A subspace is a vector space that is contained within another vector space. So every subspace is a vector space in its own right, but it is also defined relative to some other (larger) vector space. We will discover shortly that we are already familiar with a wide variety of subspaces from previous sections.

Here is the principal definition for this section.

##### DefinitionSSubspace

Suppose that $V$ and $W$ are two vector spaces that have identical definitions of vector addition and scalar multiplication, and suppose that $W$ is a subset of $V\text{,}$ $W\subseteq V\text{.}$ Then $W$ is a subspace of $V\text{.}$

Let us look at an example of a vector space inside another vector space.

In Example SC3 we proceeded through all ten of the vector space properties before believing that a subset was a subspace. But six of the properties were easy to prove, and we can lean on some of the properties of the vector space (the superset) to make the other four easier. Here is a theorem that will make it easier to test if a subset is a vector space. A shortcut if there ever was one.

##### Proof

So just three conditions, plus being a subset of a known vector space, gets us all ten properties. Fabulous! This theorem can be paraphrased by saying that a subspace is “a nonempty subset (of a vector space) that is closed under vector addition and scalar multiplication.”

You might want to go back and rework Example SC3 in light of this result, perhaps seeing where we can now economize or where the work done in the example mirrored the proof and where it did not. We will press on and apply this theorem in a slightly more abstract setting.

Much of the power of Theorem TSS is that we can easily establish new vector spaces if we can locate them as subsets of other vector spaces, such as the vector spaces presented in Subsection VS.EVS.

It can be as instructive to consider some subsets that are not subspaces. Since Theorem TSS is an equivalence (see Proof Technique E) we can be assured that a subset is not a subspace if it violates one of the three conditions, and in any example of interest this will not be the “nonempty” condition. However, since a subspace has to be a vector space in its own right, we can also search for a violation of any one of the ten defining properties in Definition VS or any inherent property of a vector space, such as those given by the basic theorems of Subsection VS.VSP. Notice also that a violation need only be for a specific vector or pair of vectors.

There are two examples of subspaces that are trivial. Suppose that $V$ is any vector space. Then $V$ is a subset of itself and is a vector space. By Definition S, $V$ qualifies as a subspace of itself. The set containing just the zero vector $Z=\set{\zerovector}$ is also a subspace as can be seen by applying Theorem TSS or by simple modifications of the techniques hinted at in Example VSS. Since these subspaces are so obvious (and therefore not too interesting) we will refer to them as being trivial.

##### DefinitionTSTrivial Subspaces

Given the vector space $V\text{,}$ the subspaces $V$ and $\set{\zerovector}$ are each called a trivial subspace.

We can also use Theorem TSS to prove more general statements about subspaces, as illustrated in the next theorem.

##### Proof

Here is an example where we can exercise Theorem NSMS.

# SubsectionTSSThe Span of a Set¶ permalink

The span of a set of column vectors got a heavy workout in Chapter V and Chapter M. The definition of the span depended only on being able to formulate linear combinations. In any of our more general vector spaces we always have a definition of vector addition and of scalar multiplication. So we can build linear combinations and manufacture spans. This subsection contains two definitions that are just mild variants of definitions we have seen earlier for column vectors. If you have not already, compare them with Definition LCCV and Definition SSCV.

##### DefinitionLCLinear Combination

Suppose that $V$ is a vector space. Given $n$ vectors $\vectorlist{u}{n}$ and $n$ scalars $\alpha_1,\,\alpha_2,\,\alpha_3,\,\ldots,\,\alpha_n\text{,}$ their linear combination is the vector \begin{equation*} \lincombo{\alpha}{u}{n}\text{.} \end{equation*}

When we realize that we can form linear combinations in any vector space, then it is natural to revisit our definition of the span of a set, since it is the set of all possible linear combinations of a set of vectors.

##### DefinitionSSSpan of a Set

Suppose that $V$ is a vector space. Given a set of vectors $S=\{\vectorlist{u}{t}\}\text{,}$ their span, $\spn{S}\text{,}$ is the set of all possible linear combinations of $\vectorlist{u}{t}\text{.}$ Symbolically, \begin{align*} \spn{S}&=\setparts{\lincombo{\alpha}{u}{t}}{\alpha_i\in\complexes,\,1\leq i\leq t}\\ &=\setparts{\sum_{i=1}^{t}\alpha_i\vect{u}_i}{\alpha_i\in\complexes,\,1\leq i\leq t}\text{.} \end{align*}

##### Proof

Let us again examine membership in a span.

Notice how Example SSP and Example SM32 contained questions about membership in a span, but these questions quickly became questions about solutions to a system of linear equations. This will be a common theme going forward.

Several of the subsets of vectors spaces that we worked with in Chapter M are also subspaces — they are closed under vector addition and scalar multiplication in $\complex{m}\text{.}$

##### Proof

That was easy! Notice that we could have used this same approach to prove that the null space is a subspace, since Theorem SSNS provided a description of the null space of a matrix as the span of a set of vectors. However, I much prefer the current proof of Theorem NSMS. Speaking of easy, here is a very easy theorem that exposes another of our constructions as creating subspaces.

One more.

##### Proof

So the span of a set of vectors, and the null space, column space, row space and left null space of a matrix are all subspaces, and hence are all vector spaces, meaning they have all the properties detailed in Definition VS and in the basic theorems presented in Section VS. We have worked with these objects as just sets in Chapter V and Chapter M, but now we understand that they have much more structure. In particular, being closed under vector addition and scalar multiplication means a subspace is also closed under linear combinations.

##### 1

Summarize the three conditions that allow us to quickly test if a set is a subspace.

##### 2

Consider the set of vectors \begin{equation*} W=\setparts{\colvector{a\\b\\c}}{3a-2b+c=5}\text{.} \end{equation*} Is the set $W$ a subspace of $\complex{3}\text{?}$ Explain your answer.

##### 3

Name five general constructions of sets of column vectors (subsets of $\complex{m}$) that we now know as subspaces.

# SubsectionExercises

##### C15

Working within the vector space $\complex{3}\text{,}$ determine if $\vect{b} = \colvector{4\\3\\1}$ is in the subspace $W\text{,}$ \begin{equation*} W = \spn{\set{ \colvector{3\\2\\3}, \colvector{1\\0\\3}, \colvector{1\\1\\0}, \colvector{2\\1\\3} }}\text{.} \end{equation*}

Solution
##### C16

Working within the vector space $\complex{4}\text{,}$ determine if $\vect{b} = \colvector{1\\1\\0\\1}$ is in the subspace $W\text{,}$ \begin{equation*} W =\spn{\set{ \colvector{1\\2\\-1\\1}, \colvector{1\\0\\3\\1}, \colvector{2\\1\\1\\2} }}\text{.} \end{equation*}

Solution
##### C17

Working within the vector space $\complex{4}\text{,}$ determine if $\vect{b} = \colvector{2\\1\\2\\1}$ is in the subspace $W\text{,}$ \begin{equation*} W = \spn{\set{ \colvector{1\\2\\0\\2}, \colvector{1\\0\\3\\1}, \colvector{0\\1\\0\\2}, \colvector{1\\1\\2\\0} }}\text{.} \end{equation*}

Solution
##### C20

Working within the vector space $P_3$ of polynomials of degree 3 or less, determine if $p(x)=x^3+6x+4$ is in the subspace $W$ below. \begin{equation*} W=\spn{\set{x^3+x^2+x,\,x^3+2x-6,\,x^2-5}} \end{equation*}

Solution
##### C21

Consider the subspace \begin{equation*} W=\spn{\set{ \begin{bmatrix} 2 & 1\\3 & -1 \end{bmatrix} ,\, \begin{bmatrix} 4 & 0\\2 & 3 \end{bmatrix} ,\, \begin{bmatrix} -3 & 1\\2 & 1 \end{bmatrix} }} \end{equation*} of the vector space of $2\times 2$ matrices, $M_{22}\text{.}$ Is \begin{equation*} C=\begin{bmatrix} -3 & 3\\6 & -4 \end{bmatrix} \end{equation*} an element of $W\text{?}$

Solution
##### C26

Show that the set $Y=\setparts{\colvector{x_1\\x_2}}{x_1\in{\mathbb Z},\,x_2\in{\mathbb Z}}$ from Example NSC2S has Property AC.

