1\\0&0&0&1&0\\12L^2&6L&2&0&0\\24L&0&0&0&0\end{bmatrix}^{-1}\begin{bmatrix}0\\0\\0\\0\\wL\end{bmatrix}$$

Therefore the equation for the displacement of the propped cantilever is

$$y=\frac{w}{24EI}x^4-\frac{5wL}{48EI}x^3+\frac{wL^2}{16EI}x^2$$

Fixed Support Beam

To solve for the coefficients of

$$EIy=ax^4+bx^3+cx^2+dx+e$$

for a fixed support beam with length $L$ and uniform load $w$ the constraints are:

• Both ends can not move vertically $$y(0)=0=e$$ $$y(L)=0=aL^4+bL^3+cL^2+dL+e$$
• Both ends can not rotate $$\frac{dy(0)}{dx}=0=d$$ and $$\frac{dy(0)}{dx}=0=4aL^3+3bL^2+2cL+d$$
• The difference in shear force (third derivative) at the ends equals the applied load $$\frac{d^3y(L)}{dx^3}-\frac{d^3y(0)}{dx^3}=wL=24aL+6b-6b$$

Assembling these constraint equations into a matrix and rearranging for the coefficients gives

$$\begin{bmatrix}a\\b\\c\\d\\e\end{bmatrix}=\begin{bmatrix}0&0&0&0&1\\L^4&L^3&L^2&L&1\\0&0&0&1&0\\4L^3&3L^2&2L&1&0\\24L&0&0&0&0\end{bmatrix}^{-1}\begin{bmatrix}0\\0\\0\\0\\wL\end{bmatrix}$$

Therefore the equation for the displacement of the fixed support beam is

$$y=\frac{w}{24EI}x^4-\frac{wL}{12EI}x^3+\frac{wL^2}{24EI}x^2$$

An inverse function is a function that you can substitute your function into to get your function’s input as explained in detail on the inverse function page. However, a quartic does not have an inverse function as a function must only have one output for every input. Any inverse for a quartic will have between two and four outputs for every input so it is still possible to find an inverse equation. To find a quartic function inverse we solve the quartic equation for $x$. This follows the exact same procedure as finding the roots excluding that we do not set $y$ to equal zero which results in the inverse equation of:

$$f^{-1}(y)=-\frac{b}{4a}+\begin{cases}S\pm\frac{1}{2}\sqrt{-4S^2-2p-\frac{q}{S}}\\-S\pm\frac{1}{2}\sqrt{-4S^2-2p+\frac{q}{S}}\end{cases}$$

where

$$p=\frac{8ac-3b^2}{8a^2}$$

$$q=\frac{8a^2d-4abc+b^3}{8a^3}$$

$$S=\frac{1}{2}\sqrt{\frac{-2p}{3}+V-\frac{P}{3V}}$$

in which

$$V=\left(\frac{-Q\pm\sqrt{Q^2+\frac{4P^3}{27}}}{2}\right)^{\frac{1}{3}}$$

$$P=\frac{3bd-12a(e-y)-c^2}{3a^2}$$

$$Q=\frac{72ac(e-y)-27ad^2-27b^2(e-y)+9bcd-2c^3}{27a^3}$$

and the case needs to be selected such that $f^{-1}(f(x))=x$ and the boundaries between the case are at the roots of the first derivative of the quartic function.

The domain of a function is all the possible input values and the range is all the possible output values. For detailed information on what domain and range are and the notation used see the domain and range page. For the quartic function discussed on this page its domain in all three notations is

$$-\infty<x<\infty$$

$$(-\infty,\infty)$$

$$\{x:x\in\mathbb{R}\}$$

and its range

$$h\le y<\infty \text{ if } a>0$$$$-\infty<y\le h \text{ if } a<0$$$$-\infty<y<\infty \text{ if } a=0$$

$$[h,\infty)\text{ if } a>0$$$$(-\infty,h]\text{ if } a>0$$$$(-\infty,\infty)\text{ if } a=0$$

$$\{y:y\in\mathbb{R},y\ge h\}\text{ if } a>0$$$$\{y:y\in\mathbb{R},y\le h\}\text{ if } a<0$$$$\{y:y\in\mathbb{R}\}\text{ if } a=0$$

where $h$ is the global minimum if $a>0$ and the global maximum if $a<0$. In choosing the domain and range I have restricted the function to real numbers $\mathbb{R}$ but there is no reason this function could not be used with complex numbers $\mathbb{C}$.