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# 16.4: Spectroscopic Properties

## Infrared Spectra

A carbonyl group in a compound can be positively identified by the strong infrared absorption band in the region $$1650$$-$$1850 \: \text{cm}^{-1}$$, which corresponds to the stretching vibration of the carbon-oxygen double bond. The position of the band within this frequency range depends on the molecular environment of the carbonyl group. As a result, we frequently can tell from the band position whether the structure is an aldehyde, ketone, carboxylic acid, ester, amide, or anhydride. The date of Table 16-3 show typical infrared absorption frequencies for specific types of carbonyl compounds. Thus aldehydes and ketones absorb at slightly lower frequencies (longer wavelengths) than carboxylic esters and anhydrides. We usually find that absorption shifts to lower frequencies $$\left( \sim 20 \: \text{cm}^{-1} \right)$$ when the carbonyl group is conjugated with other multiple bonds, as in aromatic ketones, $$\ce{C_6H_5COCH_3}$$.

Table 16-3: Characteristic Infrared Absorption Frequencies of Carbonyl Compounds$$^a$$

Aldehydes can be distinguished from ketones by a band at $$2720 \: \text{cm}^{-1}$$ which is characteristic of the $$\ce{C-H}$$ stretching vibration of an aldehyde function:

This band is unusually low in frequency for a $$\ce{C-H}$$ stretching vibration; although the band is rather weak, it occurs in a region of the spectrum where other absorptions generally are absent so it can be identified with no special difficulty.

## Electronic Absorption Spectra

Aldehydes and ketones absorb ultraviolet light in the region $$275$$-$$295 \: \text{nm}$$, and the result is excitation of an unshared electron on oxygen to a higher energy level. This is the $$n \rightarrow \pi^*$$ transition discussed in Section 9-9. A more intense $$\pi \rightarrow \pi^*$$ transition occurs about \180\)-$$190 \: \text{nm}$$, which corresponds to excitation of an electron from a $$\pi$$-bonding orbital to a $$\pi$$-antibonding orbital. Neither of these absorptions is especially useful for specific identification unless the carbonyl group is conjugated, in which case the $$n \rightarrow \pi^*$$ and $$\pi \rightarrow \pi%+^*$$ bands occur at longer wavelengths (by $$30$$-$$40 \: \text{nm}$$). For example, if you suspect that a compound is an alkenone from its infrared spectrum, you easily could tell from the $$\lambda_\text{max}$$ of the $$n \rightarrow \pi^*$$ and $$\pi \rightarrow \pi^*$$ absorptions of the compound whether it is a conjugated alkenone. The absorption frequency would be expected around $$320 \: \text{nm}$$ and $$220 \: \text{nm}$$ (see Figure 9-20).

## Mass Spectra

Aldehydes and ketones generally give moderately intense signals due to their molecular ions, $$\ce{M^+}$$. Thus the determination of the molecular weight of a ketone by mass spectroscopy usually is not difficult. Furthermore, there are some characteristic fragmentation patterns that aid in structural identification. These are:

• $$\alpha$$ cleavage

• transfer of $$\gamma$$ hydrogen with $$\beta$$ cleavage (McLafferty rearrangement)

## NMR Spectra

The character of the carbonyl bond gives rise to very low-field nmr absorptions for the proton of an aldehyde group $$\left( \ce{-CH=O} \right)$$. As Table 9-4 shows, these absorptions are some $$4 \: \text{ppm}$$ to lower fields than alkenyl hydrogens (). Some of this difference in shift can be ascribed to the polarity of the carbonyl group $$\overset{\delta \oplus}{\ce{C}} \ce{=} \overset{\delta \ominus}{\ce{O}}$$, which reduces electron density around the aldehyde hydrogen (see Section 9-10E). The effect appears to carry over in much smaller degree in the $$\alpha$$ positions, and protons of the type are about $$0.3 \: \text{ppm}$$ to lower fields than those of .


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Algebra 2 (1st Edition)

$-\dfrac{\sqrt 3}{2}$
Here, we have $\theta=-840^{\circ}$ Here, $\theta'=-840^{\circ}+(3) 360^{\circ}=240^{\circ}$ and $\theta''=\theta'-180^{\circ}=240^{\circ}-180^{\circ}=60^{\circ}$ The function $\sin$ is negative in Quadrant $II$. Now, $\sin \theta=\sin \theta'= -\sin \theta''=-\sin 60^{\circ}=-\dfrac{\sqrt 3}{2}$

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# Mapping multiple functions

This question was part of a course on Mathematica written in 1998.