##### M20

In $\complex{3}\text{,}$ the vector space of column vectors of size 3, prove that the set $Z$ is a subspace. \begin{equation*} Z=\setparts{\colvector{x_1\\x_2\\x_3}}{4x_1-x_2+5x_3=0} \end{equation*}

Solution
##### T20

A square matrix $A$ of size $n$ is upper triangular if $\matrixentry{A}{ij}=0$ whenever $i\gt j\text{.}$ Let $UT_n$ be the set of all upper triangular matrices of size $n\text{.}$ Prove that $UT_n$ is a subspace of the vector space of all square matrices of size $n\text{,}$ $M_{nn}\text{.}$

Solution
##### T30

Let $P$ be the set of all polynomials, of any degree. The set $P$ is a vector space. Let $E$ be the subset of $P$ consisting of all polynomials with only terms of even degree. Prove or disprove: the set $E$ is a subspace of $P\text{.}$

Solution
##### T31

Let $P$ be the set of all polynomials, of any degree. The set $P$ is a vector space. Let $F$ be the subset of $P$ consisting of all polynomials with only terms of odd degree. Prove or disprove: the set $F$ is a subspace of $P\text{.}$

Solution


# Distribution of ratio between two independent uniform random variables

Supppse $X$ and $Y$ are standard uniformly distributed in $[0, 1]$, and they are independent, what is the PDF of $Z = Y / X$?

The answer from some probability theory textbook is

$$f_Z(z) = \begin{cases} 1/2, & \text{if } 0 \le z \le 1 \\ 1/(2z^2), & \text{if } z > 1 \\ 0, & \text{otherwise}. \end{cases}$$

I am wondering, by symmetry, shouldn't $f_Z(1/2) = f_Z(2)$? This is not the case according to the PDF above.

• What is the domain of $X$ and $Y$? – Sobi Dec 8 '15 at 13:47
• en.wikipedia.org/wiki/Ratio_distribution – kjetil b halvorsen Dec 8 '15 at 13:54
• Why would you expect this to be true? The density function tells you how tightly packed the probability is in the neighborhood of a point, and it is clearly more difficult for $Z$ to be near $2$ than $1/2$ (consider for instance that $Z$ can always be $1/2$ no matter what $X$ is, but $Z < 2$ when $X > 1/2$). – dsaxton Dec 8 '15 at 14:06
• Possible duplicate of Distribution of a ratio of uniforms: What is wrong? – Xi'an Dec 8 '15 at 14:21
• I don't think it's a duplicate, that question is seeking the PDF, here I have the PDF, I am just questioning its correctness (perhaps rather naively). – qed Dec 8 '15 at 14:29

The right logic is that with independent $X, Y \sim U(0,1)$, $Z=\frac YX$ and $Z^{-1} =\frac XY$ have the same distribution and so for $0 < z < 1$ \begin{align} P\left\{\frac YX \leq z\right\} &= P\left\{\frac XY \leq z\right\}\\ &= P\left\{\frac YX \geq \frac 1z \right\}\\ \left.\left.F_{Z}\right(z\right) &= 1 - F_{Z}\left(\frac 1z\right) \end{align} where the equation with CDFs uses the fact that $\frac YX$ is a continuous random variable and so $P\{Z \geq a\} = P\{Z > a\} = 1-F_Z(a)$. Hence the pdf of $Z$ satisfies $$f_Z(z) = z^{-2}f_Z(z^{-1}), \quad 0 < z < 1.$$ Thus $f_Z(\frac 12) = 4f_Z(2)$, and not $f_Z(\frac 12) = f_Z(2)$ as you thought it should be.

This distribution is symmetric--if you look at it the right way.

The symmetry you have (correctly) observed is that $Y/X$ and $X/Y = 1/(Y/X)$ must be identically distributed. When working with ratios and powers, you are really working within the multiplicative group of the positive real numbers. The analog of the location invariant measure $d\lambda=dx$ on the additive real numbers $\mathbb{R}$ is the scale invariant measure $d\mu = dx/x$ on the multiplicative group $\mathbb{R}^{*}$ of positive real numbers. It has these desirable properties:

1. $d\mu$ is invariant under the transformation $x\to ax$ for any positive constant $a$: $$d\mu(ax) = \frac{d(ax)}{ax} = \frac{dx}{x} = d\mu.$$

2. $d\mu$ is covariant under the transformation $x\to x^b$ for nonzero numbers $b$: $$d\mu(x^b) = \frac{d(x^b)}{x^b} = \frac{b x^{b-1} dx}{x^b} = b\frac{dx}{x} = b\, d\mu.$$

3. $d\mu$ is transformed into $d\lambda$ via the exponential: $$d\mu(e^x) = \frac{de^x}{e^x} = \frac{e^x dx}{e^x} = dx = d\lambda.$$ Likewise, $d\lambda$ is transformed back to $d\mu$ via the logarithm.

(3) establishes an isomorphism between the measured groups $(\mathbb{R}, +, d\lambda)$ and $(\mathbb{R}^{*}, *, d\mu)$. The reflection $x \to -x$ on the additive space corresponds to the inversion $x \to 1/x$ on the multiplicative space, because $e^{-x} = 1/e^x$.

Let's apply these observations by writing the probability element of $Z=Y/X$ in terms of $d\mu$ (understanding implicitly that $z \gt 0$) rather than $d\lambda$:

$$f_Z(z)\,dz = g_Z(z)\,d\mu = \frac{1}{2}\begin{cases} 1\,dz = z\, d\mu, & \text{if } 0 \le z \le 1 \\ \frac{1}{z^2}dz = \frac{1}{z}\, d\mu, & \text{if } z > 1. \end{cases}$$

That is, the PDF with respect to the invariant measure $d\mu$ is $g_Z(z)$, proportional to $z$ when $0\lt z \le 1$ and to $1/z$ when $1 \le z$, close to what you had hoped.

This is not a mere one-off trick. Understanding the role of $d\mu$ makes many formulas look simpler and more natural. For instance, the probability element of the Gamma function with parameter $k$, $x^{k-1}e^x\,dx$ becomes $x^k e^x d\mu$. It's easier to work with $d\mu$ than with $d\lambda$ when transforming $x$ by rescaling, taking powers, or exponentiating.

The idea of an invariant measure on a group is far more general, too, and has applications in that area of statistics where problems exhibit some invariance under groups of transformations (such as changes of units of measure, rotations in higher dimensions, and so on).

• Looks like a very insightful answer. It's a pity I don't understand it at the moment. I will check back later. – qed Dec 8 '15 at 16:40

If you think geometrically...

In the $X$-$Y$ plane, curves of constant $Z = Y/X$ are lines through the origin. ($Y/X$ is the slope.) One can read off the value of $Z$ from a line through the origin by finding its intersection with the line $X=1$. (If you've ever studied projective space: here $X$ is the homogenizing variable, so looking at values on the slice $X=1$ is a relatively natural thing to do.)

