text:
$\sqrt{9-2\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}}$

I have no idea how to unnest radicals, can anyone help?

-
What are you trying to do with this, simplify it or convert it to a decimal? – Jed Mar 7 at 1:23
Simplify; I know the answer which is $\sqrt{2} +1$ but no idea how to get it – Anonymous Mar 7 at 1:24
In general, $a+b\sqrt{d}$ with $a,b,d$ integers might have a square root of the form $a'+b'\sqrt{d}$ only if $a^2-db^2$ is a square. That isn't a sufficient condition, but it is a good start. For example, you need $a+b\sqrt{d}>0$, and it doesn't help you find the square root. Finding the square root is often a trial by error with a little help from knowledge of "Pell-like equations." – Thomas Andrews Mar 7 at 1:42

Here's a hint to give you the kind of idea involved:

The innermost term $3-2\sqrt2$ can be rewritten as $1^2-2\cdot1\cdot\sqrt 2+(-\sqrt{2})^2=(1-\sqrt 2)^2$ which cancels the square root it lies inside. Now simply and apply the same strategy of recognizing squares to keep canceling square roots until you are left with a simple expression.

Edit: How do we recognize these squares? The core idea lies in the "mixed term". Our goal is to make the term inside the current innermost radical look like either $(a+b)^2=a^2+b^2+2ab$ or $(a-b^2)=a^2+b^2-2ab$ which have middle terms $2ab$ or $-2ab$ respectively. So in our example, we had $-2\sqrt 2$, so you know that we're going to get something of the form $(a-b)^2$ with $$ab=\sqrt 2.$$ However, we need $$a^2+b^2=3.$$

At this point you one of the $a$ or $b$ as $\sqrt 2$ since it shows prominently in the $ab$ term. So $a^2+(\sqrt 2)^2=3$, so $a$ needs to be $1$. This checks out with $ab=\sqrt 2$. This last bit, you have to eyeball it a little bit.

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I see that, but are there any tricks to recognize these squares? – Anonymous Mar 7 at 1:38
@Anonymous See my edit. Hope this helps! – Ravi Mar 7 at 1:47

Suppose that the term $\sqrt{3-2\sqrt2}$ can be written as $\sqrt a + b$ for some $a,b\in\mathbb Z$. We can equate these expressions then square both sides: $$\sqrt{3-2\sqrt2} = \sqrt a + b\\ 3 - 2\sqrt2 = a+b^2 + 2b\sqrt a$$ and from here you can see that the solution is $(a,b) = (2,-1)$. This gives you a methodical approach that you can apply recursively.

-

Observe that $$(\sqrt{2}-1)^2=2-2\sqrt{2}+1=3-2\sqrt{2}\qquad\implies\qquad\color{blue}{\sqrt{3-2\sqrt{2}}=\sqrt{2}-1}$$ Then, $$\color{blue}{\sqrt{10+4\sqrt{3-2\sqrt{2}}}}=\sqrt{10+4(\sqrt{2}-1)}=\sqrt{6+4\sqrt{2}}=\sqrt{4+4\sqrt{2}+2}=\color{blue}{\sqrt{(2+\sqrt{2})^2}}$$ So \begin{align} \sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}&=\sqrt{23-6(2+\sqrt{2})}\\ &=\sqrt{11-6\sqrt{2}}\\ &=\sqrt{9-6\sqrt{2}+2}\\ &=\sqrt{(3-\sqrt{2})^2}\\ &=3-\sqrt{2} \end{align} Finally \begin{align} \sqrt{9-2\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}}&=\sqrt{9-2(3-\sqrt{2})}\\ &=\sqrt{3-2\sqrt{2}}\\ &=\sqrt{2}-1 \end{align}

-

Work from the inside out.

Let $\sqrt{3 - 2\sqrt{2}} = \sqrt{a} - \sqrt{b}$, where $a$ and $b$ are rational numbers. Squaring both sides yields $$3 - 2\sqrt{2} = a + b - 2\sqrt{ab}$$ Then \begin{align*} a + b & = 3 \tag{1}\\ -2\sqrt{ab} & = -2\sqrt{2} \tag{2} \end{align*} Dividing equation by $-2$ yields $$\sqrt{ab} = \sqrt{2}$$ Squaring both sides of the equation yields $$ab = 2$$ Hence, $$b = \frac{2}{a}$$ Substituting for $b$ in equation 1 yields \begin{align*} a + \frac{2}{a} & = 3\\ a^2 + 2 & = 3a\\ a^2 - 3a + 2 & = 0\\ (a - 1)(a - 2) & = 0 \end{align*} Hence, $a = 1$ or $a = 2$. If $a = 1$, then $b = 2$. However, $\sqrt{1} - \sqrt{2} = 1 - \sqrt{2} < 0$, but $\sqrt{3 - 2\sqrt{2}} > 0$. Thus, $a = 2$ and $b = 1$. Hence, $$\sqrt{3 - 2\sqrt{2}} = \sqrt{2} - \sqrt{1} = \sqrt{2} - 1$$ Substituting $\sqrt{2} - 1$ for $\sqrt{3 - 2\sqrt{2}}$ yields $$\sqrt{10 + 4\sqrt{3 - 2\sqrt{2}}} = \sqrt{10 + 4(\sqrt{2} - 1)} = \sqrt{6 + 4\sqrt{2}}$$ Let $\sqrt{10 + 4\sqrt{2}} = \sqrt{c} + \sqrt{d}$, where $c$ and $d$ are rational numbers. Continue.

-


text:
# Displaying spatial sensor data from Arduino with R on Google Maps

February 26, 2014
By

(This article was first published on R Video tutorial for Spatial Statistics, and kindly contributed to R-bloggers)

For Christmas I decided to treat myself with an Arduino starter kit. I started doing some basic experiments and I quickly found out numerous website that sell every sort of sensor: from temperature and humidity, to air quality.

Long story short, I bought a bunch of sensors to collect spatial data. I have a GPS, accelerometer/magnetometer, barometric pressure, temperature/humidity, UV Index sensor.

Below is the picture of the sensors array, still in a breadboard version. Of course I also had to use an Arduino pro mini 3.3V and an openlog to record the data. Below the breadboard there is a 2200mA lithium battery that provides 3.7V to the Arduino pro mini.

With this system I can collect 19 columns of data: Date, Time, Latitude and Longitude, Speed, Angle, Altitude, Number of Satellites and Quality of the signal, Acceleration in X, Y and Z, Magnetic field in X, Y and Z, Temperature, Humidity, Barometric Pressure, UV Index.
With all these data I can have some fun testing different plotting method on R. In particular I was interested in plotting my data on Google Maps.

These are my results.

First of all I installed the package xlsx, because I have to first import the data into Excel for some preliminary cleaning. Sometimes the satellite sensor loses the connection and I have to delete those entries, or maybe the quality of the signal is poor and the coordinates are not reliable. In all these case I have to delete the entry and I cannot do it in R because of the way the GPS write the coordinates.
In fact, the GPS library, created by Adafruit, writes the coordinate in a format similar to this:
4725.43N        831.39E

This is read into a string format in R, so it requires a visual inspection to get rid of bad data.
This format is a combination of degrees and minutes, so it needs to be cleaned up before use. In the example above, the real coordinate in decimal degrees in calculated using the following formula:

47 + 25.13/60 N        8 + 31.39/60 E

Luckily, the fact that R recognises it as a string facilitates this transformation. With the following two lines of code I can transform each entry into a format that I can use:

data$LAT <- as.numeric(substr(paste(data[,"Lat"]),1,2)) + as.numeric(substr(paste(data[,"Lat"]),3,7))/60 data$LON <- as.numeric(substr(paste(data[,"Lon"]),1,1)) + as.numeric(substr(paste(data[,"Lon"]),2,6))/60

Then I assign these two columns as coordinate for the file with the sp package and set the projection to WGS84.

Now I can start working on visualising the data. I installed two packages: plotGoogleMaps and RgoogleMaps
The first is probably the simplest to use and let the user create a javascript page to plot the data from R to google maps through the google maps API. The plot is displayed into the default web browser and it looks extremely good.
With one line of code I can plot the temperature for each spatial point with bubble markers:

"Temp" is the name of the column in the SpatialPointsDataFrame that has the temperature data.