Each of the functions f1, f2, and f3 defined below has a root near x=2. Map an anonymous function involving FindRoot[] onto the list {f1,f2,f3} to find all three roots.

 f1[x_]= 2x-17Cos[x]
f2[x_]= x^2-3Sin[x]
f3[x_]= 2Sin[x]^2-x
Plot[{f1[x],f2[x],f3[x]},{x,0,5}]

Which produced the output:

Out[242]= 2 x - 17 Cos[x]

Out[243]= x^2 - 3 Sin[x]

Out[244]= -x + 2 Sin[x]^2

And then proceeds to draw a nice graph, showing that all of these functions have a root between @1.5 and 1.8. I then individually evaluated the roots (near two) for each function so that I would know if the answers to the "mapped" functions were correct. FindRoot[f1, {2}]

Out[287]= {1.40477}

FindRoot[f2, {2}]

Out[284]= {1.72213}

 FindRoot[f3, {2}]

Out[285]= {1.84908}

Here are some of my attempts to answer the question using Map and FindRoot along with the error messages and output that each produced. (* Note that I attempted about half a dozen other solutions, but did not want to make this post too long*)

 Map[FindRoot, {f1, f2, f3}, {1}]

During evaluation of In[296]:= FindRoot::argmu: FindRoot called with 1 argument; 2 or more arguments are expected. >>

During evaluation of In[296]:= FindRoot::argmu: FindRoot called with 1 argument; 2 or more arguments are expected. >>

During evaluation of In[296]:= FindRoot::argmu: FindRoot called with 1 argument; 2 or more arguments are expected. >>

During evaluation of In[296]:= General::stop: Further output of FindRoot::argmu will be suppressed during this calculation. >>

Out[296]= {FindRoot[f1], FindRoot[f2], FindRoot[f3]}

     Map[FindRoot, {[f1, {2}], [f2, {2}], [f3, {2}]}]

During evaluation of In[303]:= Syntax::sntxf: "{" cannot be followed by "[f1,{2}],[f2,{2}],[f3,{2}]}".

During evaluation of In[303]:= Syntax::tsntxi: "[f1,{2}]" is incomplete; more input is needed.

During evaluation of In[303]:= Syntax::sntxi: Incomplete expression; more input is needed .

     Map[FindRoot, {{f1, {2}}, {f2, {2}}, {f3, {2}}}]

During evaluation of In[303]:= FindRoot::argmu: FindRoot called with 1 argument; 2 or more arguments are expected. >>

During evaluation of In[303]:= FindRoot::argmu: FindRoot called with 1 argument; 2 or more arguments are expected. >>

During evaluation of In[303]:= FindRoot::argmu: FindRoot called with 1 argument; 2 or more arguments are expected. >>

During evaluation of In[303]:= General::stop: Further output of FindRoot::argmu will be suppressed during this calculation. >>

Out[303]= {FindRoot[{f1, {2}}], FindRoot[{f2, {2}}], FindRoot[{f3, {2}}]}

I have read the documentation on Map, FindRoot, Listable, Fold and Function. None of these gave me an idea of how to answer this question. In addition I ran a search of this site(and also looked at each of the threads that were linked when I filled in the subject), the "comp.soft-sys.math.mathematica" site, as well as a general Internet search for "Mapping multiple functions in Mathematica", and didn't find anything that gave me an idea of how to answer the question. I would appreciate someone pointing me in the right direction.