Consider a small interval of $Z$s, $(a,b)$. This interval can also be discussed on the line $X=1$ as the line segment from $(1,a)$ to $(1,b)$. The set of lines through the origin passing through this interval forms a solid triangle in the square $(X,Y) \in U = [0,1]\times[0,1]$, which is the region we're actually interested in. If $0 \leq a < b \leq 1$, then the area of the triangle is $\frac{1}{2}(1-0)(b-a)$, so keeping the length of the interval constant and sliding it up and down the line $X=1$ (but not past $0$ or $1$), the area is the same, so the probability of picking an $(X,Y)$ in the triangle is constant, so the probability of picking a $Z$ in the interval is constant.

However, for $b>1$, the boundary of the region $U$ turns away from the line $X = 1$ and the triangle is truncated. If $1 \leq a < b$, the projections down lines through the origin from $(1,a)$ and $(1,b)$ to the upper boundary of $U$ are to the points $(1/a,1)$ and $(1/b,1)$. The resulting area of the triangle is $\frac{1}{2}(\frac{1}{a} - \frac{1}{b})(1-0)$. From this we see the area is not uniform and as we slide $(a,b)$ further and further to the right, the probability of selecting a point in the triangle decreases to zero.

Then the same algebra demonstrated in other answers finishes the problem. In particular, returning to the OP's last question, $f_Z(1/2)$ corresponds to a line that reaches $X=1$, but $f_Z(2)$ does not, so the desired symmetry does not hold.

Just for the record, my intuition was totally wrong. We are talking about density, not probability. The right logic is to check that

$$\int_1^k f_Z(z) dz = \int_{1/k}^1 f_Z(z) = \frac{1}{2}(1 - \frac{1}{k})$$,

and this is indeed the case.

Yea the link Distribution of a ratio of uniforms: What is wrong? provides CDF of $Z=Y/X$. The PDF here is just derivative of the CDF. So the formula is correct. I think your problem lies in the assumption that you think Z is "symmetric" around 1. However this is not true. Intuitively Z should be a skewed distribution, for example it is useful to think when Y is a fixed number between $(0,1)$ and X is a number close to 0, thus the ratio would be going to infinity. So the symmetry of distribution is not true. I hope this help a bit.


# Displacement and Distance

#### alane1994

##### Active member
Here is my question.

The function $$v(t)=15\cos{3t}$$, $$0 \leq t \leq 2 \pi$$, is the velocity in m/sec of a particle moving along the x-axis. Complete parts (a) through (c).

a.Graph the velocity function over the given interval. then determine when the motion is in the positive direction and when it is in the negative direction.
(I have done this part)
b. Find the displacement over the given interval.
(I have done this part)
c. Find the distance traveled over the given interval.
(This is the part that I am fuzzy on)

For this, does it involve previously found information? Or is it a separate set of calculations all its own?
Any help is appreciated!

~Austin

#### Jameson

Staff member
Displacement is the integral of the velocity vector over the given time interval and results in the distance between the starting point and end point. If you move 10 meters north and 10 meters south, the displacement is 0.

To find the distance you need to figure out where the displacement is negative and count that as a positive value. I believe this is the same as taking the integral of the absolute value of velocity. The way I would do this is find the regions where v(t) is positive and negative and then breaking the calculation into multiple calculations.

When considering $$\displaystyle v(t)=15\cos{3t}$$ where $$\displaystyle 0 \le t \le 2\pi$$ at $t=0$ v(t) is positive and becomes negative when $$\displaystyle 3t=\frac{\pi}{2}$$ so when $$\displaystyle t=\frac{\pi}{6}$$.

So your first integral is $$\displaystyle \int_{0}^{\frac{\pi}{6}}v(t)dt$$ This value will be positive.

Now v(t) will be negative until until it touches the x-axis again where $$\displaystyle 3t=\frac{3\pi}{2}$$ or when $$\displaystyle t=\frac{\pi}{2}$$. The second integral is now $$\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}v(t)dt$$. This will be negative but when considering distance you don't need to take into account the sign of this so count it as positive.

If you keep repeating this over until the end of the interval you should get the final answer. It will take a while to calculate this way and there very well could be a quicker way to do it but that's how I would do this.

#### alane1994

##### Active member
Yeah, that's what I did after consulting a professor, I got the answer 60...

#### Jameson

Staff member
Yeah, that's what I did after consulting a professor, I got the answer 60...
You already found all of these intervals for part A so this calculation shouldn't have been too tedious. Wow your professor responds fast. You posted this question just a couple of hours ago

#### topsquark

##### Well-known member
MHB Math Helper
Just to put my two cents in...

The formula for distance is
$$dist = \int |v|dt$$

where v is the velocity vector.

In a practical sense this can pretty much only be calculated in the way Jameson has described.

-Dan


# Counterexample Check for Sum of Limit Points of Subsequences

Let $c$ be a limit point of a sequence of real numbers $\langle a_n \rangle$ and $d$ a limit point of $\langle b_n \rangle$. Is $c+d$ necessarily a limit point of $\langle a_n + b_n \rangle$?

My Question:

When considering this question, do I have to sum over the same index or can the indices for the different subsequences differ? My intuition is that the subsequences must be summed over identical indices, in which case I believe that the following example serves as a counterexample:

Let $\langle a_n \rangle = (-1)^n, \ \langle b_n \rangle = (-1)^{n+1}$. Then summing over even and odd indices, I get $0 \ne 2$ or $-2$, which is the sum of their limit points. Have I done this correctly?

-
Your argument is correct; but generally, the indices could differ. Take $a_n=b_n=(-1)^n$. $(a_n)$ has $1$ as a limit point and $(b_n)$ has $-1$ as a limit point but $(a_n+b_n)$ does not have $1+(-1)=0$ as a limit point. –  David Mitra Aug 14 '12 at 22:10

Your example is fine. Both $\langle(-1)^n:n\in\Bbb N\rangle$ and $\langle(-1)^{n+1}:n\in\Bbb N\rangle$ have $1$ as a limit point, but $\langle(-1)^n+(-1)^{n+1}:n\in\Bbb N\rangle=\langle 0:n\in\Bbb N\rangle$ converges to $0$ and so does not have $1+1=2$ as a limit point.
It doesn’t matter what subsequence of $\langle a_n:n\in\Bbb N\rangle$ has $c$ as limit or what subsequence of $\langle b_n:n\in\Bbb N\rangle$ has $d$ as a limit; all that matters is whether some subsequence of $\langle a_n+b_n:n\in\Bbb N\rangle$ has $c+d$ as a limit. In your example that’s not the case, so yours is a genuine counterexample to the conjecture.
@Rolando posted what looked like a proof of this conjecture, and then deleted it after you posted your answer. It looked something like this: Let $\epsilon >0$ be given. Since $a_n \to c$ and $b_n \to d$, there are $N_1, N_2$ such that for $l \ge N_1$, $k \ge N_2$ $|a_l - c| \le \epsilon/2$ and $|b_k - d| \le \epsilon/2$. Let $m = \max\{N_1, N_2\}$ and therefore we have for $n \ge m$ we have $|a_n - c + b_n - d| \le \epsilon$. Is the problem with his proof that he is not considering subsequences, and that he interpreted the limit points as limits? –  Zvpunry Aug 14 '12 at 22:14


# Sequence of functions that fails certain conditions of Arzela-Ascoli theorem

For a closed, bounded interval $[a,b]$, let $\{ f_{n}\}$ be a sequence in $C[a,b]$. If $\{f_{n}\}$ is equicontinuous, does $\{f_{n}\}$ necessarily have a uniformly convergent subsequence?