By adding an additional line of code I can set the markers as coloured text:

ic=iconlabels(data$Temp, height=12) plotGoogleMaps(data,iconMarker=ic,zcol="Temp") The package RgoogleMaps works differently, because it allows the user to plot google maps as background for static plots. This creates less stunning results but also allows more customisation of the output. It also requires a bit more work. In this example, I will plot the locations for which I have a data with the road map as background. I first need to create the map object and for centre the map in the centre of the area I visited, I used the bounding box of my spatial dataset: box<-bbox(data) Map<-GetMap(center = c(lat =(box[2,1]+box[2,2])/2, lon = (box[1,1]+box[1,2])/2), size = c(640, 640),zoom=16,maptype = c("roadmap"),RETURNIMAGE = TRUE, GRAYSCALE = FALSE, NEWMAP = TRUE) Then I need to transform the spatial coordinates into plot coordinates: tranf=LatLon2XY.centered(Map,data$LAT, data$LON, 16) x=tranf$newX
y=tranf$newY this function creates a new set of coordinates optimised for the zoom of the Map I created above. At this point I can create the plot with the following two lines: PlotOnStaticMap(Map) points(x,y,pch=16,col="red",cex=0.5) as you would do with any other plot, you can add points to the google map with the function points. As I mentioned above, this package creates less appealing results, but allows you to customise the output. In the following example I will compute the heading (the direction I was looking at) using the data from the magnetometer to plot arrows on the map. I computed the heading with these lines: Isx = as.integer(data$AccX)

Isy = as.integer(data$AccY) heading = (atan2(Isy,Isx) * 180) / pi heading[heading<0]=360+heading[heading<0] I did not tilt compensated the heading because the sensor was almost horizontal all the times. If you need tilt compensation, please look at the following websites for help: http://theccontinuum.com/2012/09/24/arduino-imu-pitch-roll-from-accelerometer/ With the heading I can plot arrows directed toward my heading with the following loop: PlotOnStaticMap(Map) for(i in 1:length(heading)){ if(heading[i]>=0 & heading[i]<90){arrows(x[i],y[i],x1=x[i]+(10*cos(heading[i])), y1=y[i]+(10*sin(heading[i])),length=0.05, col="Red")} if(heading[i]>=90 & heading[i]<180){arrows(x[i],y[i],x1=x[i]+(10*sin(heading[i])), y1=y[i]+(10*cos(heading[i])),length=0.05, col="Red")} if(heading[i]>=180 & heading[i]<270){arrows(x[i],y[i],x1=x[i]-(10*cos(heading[i])), y1=y[i]-(10*sin(heading[i])),length=0.05, col="Red")} if(heading[i]>=270 & heading[i]<=360){arrows(x[i],y[i],x1=x[i]-(10*sin(heading[i])), y1=y[i]-(10*cos(heading[i])),length=0.05, col="Red")} } I also tested the change of the arrow length according to my speed, using this code: length=data$Speed.Km.h.*10

}

However, the result is not perfect because in some occasions the GPS recorded a speed of 0 km/h, even though I was pretty sure to be walking.
The arrow plot looks like this:

format: plain


text:
NAG Library Function Document

1Purpose

nag_zgebak (f08nwc) transforms eigenvectors of a balanced matrix to those of the original complex general matrix.

2Specification

 #include #include
 void nag_zgebak (Nag_OrderType order, Nag_JobType job, Nag_SideType side, Integer n, Integer ilo, Integer ihi, const double scale[], Integer m, Complex v[], Integer pdv, NagError *fail)

3Description

nag_zgebak (f08nwc) is intended to be used after a complex general matrix $A$ has been balanced by nag_zgebal (f08nvc), and eigenvectors of the balanced matrix ${A}_{22}^{\prime \prime }$ have subsequently been computed.
For a description of balancing, see the document for nag_zgebal (f08nvc). The balanced matrix ${A}^{\prime \prime }$ is obtained as ${A}^{\prime \prime }=DPA{P}^{\mathrm{T}}{D}^{-1}$, where $P$ is a permutation matrix and $D$ is a diagonal scaling matrix. This function transforms left or right eigenvectors as follows:
• if $x$ is a right eigenvector of ${A}^{\prime \prime }$, ${P}^{\mathrm{T}}{D}^{-1}x$ is a right eigenvector of $A$;
• if $y$ is a left eigenvector of ${A}^{\prime \prime }$, ${P}^{\mathrm{T}}Dy$ is a left eigenvector of $A$.

None.

5Arguments

1:    $\mathbf{order}$Nag_OrderTypeInput
On entry: the order argument specifies the two-dimensional storage scheme being used, i.e., row-major ordering or column-major ordering. C language defined storage is specified by ${\mathbf{order}}=\mathrm{Nag_RowMajor}$. See Section 3.3.1.3 in How to Use the NAG Library and its Documentation for a more detailed explanation of the use of this argument.
Constraint: ${\mathbf{order}}=\mathrm{Nag_RowMajor}$ or $\mathrm{Nag_ColMajor}$.
2:    $\mathbf{job}$Nag_JobTypeInput
On entry: this must be the same argument job as supplied to nag_zgebal (f08nvc).
Constraint: ${\mathbf{job}}=\mathrm{Nag_DoNothing}$, $\mathrm{Nag_Permute}$, $\mathrm{Nag_Scale}$ or $\mathrm{Nag_DoBoth}$.
3:    $\mathbf{side}$Nag_SideTypeInput
On entry: indicates whether left or right eigenvectors are to be transformed.
${\mathbf{side}}=\mathrm{Nag_LeftSide}$
The left eigenvectors are transformed.
${\mathbf{side}}=\mathrm{Nag_RightSide}$
The right eigenvectors are transformed.
Constraint: ${\mathbf{side}}=\mathrm{Nag_LeftSide}$ or $\mathrm{Nag_RightSide}$.
4:    $\mathbf{n}$IntegerInput
On entry: $n$, the number of rows of the matrix of eigenvectors.
Constraint: ${\mathbf{n}}\ge 0$.
5:    $\mathbf{ilo}$IntegerInput
6:    $\mathbf{ihi}$IntegerInput
On entry: the values ${i}_{\mathrm{lo}}$ and ${i}_{\mathrm{hi}}$, as returned by nag_zgebal (f08nvc).
Constraints:
• if ${\mathbf{n}}>0$, $1\le {\mathbf{ilo}}\le {\mathbf{ihi}}\le {\mathbf{n}}$;
• if ${\mathbf{n}}=0$, ${\mathbf{ilo}}=1$ and ${\mathbf{ihi}}=0$.
7:    $\mathbf{scale}\left[\mathit{dim}\right]$const doubleInput
Note: the dimension, dim, of the array scale must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
On entry: details of the permutations and/or the scaling factors used to balance the original complex general matrix, as returned by nag_zgebal (f08nvc).
8:    $\mathbf{m}$IntegerInput
On entry: $m$, the number of columns of the matrix of eigenvectors.
Constraint: ${\mathbf{m}}\ge 0$.
9:    $\mathbf{v}\left[\mathit{dim}\right]$ComplexInput/Output
Note: the dimension, dim, of the array v must be at least
• $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{pdv}}×{\mathbf{m}}\right)$ when ${\mathbf{order}}=\mathrm{Nag_ColMajor}$;
• $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}×{\mathbf{pdv}}\right)$ when ${\mathbf{order}}=\mathrm{Nag_RowMajor}$.
The $\left(i,j\right)$th element of the matrix $V$ is stored in
• ${\mathbf{v}}\left[\left(j-1\right)×{\mathbf{pdv}}+i-1\right]$ when ${\mathbf{order}}=\mathrm{Nag_ColMajor}$;
• ${\mathbf{v}}\left[\left(i-1\right)×{\mathbf{pdv}}+j-1\right]$ when ${\mathbf{order}}=\mathrm{Nag_RowMajor}$.
On entry: the matrix of left or right eigenvectors to be transformed.
On exit: the transformed eigenvectors.
10:  $\mathbf{pdv}$IntegerInput
On entry: the stride separating row or column elements (depending on the value of order) in the array v.
Constraints:
• if ${\mathbf{order}}=\mathrm{Nag_ColMajor}$, ${\mathbf{pdv}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$;
• if ${\mathbf{order}}=\mathrm{Nag_RowMajor}$, ${\mathbf{pdv}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$.
11:  $\mathbf{fail}$NagError *Input/Output
The NAG error argument (see Section 3.7 in How to Use the NAG Library and its Documentation).

6Error Indicators and Warnings

NE_ALLOC_FAIL
Dynamic memory allocation failed.
See Section 2.3.1.2 in How to Use the NAG Library and its Documentation for further information.
On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value.
NE_INT
On entry, ${\mathbf{m}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{m}}\ge 0$.
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}\ge 0$.
On entry, ${\mathbf{pdv}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{pdv}}>0$.
NE_INT_2
On entry, ${\mathbf{pdv}}=〈\mathit{\text{value}}〉$ and ${\mathbf{m}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{pdv}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$.
On entry, ${\mathbf{pdv}}=〈\mathit{\text{value}}〉$ and ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{pdv}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
NE_INT_3
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$, ${\mathbf{ilo}}=〈\mathit{\text{value}}〉$ and ${\mathbf{ihi}}=〈\mathit{\text{value}}〉$.
Constraint: if ${\mathbf{n}}>0$, $1\le {\mathbf{ilo}}\le {\mathbf{ihi}}\le {\mathbf{n}}$;
if ${\mathbf{n}}=0$, ${\mathbf{ilo}}=1$ and ${\mathbf{ihi}}=0$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
See Section 2.7.6 in How to Use the NAG Library and its Documentation for further information.
NE_NO_LICENCE
Your licence key may have expired or may not have been installed correctly.
See Section 2.7.5 in How to Use the NAG Library and its Documentation for further information.

7Accuracy

The errors are negligible.

8Parallelism and Performance

nag_zgebak (f08nwc) makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.
Please consult the x06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this function. Please also consult the Users' Note for your implementation for any additional implementation-specific information.

The total number of real floating-point operations is approximately proportional to $nm$.


text:
# Homework Help: Simplify the following expression

1. Jan 13, 2012

### tahayassen

1. The problem statement, all variables and given/known data

http://img513.imageshack.us/img513/88/71893688.png [Broken]

2. Relevant equations

Not applicable.

3. The attempt at a solution

http://img441.imageshack.us/img441/846/14479081.png [Broken]

I'm already stuck! I calculated both expressions and they don't equal each other. :(

Last edited by a moderator: May 5, 2017
2. Jan 13, 2012

### tahayassen

Actually, don't bother helping me with this one. I think I've figured out. I needed to find the LCD between the two fractions.