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"anonymous function" - so, you'll also want to look at the documentation for Slot[] in addition to the one you already read for Function[]. In this case, since FindRoot[] takes two arguments, insert a slot as one of the arguments, and then you can use Function[]. –  Guess who it is. May 23 '13 at 18:44
The title is a bit misleading---here you are mapping a single function onto a list of functions. –  Szabolcs May 23 '13 at 19:15
@Szabolcs, Sorry, I was confused about what I was asking, looking at the answer below I now understand what you are saying. –  Clif May 23 '13 at 19:51

One way to do this:

 Map[FindRoot[#, {2}] &, {f1, f2, f3}]

 {{1.40477}, {1.72213}, {1.84908}}

The point here (as J.M. was suggesting) is that you can use the Slot (the symbol #, which is used to represent arguments or formal parameters in pure functions) to act as a placeholder that will get filled with each of the f1, f2 and f3 in turn. Note that # is always paired with &. You could also use the shortcut (the "infix" form) for Map if you wanted a slicker representation

FindRoot[#, {2}] & /@ {f1, f2, f3}

which returns the same answer. Another way to do this is to use MapThread rather than Map, for instance:

MapThread[FindRoot, {{f1, f2, f3}, {{2}, {2}, {2}}}]

which "threads" the FindRoot function through both arguments and again gives the same result. This has the advantage that you can use different starting values in both arguments (i.e., some of the {2}'s in the original problem could be changed).

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Thank You. To be sure that I understand about this the [#,{2}] is fulfilling the requirements of FindRoot[italic_function, starting point_italic] while the comma after the & is a requirement of Map[italic_function,expression_italic]. –  Clif May 23 '13 at 20:08
I might say it this way: FindRoot[#, {2}] & is the first argument of Map and {f1, f2, f3} is the second argument of Map. These need to be separated by a comma. –  bill s May 23 '13 at 20:18
code Map[FindRoot[#1, {#2}] &, [{f1, f2, f3}, {2, 0, 2}]] If I could ask an additional question, why won't this work, it would seem that #1 should be pulling from the {f1...}list and #2 should be pulling from the {2,0,2} list. However all that I get is an error message saying that more input is needed. –  Clif May 23 '13 at 20:20
I just added this approach to the answer -- you can do it this way, but it requires MapThread rather than Map. –  bill s May 23 '13 at 20:25
Yes that does work well and that way I can change the "starting from point" values –  Clif May 23 '13 at 20:31

MapThread answer I really like for individual seeds can be specified. A constant alternative to Map is Table where you cycle through a list of functions:

fun = {
2 # - 17 Cos[#] &,
#^2 - 3 Sin[#] &,
2 Sin[#]^2 - # &};

Table[FindRoot[f[x], {x, 2}], {f, fun}]

{{x -> 1.40477}, {x -> 1.72213}, {x -> 1.84908}}

Modifying Table to variable seeds:

With[{s = {1.5, 1.7, 2.}},
Table[FindRoot[fun[[i]][x], {x, s[[i]]}], {i, 3}]]

What about just Thread?

Thread[f[{f1, f2, f2}, {a, b, c}]]

{f[f1, a], f[f2, b], f[f2, c]}

Yet this doesn't work:

Thread[FindRoot[#1[x], {x, #2}] &[fun, {1.5, 1.7, 2.}]]

Because Thread first evaluates its argument (fun[x] gives error). So I try:

Thread[Unevaluated[FindRoot[#1[x], {x, #2}] &[fun, {1.5, 1.7, 2.}]]]

But it still doesn't work. Can somebody enlighten on this, anybody?

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what helps with Thread is preparing the arguments as Lists of length three (one the first level, i.e. not {x,{1.5,1.7,2.}}, neither {f1,f2,f3}[x] which no longer has Head List). Then Hold on FindRootand it works: Thread[(Hold@FindRoot)[#1, #2] &[ Through[fun[x]], {x, #} & /@ {1.5, 1.7, 2.}]] // ReleaseHold. I hope it is more or less clear what I try to say... –  Pinguin Dirk May 24 '13 at 10:21
@PinguinDirk MapThread! :) –  BoLe May 24 '13 at 10:30
just wanted to force Thread on the problem :) –  Pinguin Dirk May 24 '13 at 10:31


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Difference between $a^{2n} b^n$ and $n_a(w) = 2n_b(w)$

I have encountered two questions related to npda:

1. Construct an npda for $$L_1 = \{a^{2n} b^n \mid n \geq 0\}$$ as a language over $$\Sigma = \{a,b,c\}$$.

2. Construct an npda for $$L_2 = \{w \in \{a, b, c\}^* \mid n_a(w) = 2n_b(w)\}$$.

What is the difference between the above two languages?