I would think not, because according to the Arzela-Ascoli Theorem, $\{f_{n} \}$ also needs to be uniformly bounded. Is this all that needs to be violated in order for an equicontinuous sequence of continuous functions on a compact interval to not have a uniformly convergent subsequence?

And if so, what is an example of a sequence that illustrates this, and how to show it does not have a uniformly convergent subsequence? Thank you.

Take $f_n(x) = n$ for $x \in [0,1]$. These functions are all constant, so clearly equicontinuous, but $\| f_n - f_m \|_\infty = \lvert n - m \rvert \ge 1$ for $n \neq m$ so no subsequence can converge since no subsequence is Cauchy.
• Yes of course. If it was uniformly bounded, then we could apply Arzela Ascoli and find a convergent subsequence. A sequence of functions $f_n$ is uniformly bounded if and only if the sequence of real numbers $\| f_n\|_\infty$ is bounded. Here that clearly is not the case because $\| f_n \|_\infty = n$. – User8128 Apr 11 '16 at 23:29


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Write. linear functions by the shape of their graphs and by noting differences in their expressions. Graph 3x - 2y = 8. And here is its graph: It makes a 45° (its slope is 1) It is called "Identity" because what comes out is identical to what goes in: In. We can begin graphing by plotting the point (0, 1) We know that the slope is rise over run, $m=\frac{\text{rise}}{\text{run}}$. A function may also be transformed using a reflection, stretch, or compression. Plot the points and graph the linear function. However, in linear algebra, a linear function is a function that maps a sum to the sum of the images of the summands. In the equation $f\left(x\right)=mx+b$, $m=\frac{\text{change in output (rise)}}{\text{change in input (run)}}=\frac{\Delta y}{\Delta x}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$. PLAY. This is why we performed the compression first. The y-intercept is the point on the graph when x = 0. f(x)=b. of f is the To find the y-intercept, we can set $x=0$ in the equation. Gravity. The graph of a linear function is a line. The slope of a linear function is equal to the ratio of the change in outputs to the change in inputs. This inequality notation means that we should plot the graph for values of x between and including -3 and 3. Use the resulting output values to identify coordinate pairs. The a represents the gradient of the line, which gives the rate of change of the dependent variable. In order to write the linear function in the form of y=mx+b, we will need to determine the line's: 1. slope (m) 2. y-intercept (b) We can tell from the graph that the slope of the line is negative because the line goes down and to the right. Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. The graph below is of the function $f\left(x\right)=-\frac{2}{3}x+5$. Linear Parent Graph And Transformations. So, for this definition, the above function is linear only when c = 0, that is when the line passes through the origin. Linear functions are functions that produce a straight line graph.. The graph of f is a line with slope m and y intercept b. The graph of f is a line with slope m and y intercept Graphing Linear Equations Find the Equation of a Line. Horizontal lines are written in the form, $f(x)=b$. Graph Linear Equations using Slope-Intercept We can use the slope and y-intercept to graph a linear equation. Graph Linear Equations in Two Variables Learning Objectives. We were also able to see the points of the function as well as the initial value from a graph. The graph of a linear function is a line. f(0). The graph of the function is a line as expected for a linear function. Evaluate the function at each input value and use the output value to identify coordinate pairs. ++drag: Hold down both the key and the key, then drag the described object. For example, following order of operations, let the input be 2. The equation for the function shows that $m=\frac{1}{2}$ so the identity function is vertically compressed by $\frac{1}{2}$. Linear functions are those whose graph is a straight line. The steepness of a hill is called a slope. Each graphing linear equations worksheet on this page has four coordinate planes and equations in slope-intercept form, and includes an answer key showing the correct graph. Graphing a Linear Function Using y-intercept and Slope. Graphing Linear Functions. In this non-linear system, users are free to take whatever path through the material best serves their needs. There are three basic methods of graphing linear functions: Keep in mind that a vertical line is the only line that is not a function.). Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. Then just draw a line that passes through both of these points. The following diagrams show the different methods to graph a linear equation. Linear functions are typically written in the form f(x) = ax + b. Furthermore, the domain and range consists of all real numbers. A table of values might look as below. Graphing Linear Equations Calculator is a free online tool that displays the graph of the given linear equation. Notice that adding a value of b to the equation of $f\left(x\right)=x$ shifts the graph of f a total of b units up if b is positive and |b| units down if b is negative. How many solutions does this linear system have? Regardless of whether a table is given to you, you should consider using one to ensure you’re correctly graphing linear and quadratic functions. Is the Function Linear or Nonlinear | Table. Learn More at mathantics.comVisit http://www.mathantics.com for more Free math videos and additional subscription based content! Often, the number in front of x is already a fraction, so you won't have to convert it. A y-intercept is a y-value at which a graph crosses the y-axis. Solving Systems of Linear Equations: Graphing. The equation for a linear function is: y = mx + b, Where: m = the slope ,; x = the input variable (the “x” always has an exponent of 1, so these functions are always first degree polynomial.). The slopes are represented as fractions in the level 2 worksheets. In Linear Functions, we saw that that the graph of a linear function is a straight line. The equation can be written in standard form, so the function is linear. eval(ez_write_tag([[728,90],'analyzemath_com-medrectangle-4','ezslot_5',344,'0','0'])); Any function of the form The first is by plotting points and then drawing a line through the points. Furthermore, the domain and range consists of all real numbers. Properties. Graphing linear functions (2.0 MiB, 1,144 hits) Slope Determine slope in slope-intercept form (160.4 KiB, 766 hits) Determine slope from given graph (2.1 MiB, 834 hits) Find the integer of unknown coordinate (273.6 KiB, 858 hits) Find the fraction of unknown coordinate (418.5 KiB, 891 hits) Linear inequalities Graph of linear inequality (2.8 MiB, 929 hits) Facebook. The first characteristic is its y-intercept which is the point at which the input value is zero. Linear functions are typically written in the form f(x) = ax + b. In mathematics, the term linear function refers to two distinct but related notions: In calculus and related areas, a linear function is a function whose graph is a straight line, that is, a polynomial function of degree zero or one. Interactive, free online graphing calculator from GeoGebra: graph functions, plot data, drag sliders, and much more! Interpret solutions to linear equations and inequalities graphically. The function $y=x$ compressed by a factor of $\frac{1}{2}$. y = f(x) = a + bx. Google+. ; b = where the line intersects the y-axis. Now we have to determine the slope of the line. The function $y=\frac{1}{2}x$ shifted down 3 units. The equation for a linear function is: y = mx + b, Where: m = the slope , x = the input variable (the “x” always has an exponent of 1, so these functions are always first degree polynomial.). Because the slope is positive, we know the graph will slant upward from left to right. 1. The order of the transformations follows the order of operations. Evaluating the function for an input value of 2 yields an output value of 4 which is represented by the point (2, 4). $f\left(x\right)=\frac{1}{2}x+1$. Another option for graphing is to use transformations on the identity function $f\left(x\right)=x$. A linear function has the following form y = f (x) = a + bx A linear function has one independent variable and one dependent variable. The graph of the linear equation will always result in a straight line. The input values and corresponding output values form coordinate pairs. A linear function is a polynomial function in which the variable x has degree at most one: = +.Such a function is called linear because its graph, the set of all points (, ()) in the Cartesian plane, is a line.The coefficient a is called the slope of the function and of the line (see below).. We then plot the coordinate pairs on a grid. Free graph paper is available. The graph of a linear function is a straight line, while the graph of a nonlinear function is a curve. Students learn that the linear equation y = x, or the diagonal line that passes through the origin, is called the parent graph for the family of linear equations. Vertically stretch or compress the graph by a factor. Write the equation of a line parallel or perpendicular to a given line. Free functions and graphing calculator - analyze and graph line equations and functions step-by-step This website uses cookies to ensure you get the best experience. That line is the solution of the equation and its visual representation. Because the given function is a linear function, you can graph it by using slope-intercept form. In Linear Functions, we saw that that the graph of a linear function is a straight line.We were also able to see the points of the function as well as the initial value from a graph. Explore math with our beautiful, free online graphing calculator. The equation for the function also shows that $b=-3$, so the identity function is vertically shifted down 3 units. We will choose 0, 3, and 6. Another way to think about the slope is by dividing the vertical difference, or rise, between any two points by the horizontal difference, or run. It looks like the y-intercept (b) of the graph is 2, as represented by point (0,2). f(a) is called a function, where a … Match. We repeat until we have multiple points, and then we draw a line through the points as shown below. Learn. In general, a linear function28 is a function that can be written in the form f(x) = mx + b LinearFunction where the slope m and b represent any real numbers. Begin by choosing input values. In general, a linear function Any function that can be written in the form f ( x ) = m x + b is a function that can be written in the form f ( x ) = m x + b L i n e a r F u n c t i o n where the slope m and b represent any real … Learn more Accept. Because y = f(x), we can use y and f(x) interchangeably, and ordered pair solutions on the graph (x, y) can be written in the form (x, f(x)). eval(ez_write_tag([[728,90],'analyzemath_com-medrectangle-3','ezslot_4',320,'0','0'])); Determine the x intercept, set f(x) = 0 and The slopes in level 1 worksheets are in the form of integers. Graph a straight line by finding its x - and y-intercepts. The, of this function is the set of all real numbers. By using this website, you agree to our Cookie Policy. The graph crosses the y-axis at (0, 1). Graphing Linear Functions. There are three basic methods of graphing linear functions. No. Recall that the slope is the rate of change of the function. 