Instead, answer this: why can't I take the square root of a binomial?

edit: I'm stuck again with this question. I can't seem to get rid of the large radical even after I combine the fractions. :(

3. Jan 13, 2012

### Mentallic

You can't do that because generally,

$$\sqrt{a+b}\neq \sqrt{a}+\sqrt{b}$$

And if you want to see why, if these are equal, then their squares must be equal. Try squaring both sides and see what happens.

This is one of those problems that unless you know where you're going, it's going to be difficult finding your way there. I tried a few things out but I don't see the radicals being simplified. I believe the only way to simplify this expression is to take a common denominator and then take a factor of 4 in the denominator out of the larger surd.

4. Jan 13, 2012

### I like Serena

Wolfram comes up with a nice one:
$$\frac 1 2 \sqrt{2 + \sqrt{2 - \sqrt 3}}$$

Beyond that I also believe there's not much that is useful to simplify.

5. Jan 13, 2012

### Mentallic

If you consider further nested roots as being simpler, then sure :tongue:

I made a futile attempt to see if the larger surd can be removed by trying to find if there exist integers a,b,c,d to the following equality:

$$\left(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}\right)^2= \sqrt{6} -\sqrt{2}+4$$

but to my amazement (not really, I was expecting it) to figure this out, one would need to solve these non-linear simultaneous equations:

$$a^2+2b^2+3c^2+6d^2=4$$
$$ab+3cd=\frac{-1}{2}$$
$$ac+2bd=0$$
$$ad+bc=\frac{1}{2}$$

And I haven't a clue how to proceed - plus I'm not really feeling up for the challenge considering its pretty unlikely that the variables are all in fact integers.

Last edited: Jan 13, 2012
6. Jan 13, 2012

### I like Serena

Less work to type into a calculator.
And it looks nice!

Neither have I, but if you do proceed, please swap the signs of the 2nd and 4th line. ;)

7. Jan 13, 2012

### Curious3141

I'm pretty sure there's no way to make that surd much simpler in any meaningful sense.

8. Jan 13, 2012

### Mentallic

Agreed, it looks pretty sweet!

Uhh... How I managed to transcribe that incorrectly I'll never know :yuck:

9. Jan 13, 2012

### Ray Vickson

Well, this is probably not what is wanted in high-school or first-year college, but one _can_ get a solution using Groebner bases. Here is what happens in Maple 14:
sys:={a^2+2*b^2+3*c^2+4*d^2-4,2*a*b+6*c*d+1,a*c+2*b*d,2*a*d+2*b*c-1};
with(Groebner);

B:=Basis(sys,plex(a,b,c,d));
B := [1-256*d^2+19072*d^4-626560*d^6+10231872*d^8-85606400*d^10+351143936*d^12-621502464*d^14+359817216*d^16,

58700876744*d-6772491730030*d^3+516306245937248*d^5-16676091040182256*d^7+214982693609933568*d^9-1128998897071871232*d^11+2264133466874087424*d^13-1391829359823378432*d^15+5466782401*a,

20794688424*d-25887729161344*d^3+1613081383661048*d^5-35738765248743936*d^7+355188616828494592*d^9-1606762324171226112*d^11+2991485535250916352*d^13-1775184867367059456*d^15+5466782401*b,

127738358584*d-36353540383530*d^3+2108143299164768*d^5-47032277980064368*d^7+476144844758761728*d^9-2186167800738890496*d^11+4105190085058529280*d^13-2446662083621640192*d^15+5466782401*c]

So, d is one of the roots of the 16th degree polynomial in B[1]; then we can get a, b and c by setting B[2], B[3] and B[4] to zero. See? Nothing to it!

RGV

10. Jan 13, 2012

### I like Serena

Well, all roots of B[1] appear to be irrational (according to Wolfram), so I don't think we get a nice simplification. ;)

11. Jan 13, 2012

### SammyS

Staff Emeritus
Look at various ways to express $\displaystyle\cos\left(\frac{5\pi}{24}\right)\,,$ and/or $\displaystyle\sin\left(\frac{7\pi}{24}\right)\,.$

12. Jan 13, 2012

### I like Serena

Nice! However did you find that one?

13. Jan 13, 2012

### SammyS

Staff Emeritus
It reminded me of a problem I used to give students in which they were to find $\displaystyle\sin\left(\frac{\pi}{12}\right)\,,$ using both angle subtraction and half angle identities.

(I then had them give the approximate numerical answer to each -- mostly so they could see that the two different looking answers were equivalent.)

So I played around with that a bit for the problem in this thread.

14. Jan 13, 2012

### e^(i Pi)+1=0

I went a little nuts with this problem and got...

$\frac{(6(√6-√2)+24)\sqrt{6(√6-√2)+24}}{√48(6(√6-√2)+24)}$

Are radicals still considered nested if they're within parentheses?

edit: they would be since you could just distribute in the 6, oh well.

15. Jan 13, 2012

### Mentallic

Could you try this again but with
sys:={a^2+2*b^2+3*c^2+6*d^2-4,2*a*b+6*c*d+1,a*c+2*b*d,2*a*d+2*b*c-1};

What a relief!

Ahh yes that's where I remember seeing something similar to the expression $$\frac{\sqrt{6}-\sqrt{2}}{8}$$
Thanks for reminding me.


text:
# Question about how to speed up Mathematica code

When looking at the Minimal Goldbach prime partition point {p,q} for each n; where n=10^i and i = 2,3,4,...,10; I notice that these points reside in an interval with center n/2 and radius 250:

Clear[mgppp, lowerbound, upperbound]

mgppp[n_?EvenQ] :=
Last[ Transpose@{n - #, #} &@
Select[n - Prime@Range@PrimePi[n/2], PrimeQ]]

lowerbound[n_?EvenQ] := Module[{d}, {m = n/2, d = (m - 250)}; d]

upperbound[n_?EvenQ] := Module[{d}, {m = n/2, d = (m + 250)}; d]

hello = Table[{n, lowerbound[n], mgppp[n][[1]], n/2, mgppp[n][[2]],
upperbound[n]}, {n, {10^2, 10^3, 10^4, 10^5, 10^6, 10^7, 10^8,
10^9, 10^10}}];TableForm[hello,
TableHeadings -> {None, {"n", "(n/2)-250", "MGPPPp", "n/2", "MGPPPq",
"(n/2)+250"}}, TableAlignments -> Center]

Mathematica takes about 45 minutes to build the above table. Notice that each lower bound value (n/2)-250 is less than p and each upper bound value (n/2)+250 is greater than q.

Now I want to verify (or not) that the same happens for ALL even n's between 10^2 and 10^10. To simplify the number of evaluations; I compare only the lowerbound values with the p values as follows:

Clear[mgppp, lowerbound]

mgppp[n_?EvenQ] :=
Last[ Transpose@{n - #, #} &@
Select[n - Prime@Range@PrimePi[n/2], PrimeQ]]

lowerbound[n_?EvenQ] := Module[{d}, {m = n/2, d = (m - 250)}; d]

bag = {}; For[n = 10^2, n <= 10^10, n += 2,
If[lowerbound[n] < mgppp[n][[1]], Nothing,
AppendTo[bag, {n, lowerbound[n], mgppp[n][[1]]}]]]; TableForm[bag,
TableHeadings -> {None, {"n", "(n/2)-250", "MGPPPp"}},
TableAlignments -> Center]

The above evaluation is very slow and it takes days to be evaluated using my personal computer. I would appreciate any suggestions about how to speed up the calculations; perhaps using "Compile" or any other method to do this. Thank you.

• One of the problems, perhaps the problem, is PrimePi. Your algorithm as it is written will never be faster than PrimePi[Range[1, 10^10/2, 2]]; // AbsoluteTiming. Feb 23, 2018 at 20:27
• You are right C.E.; thank you; that version of mpppp is slow. The following version is much faster: mgppp2[n_?EvenQ] /; n > 3 := Block[{m = PrimePi[n/2], p}, While[! PrimeQ[q = n - (p = Prime[m])], m--]; {p, q}] But, I still have the same evaluation issues after using this second version. Feb 23, 2018 at 21:28
• What I am concerned about is actually the For loop. It will call mgpppp or mgppp2 once for every n between 10^2 and 10^8, so it will not be faster than PrimePi[Range[10^2, 10^10, 2]/2]; because those functions evaluate PrimePi for n divided by two. PrimePi appears to be slow for large n and that is a problem. Feb 24, 2018 at 7:52
• n is even. m=n/2 is an integer and PrimePi[m] is then calculated. These PrimePi[m] values fall in an interval with center n/2 and radius 250; so, computationally this shouldn’t be time exhausting. The problem lies someplace else C.E. Feb 25, 2018 at 15:22
• You're right, I made a mistake. I posted a more careful analysis. Feb 25, 2018 at 21:54

The combination of For and AppendTo in your code is an inefficient way of substituting Table, and although it is not a major culprit we might as well change that. What's more serious is that if mgppp[n][[1]] is larger than the lower bound then you evaluate it a second time. This is an expensive function, so it shouldn't be evaluated unless it has to be evaluated, so that should also be fixed. We end up with this code:

Clear[mgppp, lowerbound]

mgppp[n_?EvenQ] := Last[Transpose@{n - #, #} &@Select[n - Prime@Range@PrimePi[n/2], PrimeQ]]
lowerbound[n_?EvenQ] := Module[{d}, {m = n/2, d = (m - 250)}; d]

RuntimeToolsProfile[
res = Table[With[{y = First@mgppp[n]},
If[lowerbound[n] < y, Nothing, {n, lowerbound[n], y}]
], {n, 10^2, 10^5, 2}];
];

Where I have changed the upper limit from 10^10 to 10^5 and have wrapped the whole thing with a function that does profiling. You can read about profiling Mathematica code in the Q&A Profiling from Mathematica.