• Could you add a bit more context to these questions? What are "a_concatenation_n" or "na(w)", for example? – siracusa Nov 25 '19 at 14:29
• The second is a strict superset of the first. Examples of words included only in the second language are $c,aba,baa$. – Yuval Filmus Nov 25 '19 at 14:56
• First one can have strings like: aab, aaaabb, what about the possibility of 'c' in the first one? Please guide me.@siracusa: Sorry its same as a^2n b ^n. Some people says that use of power symbol not good in place of concatenation. – user2994783 Nov 25 '19 at 17:26

The usual definition is that $$n_a(w)$$ refers to the number of $$a$$s in $$w$$.

Based on this definition, $$L_1$$ consists exactly of words that begin with an even number of $$a$$s followed by half as many $$b$$s (no $$c$$s are permitted). Therefore, you have $$\varepsilon, aab, aaaabb, aaaaaabbb,... \in L_1$$.

Whereas $$L_2$$ consists of words containing an even number of $$a$$s and half as many $$b$$s (in any order and with an arbitrary number of $$c$$ in between).

Thus every word in $$L_1$$ is also in $$L_2$$ (formally $$L_1 \subseteq L_2$$), but there are words in $$L_2$$ that are not in $$L_1$$:

• For some $$n \geq 0$$, consider words of the form $$b^n a^{2n}$$. They are in $$L_2$$, because they contain an even number of $$a$$s and half as many $$b$$s, but if $$n > 0$$, then it does not match $$a^{2n} b^n$$, because the $$a$$s come after the $$b$$s. You have $$\varepsilon, baa,bbaaa,bbbaaaaaa,... \in L_2$$, but $$baa,bbaaa,bbbaaaaaa \notin L_1$$.
• The same argument applies for $$w = (aab)^n$$. You have $$\varepsilon, aab,aabaab,aabaabaab \in L_2$$, but $$aabaab,aabaabaab \notin L_1$$.
• A word can have an even number of $$a$$s and half as many $$b$$s and also contain $$c$$s. For example, $$a^{2n} b^n c \in L_2$$. But the words in $$L_1$$ by definition only contain $$a$$s and $$b$$s.

In both languages, each string must have twice as many $$a$$'s as $$b$$'s. For a string to be in the first language, it must satisfy the additional condition that all the $$a$$'s occur before any of the $$b$$'s. So $$aba$$ does not belong to the first language and does belong to the second language.


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1. Jun 26, 2005

### Diced Tofu

1. A tank holds 90 L of chemical solution. Currently the solution has a strength of 20%. How much of this should be drained and replaced with an 80% solution to have a final strength of 30%?

2. Jun 26, 2005

### HallsofIvy

Suppose you "drain and replace" X liters of the 20% solution. That leaves 90- X liters of 20% solution. The X liters of the 80% solution will contain 0.8X L of the chemical. The remaining 90-X liters will contain 0.2(90- X) L of the chemical so you will now have 0.8X+ 0.2(90-X) L of the chemical in the final result. How much chemical will there be in 90 L of 30% solution? Set those equal and solve for X.

3. Jun 26, 2005


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# How is the sun's equatorial circumference measured?

How is the sun's equatorial circumference measured by scientists?

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Suggestion to the question formulation (v1): Replace the word calculated with the word measured. – Qmechanic Jun 15 '13 at 16:07
Of interest: physics.stackexchange.com/q/29363 – dmckee Jun 17 '13 at 20:02

It is quite simple to get a reasonable measure of the sun's diameter, using a pinhole, a screen and the formula for experiment 2 on this At Home Astronomy article: "Finding the size of the sun and moon".

From the 'At Home Astronomy' article, the sun's diameter can be calculated as:

$$\frac{\text{Diameter of the sun's image}}{\text{Distance from the pinhole to the paper}}\times\text{Earth-sun distance} = \text{Diameter of the sun}$$

The first recorded attempts to determine the distances and sizes of the Earth and Moon were performed by Aristarchus (310 - 230 B.C.), these were not very accurate (this information is also in the first link I gave originally). Once the distance was known (or estimated), a measure for the size could be determined.

His method was to determine the angle to the sun when the moon was half full, hence at a right angle to the observer (a potential source of error) and measure the angle to the sun (as in the diagram below)

(Source)

I hope this clears up the inadequacies of my original post.