3. Vertical stretches and compressions and reflections on the function $f\left(x\right)=x$. This is also known as the “slope.” The b represents the y-axis intercept. These points may be chosen as the x and y intercepts of the graph for example. We can now graph the function by first plotting the y-intercept. Relating linear contexts to graph features Get 5 of 7 questions to level up! Method 1: Graphing Linear Functions in Standard Form 1. This means the larger the absolute value of m, the steeper the slope. Linear functions are those whose graph is a straight line. It is generally a polynomial function whose degree is utmost 1 or 0. Evaluating the function for an input value of 1 yields an output value of 2 which is represented by the point (1, 2). The third is applying transformations to the identity function $f\left(x\right)=x$. Write the equation for a linear function from the graph of a line. b. Solver to Analyze and Graph a Linear Function. Show Step-by-step Solutions. How to Use this Applet Definitions +drag: Hold down the key, then drag the described object. This website uses cookies to ensure you get the best experience. Use $\frac{\text{rise}}{\text{run}}$ to determine at least two more points on the line. Its graph is a horizontal line at y = b. The graph of a linear function is always a line. Graphing Linear Function: Type 2 - Level 1. Evaluate the function at each input value. Created by. We previously saw that that the graph of a linear function is a straight line. The graph slants downward from left to right which means it has a negative slope as expected. Evaluate when . Linear functions word problem: fuel (Opens a modal) Practice. We know that the linear equation is defined as an algebraic equation in which each term should have an exponents value of 1. Recall that the set of all solutions to a linear equation can be represented on a rectangular coordinate plane using a straight line through at least two points; this line is called its graph. Graphing Linear Functions. For example, Plot the graph of y = 2x – 1 for -3 ≤ x ≤ 3. Identify and graph a linear function using the slope and y-intercept. When m is negative, there is also a vertical reflection of the graph. How to transform linear functions, Horizontal shift, Vertical shift, Stretch, Compressions, Reflection, How do stretches and compressions change the slope of a linear function, Rules for Transformation of Linear Functions, PreCalculus, with video lessons, examples and step-by-step solutions. Learn more Accept. Solution : y = x + 3. In addition, the graph has a downward slant which indicates a negative slope. Note: A function f (x) = b, where b is a constant real number is called a constant function. This function includes a fraction with a denominator of 3 so let’s choose multiples of 3 as input values. 8 Linear Equations Worksheets. Usage To plot a function just type it into the function box. A graphing calculator can be used to verify that your answers "make sense" or "look right". solve for x. Graphs of linear functions may be transformed by shifting the graph up, down, left, or right as well as using stretches, compressions, and reflections. Write the equation in standard form. The slope of a linear function corresponds to the number in … For example, given the function $f\left(x\right)=2x$, we might use the input values 1 and 2. Evaluate the function at an input value of zero to find the. Graphs of linear functions may be transformed by shifting the graph up, down, left, or right as well as using stretches, compressions, and reflections. This graph illustrates vertical shifts of the function $f\left(x\right)=x$. The slope is defined as the ratio of the vertical change between two points, the rise, to the horizontal change between the same two points, the run. The Slider Area. In this video we look at graphing equations using a table of values Find the slopes and the x- and y-intercepts of the following lines. Solve a system of linear equations. The slope-intercept form gives you the y- intercept at (0, –2). Did you have an idea for improving this content? Graph horizontal and vertical lines. When it comes to graphing linear equations, there are a few simple ways to do it. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. The equation, written in this way, is called the slope-intercept form. According to the equation for the function, the slope of the line is $-\frac{2}{3}$. set of all real numbers. Flashcards. If variable x is a constant x=c, that will represent a line paralel to y-axis. Reddit. Two points that are especially useful for sketching the graph of a line are found with the intercepts. Given a real-world situation that can be modeled by a linear function or a graph of a linear function, the student will determine and represent the reasonable domain and range of the linear function … Function Grapher is a full featured Graphing Utility that supports graphing two functions together. Book The a represents the gradient of the line, which gives the rate of change of the dependent variable. Select two options. how to graph linear equations using the slope and y-intercept. Graph a linear function: a step by step tutorial with examples and detailed solutions. This inequality notation means that we should plot the graph for values of x between and including -3 and 3. Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error. Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. We generate these coordinates by substituting values into the linear equation. Test. In general we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph of the function. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. If we have two points: A=(x1,y1) B=(x2,y2) A slope (a) is calculated by the formula: a=y2−y1x2−x1 If the slope is equal to number 0, then the line will be paralel with x – axis. Functions: Hull: First graph: f(x) Derivative Integral From ... Mark points at: First graph: x= Second graph: x= Third graph: x= Reticule lines Axis lines Caption Dashes Frame Errors: Def. This is also known as the “slope.” The b represents the y-axis intercept. $\begin{array}{l}f\text{(2)}=\frac{\text{1}}{\text{2}}\text{(2)}-\text{3}\hfill \\ =\text{1}-\text{3}\hfill \\ =-\text{2}\hfill \end{array}$. Subtract x from each side. Find a point on the graph we drew in Example: Graphing by Using the y-intercept and Slope that has a negative x-value. Recognize the standard form of a linear function. Graphing a Linear Equation by Plotting Three Ordered Pairs. In Activity 1 the learners should enter the expressions one by one into the graphing calculator and classify the functions according to the shape of the graph. A linear equation is drawn as a straight line on a set of axes. But if it isn't, convert it by simply placing the value of m over 1. You need only two points to graph a linear function. First, graph y = x. Students also learn the different types of transformations of the linear parent graph. A Review of Graphing Lines. These unique features make Virtual Nerd a viable alternative to private tutoring. Although this may not be the easiest way to graph this type of function, it is still important to practice each method. What is Meant by Graphing Linear Equations? Possible answers include $\left(-3,7\right)$, $\left(-6,9\right)$, or $\left(-9,11\right)$. We were also able to see the points of the function as well as the initial value from a graph. Graph a straight line by finding three ordered pairs that are solutions to the linear equation. Examples: 1. To graph, choose three values of x, and use them to generate ordered pairs. The functions whose graph is a line are generally called linear functions in the context of calculus. The slope of a line is a number that describes steepnessand direction of the line. By the end of this section, you will be able to: Plot points in a rectangular coordinate system; Graph a linear equation by plotting points; Graph vertical and horizontal lines; Find the x- and y-intercepts; Graph a line using the intercepts ; Before you get started, take this readiness quiz. In the equation $f\left(x\right)=mx$, the m is acting as the vertical stretch or compression of the identity function. In Example: Graphing by Using Transformations, could we have sketched the graph by reversing the order of the transformations? You can move the graph of a linear function around the coordinate grid using transformations. The first … Graphing Linear Equations. It has the unique feature that you can save your work as a URL (website link). Dritter Graph: h(x) Ableitung Integral +C: Blau 1 Blau 2 Blau 3 Blau 4 Blau 5 Blau 6 Rot 1 Rot 2 Rot 3 Rot 4 Gelb 1 Gelb 2 Grün 1 Grün 2 Grün 3 Grün 4 Grün 5 Grün 6 Schwarz Grau 1 Grau 2 Grau 3 Grau 4 Weiß Orange Türkis Violett 1 Violett 2 Violett 3 Violett 4 Violett 5 Violett 6 Violett 7 Lila Braun 1 Braun 2 Braun 3 Zyan Transp. The slope of a linear function will be the same between any two points. The slope is $\frac{1}{2}$. By graphing two functions, then, we can more easily compare their characteristics. Graph $f\left(x\right)=-\frac{2}{3}x+5$ by plotting points. Linear equations word problems: tables Get 3 of 4 questions to level up! Plot the coordinate pairs and draw a line through the points. It will be very difficult to succeed in Calculus without being able to solve and manipulate linear equations. Method 1: Graphing Linear Functions in Standard Form 1. Introduction to Linear Relationships: IM 8.3.5. Example 1 Graph the linear function f given by f (x) = 2 x + 4 Solution to Example 1. Graph $f\left(x\right)=\frac{1}{2}x - 3$ using transformations. However, in linear algebra, a linear function is a function that maps a sum to the sum of the images of the summands. Graphing Linear Function: Type 1 - Level 2. Although the linear functions are also represented in terms of calculus as well as linear algebra. Linear functions are functions that produce a straight line graph. Starting from our y-intercept (0, 1), we can rise 1 and then run 2 or run 2 and then rise 1. Notice that multiplying the equation $f\left(x\right)=x$ by m stretches the graph of f by a factor of m units if m > 1 and compresses the graph of f by a factor of m units if 0 < m < 1. Tell whether each function is linear. 2. What is a Linear Function? f (x) = m x + b, where m is not equal to 0 is called a linear function. These pdf worksheets provide ample practice in plotting the graph of linear functions. For example, Plot the graph of y = 2x – 1 for -3 ≤ x ≤ 3. Free linear equation calculator - solve linear equations step-by-step. STUDY. Complete the function table, plot the points and graph the linear function. The other characteristic of the linear function is its slope, m, which is a measure of its steepness. We’d love your input. First, graph the identity function, and show the vertical compression. Graphing a Linear Function Using y-intercept and Slope. After studying this section, you will be able to: 1. From the initial value (0, 5) we move down 2 units and to the right 3 units. We encountered both the y-intercept and the slope in Linear Functions. Linear Function Graph. (See Getting Help in Stage 1.) In mathematics, a graphing linear equation represents the graph of the linear equation. y = mx + b y = -2x + 3/2. Knowing an ordered pair written in function notation is necessary too. The first characteristic is its y-intercept which is the point at which the input value is zero. Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. Let's try starting from a graph and writing the equation that goes with it. We can extend the line to the left and right by repeating, and then draw a line through the points. From our example, we have $m=\frac{1}{2}$, which means that the rise is 1 and the run is 2. Linear equations word problems: graphs Get 3 of 4 questions to level up! How to Use the Graphing Linear Equations Calculator? To find points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. m = -2 and b = -1/3 m = -2 and b = -2/3. Since the slope is 3=3/1, you move up 3 units and over 1 unit to arrive at the point (1, 1). If you have difficulties with this material, please contact your instructor. This website uses cookies to ensure you get the best experience. Convert m into a fraction. If so, graph the function. A linear function is a function which forms a straight line in a graph. Determine the y intercept, set x = 0 to find Draw Function Graphs Mathematics / Analysis - Plotter - Calculator 4.0. Draw a line which passes through the points. BYJU’S online graphing linear equations calculator tool makes the calculation faster and it displays the graph in a fraction of seconds. 8th grade students learn to distinguish between linear and nonlinear functions by observing the graphs. The output value when x = 0 is 5, so the graph will cross the y-axis at (0, 5). The y-intercept and slope of a line may be used to write the equation of a line. Examine the input(x) and output(y) values of the table inthese linear function worksheets for grade 8. Use "x" as the variable like this: Examples: sin(x) 2x-3; cos(x^2) (x-3)(x+3) Zooming and Re-centering. The same goes for the steepness of a line. Now we know the slope and the y-intercept. A linear function has the following form. Regardless of whether a table is given to you, you should consider using one to ensure you’re correctly graphing linear and quadratic functions. To zoom, use the zoom slider. Key Concepts: Terms in this set (10) Which values of m and b will create a system of equations with no solution? You can move the graph of a linear function around the coordinate grid using transformations. All linear functions cross the y-axis and therefore have y-intercepts. GeoGebra Classroom Activities. Equation for a linear function: a function. ) x = 0 visual representation the simplest way to... 5, so the function rather than plotting points y intercepts of the parent. There are three basic methods of graphing linear equations calculator is a line slope... Negative, there is also known as the x and f ( x ) = ax +.. Compressions and reflections on the identity function, you agree to our Cookie Policy left, or compression although may. 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# Probability of finding a molecule in the ground vibrational level