This code will not only run the computation but will also report (in debug mode – this is explained in the other question) on how much time each part of the code took. I changed the upper limit so that it would finish in a reasonable time. It outputs a report like this:

These reports are usually straightforward to interpret, in this case, however, I'm not sure. It says that Prime@Range@PrimePi[n/2] took 254.6 seconds, whereas Prime took 88 seconds. The difference should be in the time it took to evaluate PrimePi, but it says that PrimePi only took 0.46 seconds to evaluate. I'll ask about this in the chat room to see if anyone understands this.

In either case, I believe that the time is spent evaluating built-in functions. There isn't much overhead that can be optimized. Memoization can be used if we suspect that the code is doing the same work several times over, so I tried doing this:

Clear[prime, primeQ]

prime[x_] := prime[x] = Prime[x]
primeQ[x_] := primeQ[x] = PrimeQ[x]

RuntimeToolsProfile[
res = Table[With[{y = First@mgppp[n]},
If[lowerbound[n] < y, Nothing, {n, lowerbound[n], y}]
], {n, 10^2, 10^5, 2}];
];

But it didn't make it faster. So at this point, I'm wondering if this can really be optimized at the code level.

• I think Prime and PrimeQ cache results. My guess is that the 135,000,000 calls to Prime didn't add to the "time" of Prime when retrieving a cached result, but they still took around 135000000 * 10^-6 sec. to execute the function calls. Feb 25, 2018 at 22:06
• @MichaelE2 That makes sense, thank you. Feb 26, 2018 at 4:49
• I apologize for not not returning fast and proper feedback to all of your suggestions above but, my personal computer was attacked by a virus today at work and our IT technicians had to clean my PC and I had to re-install Mathematica and get use to a new Windows 10 Operations System. Feb 26, 2018 at 21:11
• Clear[mgppp, lowerbound, upperbound] mgppp[n_?EvenQ] /; n > 3 := Block[{m = PrimePi[n/2], p}, While[! PrimeQ[q = n - (p = Prime[m])], m--]; {p, q}] lowerbound[n_?EvenQ] := Module[{d}, {m = n/2, d = (m - 250)}; d] (** 10^2 to 10^3 range: **) hello = Table[{n, lowerbound[n], mgppp[n][[1]], n/2, mgppp[n][[2]], upperbound[n], If[lowerbound[n] < mgppp[n][[1]], "ok", "Not bounded"]}, {n, 10^2, 10^3, 2}]; TableForm[hello, TableHeadings -> {None, {"n", "(n/2)-250", "MGPPPp", "n/2", "MGPPPq", "(n/2)+250", "Bounding Status"}}, TableAlignments -> Center] Feb 26, 2018 at 21:12

The main problem is you are generating a large number of prime pairs then using Last to throw most of them away. This starts in the middle and stops as soon as we find a pair: This runs in a fraction of a second..

myp[n_ /; PrimeQ[n/2]] := {n/2, n/2}  (* special case if n/2 is prime *)
myp[n_?EvenQ] := Module[{y = Prime@(PrimePi[n/2]+1)},
While[! AllTrue[{y, n - y}, PrimeQ], y = NextPrime[y]];
{y,n-y}]
hello = Table[{n, lowerbound[n], (mm = myp[n])[[1]], n/2, mm[[2]],
upperbound[n]}, {n, {10^2, 10^3, 10^4, 10^5, 10^6, 10^7, 10^8,
10^9, 10^10}}]; // AbsoluteTiming
TableForm[hello,
TableHeadings -> {None, {"n", "(n/2)-250", "MGPPPp", "n/2", "MGPPPq",
"(n/2)+250"}}, TableAlignments -> Center]

also in your table you were calling mgppp[n] twice for every n. Take note how I fixed that.. (factor of two timing savings right there )

aside, we can easily find numbers that break your bounds..

myp[#] - #/2 &@ 12573226
mgppp[#] - #/2 &@ 12573226

{-1140, 1140}

{-1140, 1140}

If we evaluate:

Clear[mgppp, lowerbound, upperbound]

mgppp[n_?EvenQ] /; n > 3 :=
Block[{m = PrimePi[n/2], p},
While[! PrimeQ[q = n - (p = Prime[m])], m--]; {p, q}]

lowerbound[n_?EvenQ] := Module[{d}, {m = n/2, d = (m - 250)}; d]

(** 10^2 to 10^3 range: **)

hello = Table[{n, lowerbound[n], mgppp[n][[1]], n/2, mgppp[n][[2]],
upperbound[n],
If[lowerbound[n] < mgppp[n][[1]], "ok", "Not bounded"]}, {n, 10^2,
10^3, 2}];

TableForm[hello,
TableHeadings -> {None, {"n", "(n/2)-250", "MGPPPp", "n/2", "MGPPPq",
"(n/2)+250", "Bounding Status"}}, TableAlignments -> Center]

(** 10^3 to 10^4 range: **)

hello2 = Table[{n, lowerbound[n], mgppp[n][[1]], n/2, mgppp[n][[2]],
upperbound[n],
If[lowerbound[n] < mgppp[n][[1]], "ok", "Not bounded"]}, {n, 10^3,
10^4, 2}];

TableForm[hello2,
TableHeadings -> {None, {"n", "(n/2)-250", "MGPPPp", "n/2", "MGPPPq",
"(n/2)+250", "Bounding Status"}}, TableAlignments -> Center]

we find that the lower and upper bounds work for n in the above ranges.

If we evaluate the following:

(** 10^4 to 10^5 range: **)

hello3 = Table[
If[lowerbound[n] < mgppp[n][[1]],
Nothing, {n, lowerbound[n], mgppp[n][[1]], n/2, mgppp[n][[2]],
upperbound[n], "Unbounded"}], {n, 10^4, 10^5, 2}];

TableForm[hello3,
TableHeadings -> {None, {"n", "(n/2)-250", "MGPPPp", "n/2", "MGPPPq",
"(n/2)+250", "Bounding Status"}}, TableAlignments -> Center]

we find the the bounds fail for all the n values listed on the hello3 table.

The evaluations take longer when we reach the 10^5 to 10^6 range:

(** 10^5 to 10^6 range: **)

hello4 = Table[
If[lowerbound[n] < mgppp[n][[1]],
Nothing, {n, lowerbound[n], mgppp[n][[1]], n/2, mgppp[n][[2]],
upperbound[n], "Unbounded"}], {n, 10^5, 10^6, 2}];

TableForm[hello4,
TableHeadings -> {None, {"n", "(n/2)-250", "MGPPPp", "n/2", "MGPPPq",
"(n/2)+250", "Bounding Status"}}, TableAlignments -> Center]

For larger ranges the evaluations take too long.

The n values below correspond to sequence A065978 at: http://oeis.org/A065978

The lowerbound values below correspond to sequence A155764 at: http://oeis.org/A155764

 mgppp[n_?EvenQ] /; n > 3 :=
Block[{m = PrimePi[n/2], p},
While[! PrimeQ[n - (p = Prime[m])], m--]; p]

lowerbound[n_?EvenQ] := Module[{d}, {m = n/2, d = (m - mgppp[n])}; d]

bag = {}; n = 4; max = -1; While[n <= 10^5,
If[lowerbound[n] > max, AppendTo[bag, {n, lowerbound[n]}];
max = lowerbound[n]]; n++]; bag

Notice that I was attempting to write a program that verifies whether pre-selected lower and upper bounds for the MGPPP work for ALL even n between 10^2 and 10^10. The purpose of the examples with tables listing the lower and upper bounds for n's that are powers of 10, was to illustrate and give a simpler starting point to get at the ultimate task of doing this for ALL even n's in the above range. This is not the best approach to tackle this problem; a better approach is to collect the n's corresponding to the strictly-increasing distances n/2-MGPPP[n] as done in A065978 at http://oeis.org/A065978 and A155764 at http://oeis.org/A155764 where this approach is applied. Thank you all for your valuable help!


text:
# Maxwell fisheye lens

Contributor: Stas Fainer

This is a simulation of a Maxwell fish-eye lens, which is a spherical dielectric with refractive index $$n(\rho) = \frac{n_0}{1+(\frac{\rho}{R})^2}$$, where $$n_0=2$$ is the refractive index in the center of the lens, $$R=100$$ is the radius of the lens, and $$\rho$$ is the radial distance from the center of the lens.