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@DrStrangeLove the formula they use is a derivation of relative size ratios - a bit more explanation of the maths involved is here en.wikipedia.org/wiki/Pinhole_camera_model – user24901 Jun 15 '13 at 12:08
@DamienIgoe While links to more info are good, we discourage link-only answers, since they lose all value when the link inevitably dies. It would help if you provided a brief summary of the technique. – Chris White Jun 15 '13 at 15:28
Okay, added a relevant image. – user24901 Jun 15 '13 at 23:41
Doesn't answer the question because we don't know the object distance. This equation correlates size and distance, and this leaves us needing one more equation. – Alan Rominger Jun 16 '13 at 0:36
No one is going to delete your question on you (and indeed no one has even downvoted it), I just suspected you didn't understand Alan's objection because it was rather minimally stated. – dmckee Jun 17 '13 at 19:53

It should be simply as below: $C = \pi d$

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Great, now what is $d$? – Pricklebush Tickletush Jun 16 '13 at 0:00
d = diameter, can be determined using the 'At home astronomy article' in my reply. – user24901 Jun 16 '13 at 0:45
The sun area is assumed circle! and cicumference of a circle is school math and $\pi d$. However glad to help $d$ is the diameter. – al-Hwarizmi Jun 16 '13 at 8:00


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# Tag Info

4

If You have only females in your dataset, adding gender feature to the model input will not improve it. The technical explanation on why it won't help changes between models, but the intuition is simple - the model tries to find correlation between the features and the labels, and the correlation between any variable and a fixed-value variable is zero. You ...

2

Psychology was mentioned because psychology has a long history of assigning numeric scores to subjective topics. One of the most important concepts is inter-rater reliability, how much do different people agree on an interpretation. Other concepts that are useful are the degree of subjectivity and the degree of polarity (vs. assigning binary polarity labels)....

1

Keep subject_ID and after train_test_split pass to the model dataframes without the ID column, as in: df.loc[ : , df.columns != 'subject_ID'] Unless you are explicitly shuffling datapoints during prediction, I believe that commonly returned predictions persist the initial order. Definitely worth checking with the particular model you are using. EDIT: See an ...

1

I completely agree with the answer on statsSE, I don't have much to add to it: Essentially this is a business decision: you can voice your concerns if you think that the company is making a bad decision, but at the end of the day this is their choice to make. There's one point in particular that I think is worth making clear to the company, it's what it ...

1

You are correct. They are not independent and they are positively correlated. Let $A$ and $B$ (or $X$ and $Y$) be two events for stating general theorems and $M$ and $T$ be the events "customer purchases mozzarella" and "customer purchases tomatoes" in this specific example. We will use $\wedge$ to mean "and" so that $M \wedge T$...

1

Given you have two categorical variables and the associated contingency table, one option is to calculate the joint and marginal probabilities: $$P = \frac{count}{total}$$ Probabilities Tomatoes No Tomatoes Row Mozzarella 0.4 0.1 0.5 No Mozzarella 0.2 0.3 0.5 Column 0.6 0.4 0.1 The probabilities then can be used to answer questions about the data - If ...

1

Market Basket Analysis is the way to go about it. Market Basket Analysis is a technique which identifies the strength of association between pairs of products purchased together and identify patterns of co-occurrence. A co-occurrence is when two or more things take place together. Market Basket Analysis creates If-Then scenario rules, for example, if item A ...

1

Market Basket Analysis is a technique which identifies the strength of association between pairs of products purchased together and identify patterns of co-occurrence. A co-occurrence is when two or more things take place together. Market Basket Analysis creates If-Then scenario rules, for example, if item A is purchased then item B is likely to be ...

1

Provided you have the data already, and the data is labelled (i.e., split into the two classes $A$ and $B$), it makes sense to produce a number of visualisations to gauge what the model output would be. If you start with traditional classification algorithms like logistic regression, then the model output is going to be the probability of belonging to a ...

1

You could simulate data and fit a model to it as if it were real data. there are packages and functions in R and Python to do this. You'd have to be very clear that the data is faked. You could then examine the model and produce graphs as if it were a real one. This has the downside that it involves writing all the code and writing code to sim data, which ...

1

Look at your past experience. Even though you're a novice, you were hired as a data scientist, so you'll probably have some experience with data science projects. A simple binary classification problem with a few hundred datapoints can be solved in a productive afternoon, whereas a large project that requires significant upfront engineering for the ...