I wanted to estimate the probability of finding a molecule in the ground vibrational level using the Boltzmann distribution:

$$p_i = \frac{e^{-\epsilon_i/kT}}{\sum_{i=0}^{N}e^{-\epsilon_i/kT}}$$

Using the quantum harmonic oscillator as a model for the energy

$$\epsilon_i = h\nu (i+1/2) =/i=0/=\frac{h\nu}{2}$$

In the Boltzmann distribution, we have the state of interest divided by the sum of all possible states. But how should I treat the denominator?

Searching a bit I found that the analytical expression for this geometric series is ($i$ not imaginary number)

$$\sum_{i=0}^{N}e^{-i h\nu/kT} = \frac{1}{1 - e^{h\nu/kT}}$$

However, is this using a shifted energy scale for the harmonic potential? In that the vibrational energies are $0$, $h\nu$, $2h\nu$, ..., and not $\frac{1}{2}h\nu$, $\frac{3}{2}h\nu$, $\frac{5}{2}h\nu$, ...? Should I make sure I use the same energy scale for the nominator and denominator in the Boltzmann distribution?

Doing what porphyrin suggested, I get

$$\sum_{i=0}^{\infty} e^{-h\nu(i+\frac{1}{2})/kT} = e^{-h\nu/kT} \sum_{i=0}^{\infty} e^{- ih\nu/kT}$$

Expanding the four first terms

$$e^{-h\nu/kT} \sum_{i=0}^{\infty} e^{- ih\nu/kT} = (e^{-h\nu/kT} \cdot 1) + (e^{-h\nu/kT} \cdot e^{-h\nu/kT}) + (e^{-h\nu/kT} \cdot e^{-2h\nu/kT}) + (e^{-h\nu/kT} \cdot e^{-3h\nu/kT}) \\ = e^{-h\nu/kT} + e^{-2h\nu/kT} + e^{-3h\nu/kT} + e^{-4h\nu/kT} = \sum_{1}^{\infty}e^{-n\cdot h\nu/kT}$$

which has an analytical expression for the converged value, right?