This simulation shows a Maxwell fish-eye lens that is composed of $$N=20$$ concentric spherical lenses with radius $$R_i=5(N+1-i)$$ and refractive index $$n_i = \frac{n_0}{1+(\frac{\rho}{R})^2}$$, where $$i=1,...,N$$. However, since this simulator calculates the effective refractive index of an optical element by multiplying the element's numerical refractive index with the numerical refractive indices of the optical elements which are embedded within it, the numerical refractive index of the $$i$$th concentric spherical lens is given by $$n_{i}^\text{numerical}=\frac{n_i}{\prod\limits_{k=i+1}^{N}n_k}$$.


text:
# Prove that an odd degree polynomial must converge to $\pm\infty$ when $x\to\pm\infty$

While what I'm trying to prove is related to this topic but alas I cannot just assume that given: $p(x)$ where $\deg(p(x))=2n+1$ where $n\in\mathbb{N}$ that the only two cases are:

• $\lim \limits_{x\to -\infty}f(x)=-\infty$ and $\lim \limits_{x\to \infty}f(x)=\infty$

or

• $\lim \limits_{x\to -\infty}f(x)=\infty$ and $\lim \limits_{x\to \infty}f(x)=-\infty$

I've tried proving that given that

$x^{2n+1}$ and $\sum \limits_{k=0}^{2n}{a_{k}x^k}$, $n\in\mathbb{N}$ and $a_k>1$

that

$\exists x_1\in\mathbb{R} , \forall x>x_1$

$x^{2n+1} > \sum \limits_{k=0}^{2n}{a_{k}x^k}$

And I think I proved it but it's an ugly inductive proof: for $n=1$

$x^{3} > a_2x^2+a_1x+a_0$

then I've tried choosing $x_1 = (a_2a_1a_0)^2$

then assuming $x^{2n+1} > \sum \limits_{k=0}^{2n}{a_{k}x^k}$ proving that $x^{2n+3} > \sum \limits_{k=0}^{2n+2}{a_{k}x^k}$

Which is the same as : $x^{2n+3} > \sum \limits_{k=0}^{2n}{a_{k}x^k} + a_{2n+1}x^{2n+1}+ a_{2n+2}x^{2n+2}$ and then replace $\sum \limits_{k=0}^{2n}{a_{k}x^k}$ with $x^{2n+1}$

meaning : $x^{2n+3} > x^{2n+1} + a_{2n+1}x^{2n+1}+ a_{2n+2}x^{2n+2}$ and assuming $x\neq0$ dividing by $x^{2n+1}$ will mean that it's sufficient to show that:

$x^{2} > 1 + a_{2n+1} + a_{2n+2}x$ which is too fairly easy to show.

The problem is I don't know whether showing this is really sufficient ( I intended to use that to show that if for example the coefficient of the highest degree variable is positive that $\forall M\in\mathbb{R} \exists X_1 , x>X_1 \Rightarrow f(x)>M$ ).

1. Whether it makes sense at all because it feels like I have holes in my proof.
2. Is there a more mathematically elegant way to prove this not using derivatives.

Thanks and I deeply apologize if I broke any rule, I've really tried not to.

-

$$\lim_{x \to +\infty} 1 + 2x + x^2 = \lim_{x \to +\infty} x^2 \left(\frac{1}{x^2} + \frac{2}{x} + 1 \right)$$
$$\lim_{x \to 1} 1 - x - x^2 + x^3 = \lim_{x \to 1} (x-1)^2 \left((x-1) + 2 \right)$$
So you propose to write that given the previously mentioned $p(x)$ I can always describe it as $a_{2n+1}x^{2n+1} + a_{2n}x^{2n} + \cdots + a_{2}x^{2} + a_{0}$ and then just factor it into $\lim\limits_{x \to +\infty} x^{2n+1} \left(\frac{a_0}{x^{2n+1}} + \cdots + \frac{a_{2n}}{x} + a_{2n+1} \right)$ ? And then using limit properties show that this limit is indeed $\pm\infty$ depending on the sign of $a_{2n+1}$ when $x\to\pm\infty$ ? –  Scis Jan 12 '13 at 9:50


text:
1. ## Optimization problem

How would I start a problem like the one below?

A poster is to have an area of 220 $in^2$ with 3 inch margins at the bottom and sides and a 5 inch margin at the top. What dimensions will give the largest printed area? (Give your answers correct to one decimal place.)

Here are my steps so far:

220 = w * h
220 / h = w

A = (w - (3 + 5)) * (h - (3 + 5))
A = (w - 8) * (h - 8)
A = wh - 8h - 8w + 64

A = (220 / h)h - 8h - 8(220 / h) + 64
A = 220 - 8h - 1760 / h + 64
A = 156 - 6h - 1760 / h

A' = -8 - -1*1760 / h^2
A' = -8 + 1760 / h^2

0 = -8 + 1760 / h^2
0 = -8h^2 + 1760
8h^2 = 1760
h^2 = 220
h = 14.8

220 / h = w
220 / 14.8 = 14.8 = w

2. Originally Posted by ioke09
A = (w - (3 + 5)) * (h - (3 + 5))
This seems incorrect. There is a 3 inch margin on both sides, so it should be: $A = [w - (3 + 3)] \cdot [h - (3 + 5)]$

3. I got 12.8 in for the width and 17.1 in for the height as the final answers, is this correct?

4. Yep, that's what I got.

5. I'm stuck with solving the problem below as well, any help would be appreciated.

Here's what I have:

A(x)= (x/4)^2 + pi(18-x^2/2pi)= x^2/16 + (18-x)^2/4pi

0<= x <= 18

A'(x)= x/8 - 18-x/2pi

= (1/2pi + 1/8)x - 9/pi= 0

A piece of wire 18 m long is cut into two pieces. One piece is bent into a square and the other is bent into a circle. (Give your answers correct to two decimal places.)

(a) How much wire should be used for the circle in order to maximize the total area?

(b) How much wire should be used for the circle in order to minimize the total area?


text:
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To use the online Pythagorean tool you simply Pythagorean Theorem Calculator: This calculator performs either of 2 items: 1) If you are given a right triangle with 2 sides and 1 hypotenuse, the calculator will determine the missing item. Use the Pythagorean theorem to calculate the hypotenuse from right triangle sides. - Best calculator to learn and solve Algebra and Geometry. - Pythagorean theorem calculator do all the hard work of calculation for you. This calculator works on every device, you can use Triangle. Pythagoras theorem can be used to calculate the length of any side in a right-angled triangle.Pythagoras theorem can be applied to solve 3-dimensional problems. What is The Formula of Pythagorean Theorem? Pythagorean theorem calculator. en. Therefore 9, 12, 15 forms a right angled triangle. Enter the values of two sides in the input fields and enter x for the unknown side value. Using the Pythagorean Theorem to prove that a triangle is not right. Assign tasks. 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For right triangle: the square value the hypotenuse (c) is equal to the sum of the square value of leg (a) and the square value of leg (b): Pythagorean theorem Application of Pythagoras Theorem in Real Life. a2 + One angle is equal to 90 degrees. Because a and b are legs and c is hypotenuse, by Pythagorean Theorem, we have. Step 1: Enter the Equation you want to solve into the editor. Weekly Subscription$2.99 USD per week until cancelled. Back to: All Pythagorean Theorem Solver, Cartoon Lesson, and practice. Pythagorean theorem: find the perimeter 5. Step 2: Click the button Solve to get the unknown side measure. It was given by Pythagoras of Samos, an ancient Ionian Greek philosopher who also Calculate hypotenuse, measurements and ratio easily with our 45 45 90 triangle calculator. Weekly Subscription $2.99 USD per week until cancelled. Pythagorean Theorem Calculator. By Pythagorean theorem problem solver Comentarios 381388) Learning Target: Understand the Pythagorean Theorem. 2. 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Given : A circle with center at O There are different types of questions, some of which ask for a missing leg and some that ask for the hypotenuse Example 3 : Supplementary angles are ones that have a sum of 180 Ptolemy's theorem states the relationship between the diagonals and the sides of a cyclic quadrilateral Ptolemy's The Pythagorean theorem calculator will find b if we require to find the said value, or the Pythagorean theorem calculator will find c or the Pythagorean theorem calculator will even Fill in 3 of the 6 fields, with at least one side, and press the 'Calculate' button. The Pythagorean Theorem states that In a right-angled triangle, the square of the hypotenuse side is equal to the sum of the squares of the other two sides. Lets say you know a 225 = 225. And it's really important that you realize that it's not 9 squared plus 14 squared is going to be equal to a squared. It is called "Pythagoras' Theorem" and can be written in one short equation: a 2 + b 2 = c 2. \cos^2\theta+\sin^2\theta=1. zero property of multiplication. This tool is a web-based tool that can be used from anywhere. zepto-zero. On This Page: ; Pythagorean theorem was proven by an acient Greek named Pythagoras and says that for a right triangle with legs A and B, and hypothenuse C. See this lesson on Pythagorean Theorem, animated proof. The length of the vertical leg is 4 units. The Pythagorean theorem calculator will find b if we require to find the said value, or the Pythagorean theorem calculator will find c or the Pythagorean theorem calculator will even solve for b according to your requirement. It is one of the most basic geometric tools in mathematics. The online Pythagorean Theorem calculator helps to calculate the length of any missing sides (a, b, or c) of a right triangle. The formula is given as: Where a and b are the legs and c is the hypotenuse. Solve for x Calculator. Answer (1 of 5): The Pythagorean Theorem states that a + b = c, where a and b are the lengths of the 2 legs, and c is the length of the hypotenuse. Sum of all The lengths of sides A and B must be longer than side C. There are two textboxes for entering the side lengths of the triangle. The three sides of the triangle are hypotenuse, base and perpendicular. Solution : Let x be the length of the rectangle. In 3D. 3 sides. Click the button Solve to The tool which is used to find the long side of the right triangle is the hypotenuse calculator. One Time Payment$19.99 USD for 3 months. Where A , B, and C are the internal angles of a triangle. Pythagorean Theorem Examples & Solutions. - Visual solutions to clearly understand the process. <--- Side 2. A2_1.04 Solving Features of Pythagoras Calculator. Step no 3: Simplify the Perpendicular side length (b) = 12 cm.