Only top voted, non community-wiki answers of a minimum length are eligible


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Online Calculators

Wavelength Calculator

Wavelength Calculator
$$\lambda = \dfrac{v}{f}$$
λ =
wavelength
units

Calculator Use

Choose a calculation for wavelength λ, velocity v or frequency f. Enter the other two values and the calculator will solve for the third in the selected units. You can also enter scientific notation such as 3.45e22.

Wavelength Equation for these Calculations:

$$\lambda = \dfrac{v}{f}$$

Where:
$$\lambda$$ = wavelength
$$v$$ = velocity
$$f$$ = frequency

Speed of Light = 299,792,458 m/s

The Wavelength Calculator uses the formula λ = v/f, or wavelength (λ) is equal to velocity (v) divided by frequency (f). The calculator can use any two of the values to calculate the third. Along with values, enter the known units of measure for each and this calculator will convert among units.

Significant Figures

For a value of 83456, selecting 4 significant figures will return 83460. For a value of 0.083456, selecting 4 significant figures will return 0.08346. See also our help notes on significant figures.

Wavelength calculations used in this calculator:

Solving for wavelength, velocity or frequency we can use the following formulas where wavelength is inversely proportional to frequency of the wave:

Calculate λ Given v and f
Calculate wavelength given velocity and frequency.

$$\lambda = \dfrac{v}{f}$$

Calculate v Given λ and f
Calculate velocity given wavelength and frequency. Calculation of the wave speed.

$$v = \lambda f$$

Calculate f Given v and λ
Calculate frequency given velocity and wavelength.

$$f = \dfrac{v}{\lambda}$$

Cite this content, page or calculator as:

Furey, Edward "Wavelength Calculator λ = v/f" at https://www.calculatorsoup.com/calculators/physics/wavelength.php from CalculatorSoup, https://www.calculatorsoup.com - Online Calculators


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# garchma

Convert ARMA model to MA model

`garchma` will be removed in a future release. Use `arma2ma` instead.

## Syntax

`InfiniteMA = garchma(AR,MA,NumLags)`

## Description

`InfiniteMA = garchma(AR,MA,NumLags)` computes the coefficients of an infinite-order MA model, using the coefficients of the equivalent univariate, stationary, invertible, finite-order ARMA(R,M) model as input. `garchma` truncates the infinite-order MA coefficients to accommodate the number of lagged MA coefficients you specify in `NumLags`.

This function is useful for calculating the standard errors of minimum mean square error forecasts of univariate ARMA models.

## Arguments

 `AR` R-element vector of autoregressive coefficients associated with the lagged observations of a univariate return series modeled as a finite-order, stationary, invertible ARMA(R,M) model. `MA` M-element vector of moving-average coefficients associated with the lagged innovations of a finite-order, stationary, invertible, univariate ARMA(R,M) model. `NumLags` (optional) Number of lagged MA coefficients that `garchma` includes in the approximation of the infinite-order MA representation. `NumLags` is an integer scalar and determines the length of the infinite-order MA output vector. If `NumLags = []` or is unspecified, the default is `10`.

## Output Arguments

 `InfiniteMA` Vector of coefficients of the infinite-order MA representation associated with the finite-order ARMA model specified by `AR` and `MA`. `InfiniteMA` is a vector of length `NumLags`. The jth element of `InfiniteMA` is the coefficient of the jth lag of the innovations noise sequence in an infinite-order MA representation. Box, Jenkins, and Reinsel refer to the infinite-order MA coefficients as the "ψ weights."

In the following ARMA(R,M) model,{yt} is the return series of interest and {εt} the innovations noise process.

${y}_{t}=\sum _{i=1}^{R}{\varphi }_{i}{y}_{t-1}+{\epsilon }_{t}\sum _{j=1}^{M}{\theta }_{j}{\epsilon }_{j-1}$

If you write this model equation as

${y}_{t}={\varphi }_{1}{y}_{t-1}+...+{\varphi }_{R}{y}_{t-R}+{\epsilon }_{t}+{\theta }_{1}{\epsilon }_{t-1}+...+{\theta }_{M}{\epsilon }_{t-M}$

you can specify the `garchma` input coefficient vectors, `AR` and `MA`, as you read them from the model. In general, the jth elements of `AR` and `MA` are the coefficients of the jth lag of the return series and innovations processes yt-j and εt-j, respectively. `garchma` assumes that the current-time-index coefficients of yt and εt are `1` and are not part of `AR` and `MA`.