• The denominator should be calculated exactly. Ever heard of a geometric progression and its sum? Oct 17 '16 at 12:30
• Ah, yes. Looking at it, it should converge.
– Yoda
Oct 17 '16 at 12:32
• The summation is called the partition function and it always extends over all possible levels $n=0 ..,\infty$. You are assuming a harmonic oscillator which has an infinite number of levels. Its easier if the 1/2 is separated out first and treated separately, this just adds a constant to the energy as you realise. Expand the summation as $1+e^{-a}+e^{-2a}+....$ ($a=h\nu /(kT)$ which converges to the result you quote. Oct 17 '16 at 12:54

You're on the right track.

Also, using $i$ as an index can be confusing some times because it can be confused with the imaginary number; however, here it should not present a problem. As a matter of habbit however, I like to use $j$ or $n$ or something else..there are only so many letters in the alphabet.

The sum in the denominator is called the partition function, and has the form

$$Z = \sum_{j}e^{-\frac{\epsilon_j}{kT}}$$

For the harmonic, oscillator $\epsilon_j = (\frac{1}{2}+j)\hbar \omega$ for $j \in \{ 0,1,2.. \}$

Note that $\epsilon_0 \neq 0$ there exists a zero point energy.

Let's write out a few terms $$Z = e^{-\frac{\hbar \omega/2}{kT}} + e^{-\frac{\hbar \omega3/2}{kT}} + e^{-\frac{\hbar \omega5/2}{kT}} +.....$$

factoring out $e^{-\frac{\hbar \omega/2}{kT}}$

$$Z = e^{-\frac{\hbar \omega/2}{kT}} \left( 1+ e^{-\frac{\hbar \omega}{kT}} + e^{-\frac{2\hbar \omega}{kT}} +.....\right)$$

The sum in the bracket takes the form of a geometric series whose sum converges as shown below $$1+x+x^2+... = \frac{1}{(1-x)}$$ herein, $x \equiv e^{-\frac{\hbar\omega}{kT}}$

Putting all of this together

$$Z = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{(1-e^{-\frac{\hbar\omega}{kT}})}$$

Now, $$p_0 = \frac{e^{-\epsilon_0/kT}}{Z} = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{Z} = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{\frac{e^{-\frac{\hbar \omega/2}{kT}}}{(1-e^{-\frac{\hbar\omega}{kT}})}} = (1-e^{-\frac{\hbar\omega}{kT}})$$

• Great answer. But I strongly disagree with not using i as an index variable. The possible confusion with the imaginary unit only arises from typographic sloppiness. Variables should be italic and mathematical constants should be upright set. This includes e.g. the euler number, but no one would think that you exponented the variable e. From context it is clear that you exponented the euler number. The same holds true for sum symbols, it would make no sense to write the imaginary unit underneath them and even with typographic sloppiness it is clear from context that you mean the variable i. Oct 18 '16 at 13:04


# Dimension of Image/Kernel of Linear Transformation

Consider the vector spaces $P_n$ (polynomials of degree no more than n). The differentiation D gives a linear transformation from $P_n$ to $P_{n-1}$. What is the dimension of the image and the kernel of D? Is D a valid linear transformation from $P_n$ to $P_n$? If it is, what is the dimension of the kernel and image in this case? It seems like this uses the rank-nullity theorem, but I'm not sure. Thanks

The transformation is a linear transformation. In order for a transformation $T$ to be a linear transformation, it has to implement 2 conditions: $$(1) \hspace{0.2cm} T(v + w) = T(v) + T(w) \\ (2) \hspace{0.2cm} T(\alpha v) = \alpha T(v)$$ for any $v, w \in V$.

The differentiation transformation $D$ satisfies those constraints because of the derivatives' properties.

As for dimensions:

$Im(D) = P_{n-1}$, because for any $p^{\prime}(x) \in P_{n-1}$ you can find some $p(x) \in P_n$ s.t. $D(p(x)) = p^{\prime}(x)$, namely $\int f^{\prime}(x) dx$. Therefore $dim(Im(D)) = dim(P_{n-1}) = n$.

By rank nullity theorem, $dim(Im(D)) + dim(Ker(D)) = dim(P_n)$, and therefore $n + dim(Ker(D)) = n+1$ and $dim(Ker(D)) = 1$

• So what would be the dimension of Im(D)? Would it just be n? – JanoyCresva Apr 15 '17 at 19:47
• Had a typo, edited. The dimension of $Im(D)$ is $n$ as explained in my answer – AsafHaas Apr 15 '17 at 19:49
• Ok I got it. I'm still slightly confused about the reasoning on why it is a valid linear transformation, though – JanoyCresva Apr 15 '17 at 19:53
• Edited my answer to contain an explanation of this as well. – AsafHaas Apr 15 '17 at 20:03

Yes, it's a valid linear transformation. If $f, g \in P_n$, $c \in \mathbb{R}$ or whatever field you're working over, $D(f + cg) = D(f) + cD(g)$.

$D$ is surjective. For any $f \in P_{n-1}$, you can integrate, $\int f \in P_n$, and then $D(\int f) = f$.

The kernel has dimension 1. You could do this by rank nullity, or just note that if $D(f) = 0$, $f$ is constant.

The mapping $D$ is a linear transformation:

Proof: Let $f,g \in P_n$, let $a,b \in \mathbb{R}$

$$(af+bg)'(c) = \lim\limits_{x\to c} \frac{(af+bg)(x) - (af + bg)(c)}{x-c}$$ $$= \lim\limits_{x\to c} \frac{af(x)+bg(x) - af(c) - bg(c)}{x-c}$$ $$= \lim\limits_{x\to c} \frac{af(x) - af(c) +bg(x) - bg(c)}{x-c}$$ $$= \lim\limits_{x\to c} \frac{af(x) - af(c) }{x-c} + \lim\limits_{x\to c} \frac{ bg(x) - bg(c)}{x-c}$$ $$= a\lim\limits_{x\to c} \frac{f(x) - af(c) }{x-c} + b\lim\limits_{x\to c} \frac{ g(x) - g(c)}{x-c}$$ $$=af'(c) + bf'(c)$$

Hence: $D(af + bg) = aD(f) + bD(g) \quad \triangle$

$Im(D) = P_{n-1}$

Proof:

Let $Q \in Im(D)$ Then, there is a $P \in P_n$ such that $D(P_n) = Q$. Because differentiation is an operation that reduces the exponent of a power function by $1$ (for example polynomials), $Q \in P_{n-1}$. Hence, $Im(D) \subset P_{n-1}$

Now, let $Q \in P_{n-1}$. Then, there is a $P$ in $P_n$ such that $D(P) = Q$. Simply take $$P = \int Q dx$$ Hence, $Q \in Im(D)$

We conclude that $P_{n-1} \subset Im(D)$

obtaining $P_{n-1} = Im(D) \quad \triangle$

From this, it follows that $dim(Im(D)) = n$, and by rank nullity theorem $dim(ker(D)) = (n+1)-n = 1$

One can also prove directly the kernel has dimension 1 (which is, in fact, easier), by showing $ker(D) = P_0$ ($P_0$ might be bad notation but I mean the constant polynomials)

Proof:

$P \in ker(D) \iff D(P) = 0\iff P \text{ constant}$ We deduce that $ker(D) = P_0 \quad \triangle$

Alternatively, by rank nullity theorem, it would follow that $dim(Im(D)) = n$


# Does this double integral exists?

I've calculated $$\iint_D\frac{y}{x^2+y^2}\,dA$$ where D is bounded by $$y=x$$, $$y=2x$$, $$x=2$$, in this way: $$\iint_D\frac{y}{x^2+y^2}\,dA=\int_0^2\int_x^{2x}\frac{y}{x^2+y^2}\,dy\,dx=\cdots = \ln \frac{5}{2}.$$ However, I wonder if the fact that the $$\frac{y}{x^2+y^2}$$ is not bounded on D invalidates all the calculations and in fact the double integral does not exist. All the theorems that I consult have as a hypothesis that the integrand is bounded on the region of integration and hence my doubt regarding this double integral. Can anyone help me? Does this integral exist?