This calculator solves the Pythagorean Theorem equation for sides a or b, or the hypotenuse c. The hypotenuse is the side of the triangle opposite the right angle. Side A: Side B: Hypotenuse: Calculate Reset: Download as txt-file According to this theorem, the square of See How to generate triples of sizes that are natural. In PQR, angle PQR= 45, and angle QPR = 72 Complex numbers in the angle notation or phasor (polar coordinates r, ) may you write as rL where r is Search: Angle Sum Theorem Calculator. It states that the square of the longest side of a right triangle (the hypotenuse) is equal to the sum of the squares of the other two sides. The following are the applications of the Pythagoras theorem: Pythagoras theorem is used to check if a given triangle is a right-angled triangle or not. Given a right triangle with a side 28 Abr. Scientific Notation Calculator (Maths solver) The Pythagorean Theorem states that a+b=c and can be used to find Trigonometric pythagorean "solver", example of math trivia, algebra problems on slope and y intercept, ti89 square, multiplying with scientific notation worksheet. This website uses cookies to ensure you get the best experience. You will likely come across many problems in school and in real life that require using the theorem to solve. This calculator will use the Pythagorean Theorem to solve for the missing length of a right triangle given the lengths of the other two sides. Pythagoras's Theorem is a formula you can use to find an unknown side length of a right triangle. Pythagoras theorem could be written in its shortest form: a 2 + b 2 = c 2. See the solution with steps using the Pythagorean Theorem formula. The three sides of the triangle are hypotenuse, base and perpendicular. My code seems a bit sloppy to me, but I don't know how to improve it to make it cleaner and more understandable. Example: Find the Distance Between 2 Points. The procedure to use the Pythagorean Theorem calculator is as follows: Step 1: Enter the values of two sides in the input fields and enter x for the unknown side value. Sum of all three digit numbers divisible by 8. Pythagorean identities are identities in trigonometry that are extensions of the Pythagorean theorem. When you know the value of a and b of a right angled triangle our Pythagorean theorem calculator comes in handy. This tool is a web-based tool that can be used from anywhere. Enter 2 values and leave one blank to solve for that value. Triangle Inequality Theorem Calculator. Identify the legs and the hypotenuse of the triangle. Sample of problem solving skills in nursing how to write your common app essay, capstone project ideas stem.


text:
The  Weizmann  Institute  of  Science
Faculty of Mathematics and Computer Science

Vision and Robotics Seminar

Tal Hassner

will speak on

What Does the Scene Look Like from a Scene Point?

Abstract:
Consider a sporting event with several cameras taking pictures of the game.
What does the scene look like from the point of view of one of the players in the field?  What does the ball see"?  In this talk we will show how we can synthesize virtual views from scene points located {\it inside} the imaged scene. On one hand this provides a simple way of defining the position of the virtual camera in an uncalibrated setting. On the other hand, it implies {\it extreme} changes in viewpoint between the virtual camera and the real cameras.
Such extreme changes in viewpoint are not typical of most New-View-Synthesis (NVS) problems. Our approach does not require prior correspondence estimation nor any explicit 3D reconstruction. It can handle any number of input images while simultaneously using the information from all of them. However, very few images are usually enough to provide reasonable synthesis quality.

* This is joint work with Michal Irani and P. Anandan.

The lecture will take place in the
Lecture Hall, Room 1, Ziskind Building
on Thursday, January 31, 2002
at noon


text:
# zbMATH — the first resource for mathematics

Littlewood-Richardson rules for Grassmannians. (English) Zbl 1053.05121
One of the equivalent forms of the classical Littlewood-Richardson rule describes the structure constants obtained when the cup product of two Schubert classes in the cohomology ring of a complex Grassmannian is written as a linear combination of Schubert classes. In the paper under review the authors give a short and self-contained argument which shows that this rule is a direct consequence of the Pieri formula [M. Pieri, Lomb. Ist. Rend. (2) XXVI. 534–546 (1893), XXVII. 258–273 (1894; JFM 25.1038.02)] for the product of a Schubert class with a special Schubert class.
The Littlewood-Richardson rule has a generalization [J. R. Stembridge, Adv. Math. 74, No. 1, 87–134 (1989; Zbl 0677.20012)] for the Grassmannians which parametrize maximal isotropic subspaces of $${\mathbb C}^n$$ equipped with a symplectic or orthogonal form. The same arguments work equally well in more difficult cases and give a simple derivation of the rule of Stembridge from the analogues of the Pieri formula in [H. Hiller and B. Boe, Adv. Math. 62, 49–67 (1986; Zbl 0611.14036)].

##### MSC:
 05E10 Combinatorial aspects of representation theory 05E05 Symmetric functions and generalizations 14M15 Grassmannians, Schubert varieties, flag manifolds
Full Text:
##### References:
 [1] W. Fulton, Young Tableaux, L.M.S. Student Texts, Vol. 35, Cambridge University Press, Cambridge, 1997. [2] Hiller, H.; Boe, B., Pieri formula for SO2n+1/un and spn/un, Adv. math., 62, 49-67, (1986) [3] Hoffman, P.N.; Humphreys, J.F., Projective representations of the symmetric groups; Q-functions and shifted tableaux, (1992), Oxford University Press New York · Zbl 0777.20005 [4] A. Knutson, T. Tao, Puzzles and (equivariant) cohomology of Grassmannians, Duke Math. J., to appear. · Zbl 1064.14063 [5] A. Knutson, T. Tao, C. Woodward, The honeycomb model of GLn tensor products II: puzzles determine facets of the Littlewood-Richardson cone, J. Amer. Math. Soc., to appear. · Zbl 1043.05111 [6] Littlewood, D.E.; Richardson, A.R., Group characters and algebra, Phil. trans. R. soc. A, 233, 99-141, (1934) · Zbl 0009.20203 [7] Pieri, M., Sul problema degli spazi secanti. nota 1^a, Rend. ist. lombardo (2), 26, 534-546, (1893) · JFM 25.1038.02 [8] P. Pragacz, Algebro-geometric applications of Schur S- and Q-polynomials, Topics in Invariant Theory (Paris, 1989/1990), Lecture Notes in Mathematics, Vol. 1478, Springer, Berlin, 1991, pp. 130-191. · Zbl 0783.14031 [9] Remmel, J.B.; Shimozono, M., A simple proof of the Littlewood-Richardson rule and applications, Discrete math., 193, 1-3, 257-266, (1998) · Zbl 1061.05507 [10] Sagan, B., Shifted tableaux, Schur Q-functions, and a conjecture of R. Stanley, J. combin. theory ser. A, 45, 62-103, (1987) · Zbl 0661.05010 [11] Schur, I., Über die darstellung der symmetrischen und der alternierenden gruppe durch gebrochene lineare substitutionen, J. reine angew. math., 139, 155-250, (1911) · JFM 42.0154.02 [12] Shimozono, M., Multiplying Schur Q-functions, J. combin. theory ser. A, 87, 198-232, (1999) · Zbl 0978.05073 [13] Stembridge, J.R., Shifted tableaux and the projective representations of symmetric groups, Adv. math., 74, 87-134, (1989) · Zbl 0677.20012 [14] M.A.A. van Leeuwen, The Littlewood-Richardson rule, and related combinatorics, in: Interactions of Combinatorics and Representation Theory, MSJ Memoirs Vol. 11, Math. Soc. Japan, Tokyo, 2001, pp. 95-145. · Zbl 0991.05101 [15] D.R. Worley, A theory of shifted Young tableaux, Ph.D. Thesis, M.I.T., 1984.
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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## integrate dy/dx

can some one explain to me how the set of all solutions for dy/dx = 3y
is.
y= Ce^3x
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
 PhysOrg.com science news on PhysOrg.com >> Intel's Haswell to extend battery life, set for Taipei launch>> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens
 Recognitions: Homework Help Simply separate the variables. \begin{align*} \frac{dy}{dx} & = 3y \\ \frac 1 y \frac dy dx & = 3 \\ \int \left(\frac 1 y\right) \left(\frac{dy}{dx}\right) \, dx & =\int 3 \, dx \end{align*} You should be able to finish from here.
 Recognitions: Homework Help Try rearranging your equation to get it in the form f(y)dy=g(x) dx. Then integrate both sides. This type of differential equation technique is called 'separation of variables'

## integrate dy/dx

i got y= e^3x + e^c

how does that become y= Ce^3x
 Mentor Instead of e3x + eC, you should have gotten e3x + C = e3xeC = C' e3x (Here, C' = eC. After all, eC is just a constant.)
 o ok thanks got it.
 this is variable separable. c is constant, then you can make it In c so that when using the e function, In c become only c.

Mentor
 Quote by darkmagic this is variable separable. c is constant, then you can make it In c so that when using the e function, In c become only c.
The function is LN, not IN. The letters come from Latin: logarithmus naturalis.
 Here's a proof of the general case that I got from Courant's Introduction to Calculus and Analysis. This is one of my favourite proofs: "If a function y = f(x) satisfies an equation of the form $$y' = \alpha y$$ where $$\alpha$$ is a constant, then y has the form $$y = f(x) = ce^{\alpha x}$$ where c is also a consant; conversely, every function of the form $$ce^{\alpha x}$$ satisfies the equation $$y' = \alpha y$$. It is clear that $$y = ce^{\alpha x}$$ satisfies this equation for any arbitrary constant c. Conversely, no other function satisfies the differential equation $$y' - \alpha y = =$$. For if y is such a function, we consider the function $$u = ye^{-\alpha x}$$. We then have $$u' = y'e^{-\alpha x} - \alpha y e^{-\alpha x} = e^{-\alpha x}(y' - \alpha y}) .$$ However, the right-hand side vanishes, since we have assumed that $$y' = \alpha y$$; hence $$u' = 0$$ so that u is a constant c and $$y = ce^{\alpha x}$$ as we wished to prove."
 yes its ln but I type In. Sorry.


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Continue reading "Theorem on Properties of Triangle | p/sin P = q/sin Q = r/sin R = 2K"

### The Law of Sines | The Sine Rule | The Sine Rule Formula | Law of sines Proof

We will discuss here about the law of sines or the sine rule which is required for solving the problems on triangle. In any triangle the sides of a triangle are proportional to the sines

Continue reading "The Law of Sines | The Sine Rule | The Sine Rule Formula | Law of sines Proof"

###### Jan 20, 2018

In trigonometry we will discuss about the different properties of triangles. We know any triangle has six parts, the three sides and the three angles are generally called the elements of the triangle.

### Problems on Inverse Trigonometric Function | Inverse Circular Function Problems

We will solve different types of problems on inverse trigonometric function. 1. Find the values of sin (cos$$^{-1}$$ 3/5)

Continue reading "Problems on Inverse Trigonometric Function | Inverse Circular Function Problems"

### Principal Values of Inverse Trigonometric Functions |Different types of Problems

We will learn how to find the principal values of inverse trigonometric functions in different types of problems. The principal value of sin$$^{-1}$$ x for x > 0, is the length of the arc of a unit

Continue reading "Principal Values of Inverse Trigonometric Functions |Different types of Problems"

### Inverse Trigonometric Function Formula | Inverse Circular Function Formula

We will discuss the list of inverse trigonometric function formula which will help us to solve different types of inverse circular or inverse trigonometric function.

Continue reading "Inverse Trigonometric Function Formula | Inverse Circular Function Formula"

### 3 arctan(x) | 3 tan$$^{-1}$$ x |3 tan inverse x | Inverse Trigonometric Function

We will learn how to prove the property of the inverse trigonometric function 3 arctan(x) = arctan($$\frac{3x - x^{3}}{1 - 3 x^{2}}$$) or, 3 tan$$^{-1}$$ x = tan$$^{-1}$$

Continue reading "3 arctan(x) | 3 tan$$^{-1}$$ x |3 tan inverse x | Inverse Trigonometric Function"

### 3 arccos(x) | 3 cos$$^{-1}$$ x |3 cos inverse x | Inverse Trigonometric Function

We will learn how to prove the property of the inverse trigonometric function 3 arccos(x) = arccos(4x$$^{3}$$ - 3x) or, 3 cos$$^{-1}$$ x = cos$$^{-1}$$ (4x$$^{3}$$ - 3x)

Continue reading "3 arccos(x) | 3 cos$$^{-1}$$ x |3 cos inverse x | Inverse Trigonometric Function"

### 3 arcsin(x) | 3 sin$$^{-1}$$ x |3 sin inverse x | Inverse Trigonometric Function

We will learn how to prove the property of the inverse trigonometric function 3 arcsin(x) = arcsin(3x - 4x$$^{3}$$) or, 3 sin$$^{-1}$$ x = sin$$^{-1}$$ (3x - 4x$$^{3}$$)

Continue reading "3 arcsin(x) | 3 sin$$^{-1}$$ x |3 sin inverse x | Inverse Trigonometric Function"

### 2 arctan(x) | 2 tan$$^{-1}$$ x | 2 tan inverse x |Inverse Trigonometric Function

We will learn how to prove the property of the inverse trigonometric function, 2 arctan(x) = arctan($$\frac{2x}{1 - x^{2}}$$) = arcsin($$\frac{2x}{1 + x^{2}}$$)

Continue reading "2 arctan(x) | 2 tan$$^{-1}$$ x | 2 tan inverse x |Inverse Trigonometric Function"

### 2 arccos(x) | 2 cos$$^{-1}$$ x | 2 cos inverse x |Inverse Trigonometric Function

We will learn how to prove the property of the inverse trigonometric function 2 cos$$^{-1}$$ x = cos$$^{-1}$$ (2x$$^{2}$$ - 1) or, 2 arccos(x) = arccos(2x$$^{2}$$ - 1).

Continue reading "2 arccos(x) | 2 cos$$^{-1}$$ x | 2 cos inverse x |Inverse Trigonometric Function"

### 2 arcsin(x) | 2 sin$$^{-1}$$ x | 2 sin inverse x |Inverse Trigonometric Function

We will learn how to prove the property of the inverse trigonometric function 2 arcsin(x) = arcsin(2x$$\sqrt{1 - x^{2}}$$) or, 2 sin$$^{-1}$$ x = sin$$^{-1}$$ (2x$$\sqrt{1 - x^{2}}$$)

Continue reading "2 arcsin(x) | 2 sin$$^{-1}$$ x | 2 sin inverse x |Inverse Trigonometric Function"

### arccos(x) - arccos(y) | cos^-1 x - cos^-1 y | Inverse Trigonometric Function

We will learn how to prove the property of the inverse trigonometric function arccos(x) - arccos(y) = arccos(xy + $$\sqrt{1 - x^{2}}$$$$\sqrt{1 - y^{2}}$$)

Continue reading "arccos(x) - arccos(y) | cos^-1 x - cos^-1 y | Inverse Trigonometric Function"

### arccos(x) + arccos(y) | cos^-1 x + cos^-1 y | Inverse Trigonometric Function

We will learn how to prove the property of the inverse trigonometric function arccos (x) + arccos(y) = arccos(xy - $$\sqrt{1 - x^{2}}$$$$\sqrt{1 - y^{2}}$$)

Continue reading "arccos(x) + arccos(y) | cos^-1 x + cos^-1 y | Inverse Trigonometric Function"

### arcsin x - arcsin y |sin$$^{-1}$$ x - sin$$^{-1}$$ y|sin inverse x-sin inverse y

We will learn how to prove the property of the inverse trigonometric function arcsin (x) - arcsin(y) = arcsin (x $$\sqrt{1 - y^{2}}$$ - y$$\sqrt{1 - x^{2}}$$)

Continue reading "arcsin x - arcsin y |sin$$^{-1}$$ x - sin$$^{-1}$$ y|sin inverse x-sin inverse y"

### arcsin(x) + arcsin(y) |sin$$^{-1}$$ x+sin$$^{-1}$$ y|sin inverse x+sin inverse y

We will learn how to prove the property of the inverse trigonometric function arcsin (x) + arcsin(y) = arcsin (x $$\sqrt{1 - y^{2}}$$ + y$$\sqrt{1 - x^{2}}$$)

Continue reading "arcsin(x) + arcsin(y) |sin$$^{-1}$$ x+sin$$^{-1}$$ y|sin inverse x+sin inverse y"

### arccot(x) - arccot(y) | cot^-1 x - cot^-1 y | Inverse Trigonometric Function

We will learn how to prove the property of the inverse trigonometric function arccot(x) - arccot(y) = arccot($$\frac{xy + 1}{y - x}$$) (i.e., cot$$^{-1}$$ x + cot$$^{-1}$$ y = cot$$^{-1}$$

Continue reading "arccot(x) - arccot(y) | cot^-1 x - cot^-1 y | Inverse Trigonometric Function"

### arccot(x) + arccot(y) | cot^-1 x + cot^-1 y | Inverse Trigonometric Function

We will learn how to prove the property of the inverse trigonometric function arccot(x) + arccot(y) = arccot($$\frac{xy - 1}{y + x}$$) (i.e., cot$$^{-1}$$ x - cot$$^{-1}$$ y = cot$$^{-1}$$

Continue reading "arccot(x) + arccot(y) | cot^-1 x + cot^-1 y | Inverse Trigonometric Function "

### arctan(x) + arctan(y) + arctan(z) | tan^-1 x + tan^-1 y + tan^-1 z |Inverse Trig

We will learn how to prove the property of the inverse trigonometric function arctan(x) + arctan(y) + arctan(z) = arctan$$\frac{x + y + z – xyz}{1 – xy – yz – zx}$$ (i.e., tan$$^{-1}$$ x

Continue reading "arctan(x) + arctan(y) + arctan(z) | tan^-1 x + tan^-1 y + tan^-1 z |Inverse Trig"

### arctan x + arccot x = π/2 | arctan(x) + arccot(x) = $$\frac{π}{2}$$ | Examples

We will learn how to prove the property of the inverse trigonometric function arctan(x) + arccot(x) = $$\frac{π}{2}$$ (i.e., tan$$^{-1}$$ x + cot$$^{-1}$$ x = $$\frac{π}{2}$$).

Continue reading "arctan x + arccot x = π/2 | arctan(x) + arccot(x) = $$\frac{π}{2}$$ | Examples"

### arctan x - arctan y | tan^-1 x - tan^-1 y | Inverse Trigonometric Function

We will learn how to prove the property of the inverse trigonometric function arctan(x) - arctan(y) = arctan($$\frac{x - y}{1 + xy}$$) (i.e., tan$$^{-1}$$ x - tan$$^{-1}$$ y

Continue reading "arctan x - arctan y | tan^-1 x - tan^-1 y | Inverse Trigonometric Function"

### arctan(x) + arctan(y) = arctan($$\frac{x + y}{1 - xy}$$) | tan^-1 x + tan^-1 y

We will learn how to prove the property of the inverse trigonometric function arctan(x) + arctan(y) = arctan($$\frac{x + y}{1 - xy}$$), (i.e., tan$$^{-1}$$ x + tan$$^{-1}$$ y = tan$$^{-1}$$

Continue reading "arctan(x) + arctan(y) = arctan($$\frac{x + y}{1 - xy}$$) | tan^-1 x + tan^-1 y"

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# Anyone deal with speeduino?