• You didn't bother to show the computation, but probably it's valid. The function you integrate is positive and for such you can carelessly change the order of integration due to Tonelli's theorem. Also, a function doesn't have to be bounded for an integral to exist, think for simplicity about one-dimensional integral $\int_0^1 x^{-1/2}\,dx$. Jun 10 '20 at 21:29
• Why is the fraction not bounded on the region $D$? Jun 10 '20 at 23:27

Changing to polar coordinates reveals why the integral is finite even though the integrand is $$\mathcal{O}(r^{-1})$$ as $$r \to 0$$.

For a quick check, note that the integrand is nonnegative and $$D$$ is a subset of the sector

$$S=\{(r,\theta): 0 \leqslant r \leqslant 2\sqrt{5},\, \frac{\pi}{4} \leqslant \theta \leqslant \arctan (2) \}$$

Thus,

$$\int\int_D \frac{y}{x^2+ y^2} dA \leqslant \int_{\frac{\pi}{4}}^{\arctan(2)}\int_0^{2\sqrt{5}} \frac{r \sin \theta}{r^2} r \, dr \, d\theta = 2\sqrt{5}\int_{\frac{\pi}{4}}^{\arctan(2)}\sin \theta \, d\theta,$$

where the integral on the RHS is finite since the sine function is bounded.

$$\int \frac{ydy}{x^2+y^2} = \frac{1}{2} \int \frac{d(y^2+x^2)}{x^2+y^2} =\frac{1}{2} \ln(x^2+y^2) + C$$ So $$\int_0^2\int_x^{2x}\frac{y}{x^2+y^2}\,dy\,dx= \frac{1}{2} \int_{0}^{2}\ln(x^2+y^2)|_{x}^{2x} dx$$


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# Find value of $a_1$ such that $a_{101}=5075$

Let $$\{a_n\}$$ be a sequence of real numbers where $$a_{n+1}=n^2-a_n,\, n=\{1,2,3,...\}$$

Find value of $$a_1$$ such that $$a_{101}=5075$$.

I have

$$a_2=1^2-a_1$$

$$a_3=2^2-a_2=2^2-1^2+a_1$$

$$a_4=3^2-2^2+1^2-a_1$$

$$a_5=4^2-3^2+2^2-1^2+a_1$$

$$\vdots$$

$$a_{101}=100^2-99^2+98^2-97^2+\ldots+2^2-1^2+a_1.$$

Therefore, $$a_{101}=\sum_{i=1}^{50}(2i)^2-\sum_{i=1}^{50}(2i-1)^2.$$

Thus, $$5075=\sum_{i=1}^{50}(4i^2-4i^2+4i-1)+a_1,$$

and, $$a_1=5075-4\sum_{i=1}^{50}(i)+\sum_{i=1}^{50}(1).$$

Hence, $$a_1=5075-4(\frac{50}{2})(51)+50=25,$$

and $$a_1=25$$.

Is it correct? Do you have another way? Please check my solution, thank you.

Yes, your solution is correct. Another method of solution is to note that if $$a_{n+1} = n^2 - a_n,$$ we want to find some (possibly constant) function of $$n$$ such that $$a_{n+1} - f(n+1) = -(a_n - f(n)).$$ This of course implies $$f(n+1) + f(n) = n^2.$$ A quadratic polynomial should do the trick: suppose $$f(n) = an^2 + bn + c,$$ so that $$n^2 = f(n+1) + f(n) = 2a n^2 + 2(a+b)n + (a+b+2c).$$ Equating coefficients in $$n$$ gives $$a = 1/2$$, $$b = -1/2$$, $$c = 0$$, hence $$f(n) = \frac{n^2 - n}{2}.$$ It follows that if $$b_n = a_n - f(n) = a_n - \frac{n^2-n}{2},$$ then $$b_{n+1} = - b_n.$$ This gives us $$b_1 = b_{101}$$ which in terms of $$a_n$$, is $$a_1 = a_1 - \frac{1^2 - 1}{2} = a_{101} - \frac{(101)^2 - 101}{2} = 5075 - 5050 = 25.$$ This solution seems to come out of nowhere, but it is motivated by the idea that if we can transform the given recurrence into a corresponding recurrence for a sequence that is much simpler to determine, we can use this to recover information about the original sequence.

What if at the point where:

$$a_{101} = 100^2 - 99^2 + 98^2 - 97^2 \cdots 2^2 - 1^2 + a_1$$

You opted for a clever factorization of squares:

$$a_{101} = (100 - 99)(100 + 99) + (98 - 97)(98 + 97) \cdots (2 - 1)(2 + 1) + a_1$$

$$a_{101} = 199 + 195 + 191 \cdots 3 + a_1$$

Therefore those numbers become a simple arithmetic series with the first term as 3 and the common ratio as 4.

But how many terms exactly?

$$3 + 4(n - 1) = 199$$

$$4(n - 1) = 196$$

$$n - 1 = 49$$

$$n = 50$$

Therefore with the knowledge of evaluating the sum this comes down to:

$$a_{101} = 25(199 + 3) + a_1$$

$$a_{101} = 5050 + a_1$$

And by substituting the choice for a_101:

$$5075 = 5050 + a_1$$

$$a_1 = 25$$

Thanks to @Zera for recommending the edit. I'm learning how to use these markup languages

• Question to @Zera, is my reasoning the same as yours. I feel like it is – Nεo Pλατo Dec 21 '19 at 16:59

In the most simple way:

$$A_{n+1}+A_n=n^2~~~(1)$$ is a non-homogeneous recurrence equation. Its homogeneous part $$A_{n+1}+A_n=0 \implies A_{n+1}=-A_n \implies A_n= (-1)^n S~~~(2)$$ In the RHS of (1) being $$n^2)$$, we can take $$A_n =P n^2+Q n+R$$; inserting this in (1), we get $$2P=1,P+Q=0,P+Q+2R=0 \implies P=1/2, Q=1/2, R=0.$$ Then the total finally solution of (1) is $$A_n=\frac{n(n-1)}{2}+ (-1)^n S.$$ Given that $$A_{101}=5075$$, finally we get $$A_n=\frac{n(n-1)}{2}+(-1)^{n+1}~ 25.$$

Another way :

$$a_{n+1}=n^2-a_n=n^2-((n-1)^2-a_{n-1})=a_{n-1}+2n-1$$

$$a_{n-1}=a_{n-3}+2(n-1)-1=a_{n-3}+2(n-2)+1$$

$$a_{n+1}=a_{n+1-2r}+\underbrace{2n-1+2n-3+\cdots}_{r\text{ terms}}$$

Here $$n+1=101,n+1-2r=1$$

You could also have notice the pattern $$a_n=(-1)^{n+1}a_1+\frac{n(n-1) }{2}$$