### Help Support HomeBuiltAirplanes.com:

#### pfarber

##### Well-Known Member
HBA Supporter
Evidently there is ECU developed for the Arduino chip. Uses the exact same programming interface, TunerStudio, so I'm thinking the same firmware?

Specs are a little lighter than Megasquirt, but its also half the cost. So far I have not read much bad about the system.

Anyone have and thoughts??

#### Chris Matheny

##### Well-Known Member
HBA Supporter
Slower processor, if your planning on keeping the vvt and all sensors in tact on your Pentastar, I think a Megasquirt would probably be an advantage. If locking everything out and only running basic sensors it would probably suffice.

#### Hephaestus

##### Well-Known Member
Got one in the drawer all setup and ready to run on another TBI conversion... But we haven't done one in a while, old carb'd monsters are getting harder to find so no real practical experience yet.

HBA Supporter

#### FinnFlyer

##### Member
Tracy Crook of RWS recommends it. Likes that it interfaces with Tuner Studio and is open source. Of course you'll need to modify the code if you want manual mixture control. And of course I'd double up on the hardware.
I was lucky enough to get an RWS EC3 board. I still have to program it (software is now open source) and run it in the RV-4 I'm building.

#### proppastie

##### Well-Known Member
Log Member
Might want a magneto for 1/2 the plugs if you are going to develop an ECU for an aircraft......There are proven electronic ECU for aircraft but they are not cheap.....careful here.

#### Flyguyeddy

##### Active Member
Might want a magneto for 1/2 the plugs if you are going to develop an ECU for an aircraft......There are proven electronic ECU for aircraft but they are not cheap.....careful here.
The rv guys seem to be having good luck with the ford edis conversion setup and a megajolt lite to control the timing. Perhaps if the speeduino can control an edis with a SAW signal that would be the way to go.

#### proppastie

##### Well-Known Member
Log Member
The rv guys seem to be having good luck with the ford edis conversion setup and a megajolt lite to control the timing. Perhaps if the speeduino can control an edis with a SAW signal that would be the way to go.

The issue should be how reliable is it, is it difficult to program or easy to screw up the programming,,,,etc. The new Lycoming factory solution is a legacy mag and 1 electronic ignition which can be set up to go to advanced timing for certain conditions. The electronic ignition is almost plug and play but there is more to hook up than the original mag. (battery, fuse, manifold pressure,). More bells and whistles is not necessarily safer or better especially at 200 ft. on climb out.

#### TFF

##### Well-Known Member
The legacy mag allows Lycoming to beat the system and not have to send the engine through recertification. The other certified electronic mags have done this for certified.

#### proppastie

##### Well-Known Member
Log Member
The other certified electronic mags have done this for certified.
Which is why they are so expensive.....Might be worth it though.

#### TFF

##### Well-Known Member
Their on the same order as buying a new mag. More than a rebuilt one as long as you have a core.

A real independent test needs to be done. Once you have the timing advance, it’s essentially a single ignition system running the engine. Reliability is still duel, but flame front efficiency and detonation do change. Spark plugs are placed expecting two to fire the chamber. My Alfa Romeos have similar chambers. The racier ones had duel plugs To get the fuel to burn. My moms old 74 Grand Torino wagon, emphasis on Grand, had the 351 C two barrel; open chamber heads. Detonation city if it was not tuned perfect. The four barrel heads did not have the open chamber and did not have the same running problems.

The way I see these single ignitions is if you do fly 2000 hours you could go to your ledger and prove you saved money on fuel, just like lean of peak. If you are like my boss and twittle the mixture when the EGT changes 2 deg, you will be tuned in enough to complain you are burning a pound of fuel more. Everyone else, it would be better to put the $1500 in the fuel tank, because most are not flying if they have to spend$50, less $1500 to only see the value 2000 hours later. If 2000 is every five years, it’s worth it on the next overhaul, because it’s just around the corner. Twenty years, not worth it. My friends flying lean of peak with two electronic ignitions on their RVs can tell. But when they are at the pump, the price pain on their face is still the same. #### pfarber ##### Well-Known Member HBA Supporter More bells and whistles is not necessarily safer or better especially at 200 ft. on climb out. As no magneto engine has ever cut out at that altitude. Come on guys, if your position is 'its not reliable' the you can't hold up an Lyc or Cont as your example. I'll take a solid state timing source over spinning metal any day of the week. Seeing that a speeduino kit as almost as much as a Megasquirt 3 (calm down, I said ALMOST) I'd rather go with MS as I think finding a speed shop that has more experience with MS would be easier. The software being the same might help, but looking at the timing options on the Speeduino the MS has it beat hands down. #### jbiplane ##### Well-Known Member seems to have a lot of experience with megasquirt/Speednuino Yes, we build about 16 Speeduino-based ECUs before finally abandon any Arduino stuff. Now deliver about 50 test on Motorolla processor (same as MS2, MS3). Now we make 10 experimental ECU on Stm32 processor. I made new site www.flame-power.com where will post some info. #### pfarber ##### Well-Known Member HBA Supporter Interersting... I look forward to reading what you put up. But you have a pretty high bar to meet. Megasquirt has a LOT of talent out there for me to rely on. I've found three local shops (with dyno) that will dial in a MS3 config for$500 (after I get base maps set and engine running).

Tunerstudio software is pretty good. I'm not really looking for an ECU that I can't tinker with. Also CAN bus is a huge deal. One reason I didn't go with SDS, there is no way to get real time data (via CAN) in/out of the box. If SDS would add CAN and then a dashboard for a tablet to make adjustments? YES PLEASE.


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# If electric charges accelerate towards lower potential energies, why do opposite charges attract?

I know my logic must be wrong but I can't figure out why. I know that charges must accelerate towards lower potential energies simply because that's a general rule of nature. However, when you release a positive and negative charge close to each other, they will move towards each other, and the distance between will decrease causing the potential energy to increase (1/r^2 relationship). So aren't the two charges gaining potential energy? Does it have to do with the sign of the charges being different? I feel like that explanation doesn't fit entirely because it seems like the magnitude of the potential energy should not be increasing.

-

The potential energy between two oppositely charged particles goes like $-1/r$. First of all, the potential energy doesn't scale as $1/r^2$; you must have confused it with the force which is the gradient (spatial derivative) of the potential energy. Second of all, the potential energy of a bound state – a pair or collection of particles/objects that attracted each other – is negative.

You may see it's the right sign because the total energy is conserved. The total energy is the kinetic one plus the potential one. When the oppositely charged particles get closer, they accelerate so the kinetic energy goes up. Because the total energy has to stay constant, the potential energy has to go down; and indeed, a negative number getting even more negative (the absolute value increases) is going down. Because it was zero at infinity, it must be negative when these particles are very close.

-

first of all the energy varies proportional to 1/r not 1/r^2. that's important.

let's take potential energy = 0 when separation distance r = infinity.

secondly, the energy is NEGATIVE for a positive and negative charge so that that it become a larger NEGATIVE NUMBER as they get closer. a larger negative number is actually LESS energy corresponding to a bound state.

I know that's slightly confusing but that's how you have to think about it.

-


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## Educators

Problem 1

Now find the derivative of each of the following functions. The answers are in Chapter $19 .$
$$f(x)=\ln \left(x^{4}+8\right)$$

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Problem 2

Now find the derivative of each of the following functions. The answers are in Chapter $19 .$
$$f(x)=\ln (3 x \sqrt{3+x})$$

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Problem 3

Now find the derivative of each of the following functions. The answers are in Chapter $19 .$
$$f(x)=\ln (\cot x-\csc x)$$

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Problem 4

Now find the derivative of each of the following functions. The answers are in Chapter $19 .$
$$f(x)=\ln \left(\frac{5 x^{2}}{\sqrt{5+x^{2}}}\right)$$

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Problem 5

Now find the derivative of each of the following functions. The answers are in Chapter $19 .$
$$f(x)=e^{x \cos x}$$

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Problem 6

Now find the derivative of each of the following functions. The answers are in Chapter $19 .$
$$f(x)=e^{-3 x} \sin 5 x$$

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Problem 7

Now find the derivative of each of the following functions. The answers are in Chapter $19 .$
$$f(x)=e^{\pi x}-\ln e^{\pi x}$$

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Problem 8

Now find the derivative of each of the following functions. The answers are in Chapter $19 .$
$$f(x)=\log _{12}\left(x^{3}\right)$$

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Problem 9

Now find the derivative of each of the following functions. The answers are in Chapter $19 .$
$$f(x)=\frac{\log _{4} x}{e^{4 x}}$$

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Problem 10

Now find the derivative of each of the following functions. The answers are in Chapter $19 .$
$$f(x)=\log \sqrt{10^{3 x}}$$

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Problem 11

Now find the derivative of each of the following functions. The answers are in Chapter $19 .$
$$f(x)=e^{3 x}-3^{e x}$$

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Problem 12

Now find the derivative of each of the following functions. The answers are in Chapter $19 .$
$$f(x)=10^{\sin x}$$

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Problem 13

Now find the derivative of each of the following functions. The answers are in Chapter $19 .$
$$f(x)=\ln \left(10^{x}\right)$$

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Problem 14

Now find the derivative of each of the following functions. The answers are in Chapter $19 .$
$$f(x)=x^{5} 5^{x}$$

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