In theory, you can use the ψ weights returned in `InfiniteMA` to approximate yt as a pure MA process.

${y}_{t}={\epsilon }_{t}+\sum _{i=1}^{\infty }{\psi }_{i}{\epsilon }_{t-i}$

The jth element of the truncated infinite-order moving-average output vector, ψj or `InfiniteMA(j)`, is consistently the coefficient of the jth lag of the innovations process, εt-j, in this equation. See Box, Jenkins, and Reinsel [15], Section 5.2.2, pages 139-141.

## Examples

### Convert an ARMA Model to an MA Model

Calculate a forecast horizon of 10 periods for the following ARMA(2,2) model:

${y}_{t}=0.5{y}_{t-1}-0.8{y}_{t-2}+{\epsilon }_{t}-0.6{\epsilon }_{t-1}+0.08{\epsilon }_{t-2}$

To obtain probability limits for these forecasts, use `garchma` to compute the first 9 (that is, `10 - 1`) weights of the infinite order MA approximation.

```PSI = garchma([0.5 -0.8], [-0.6 0.08], 9)' ```
```Warning: GARCHMA will be removed in a future release. Use ARMA2MA instead. PSI = -0.1000 -0.7700 -0.3050 0.4635 0.4758 -0.1329 -0.4471 -0.1172 0.2991 ```

From the model, `AR = [0.5 -0.8]` and ```MA = [-0.6 0.08]```.

 Note:   Since the current-time-index coefficients of yt and εt are `1`, the example omits them from `AR` and `MA`. This saves time and effort when you specify parameters via dot notation on a `garch` model.

## References

[1] Box, G. E. P., G. M. Jenkins, and G. C. Reinsel. Time Series Analysis: Forecasting and Control. 3rd ed. Englewood Cliffs, NJ: Prentice Hall, 1994.


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UV Coordinates in MaxScript

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Hey all, I''m writing a maxscript file to export Max files to the format I want them in, it''s going to be pretty standard, but when exporting the texture coordinates for every face''s vertices, I used something like this: ... in somefunction ExportMesh format "uvcoords %\n" m.numtverts to: outstream for i=1 to m.numtverts do ( local v = gettvert m i format " % % 0 \n" v.x v.y to: outstream ) .. so this is done for every face in the current mesh being exported... I''m only doing UV coordinates, so no W, which is why i dont use the v.z, so... my problem is, the coordinates that are being output to file aren''t in the range of 0.0f to 1.0f.. they''re actually arn''t anywhere near that range. So i figure MAX does things a little differently... (btw this is 3d studio max 4).. does anyone know how I can convert these texture coordinates to the DirectX 0..1 range? To simplify things, faces are only allowed 3 vertices in my engine, so basically triangles only... my structure goes something like this: struct sface { int m_iInds[3]; //indices to vertex list held one level up float m_fU[3], m_fV[3]; // *snip* everything else like diffuse, normals, etc.. }; Everything else in my script works perfectly, my engine loads the geometry correctly and can render it w/ the diffuse/specular colors... just w/o texture info, and if I feed it texture coords or use directx autogenerate UV coords, the textures are mapped, they just don''t display the way I want them to... (w/ the coordinates from MAX).... so any help would be much appreciated thanks, rj

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Been awhile since I wrote a max script..but here is part of an exporter I made...

Max has like a buffer of UV values..and each triangle has 3 indexes to the buffer....Kind of like Vertex and Index buffers..

				try (					print "<TEXCOORDS>" to:file					print o[i].NumTVerts to:file					for v = 1 to o[i].NumTVerts do (						u = gettvert o[i] v						print u to:file					)				)				catch (print "Error...Texcoords")							try (					if o[i].NumTVerts != 0 then (					print "<TEXFACES>" to:file						for f = 1 to o[i].NumFaces do (						ti = gettvface o[i] f						print ti to:file					)					)				)				catch (print "Error...TexFaces")

Here is the output for a simple box from my exporter...

"<MESH>""Box01"[184.0,228.0,153.0]"<VERTS>"8[-10.8372,-12.1117,0][9.1817,-12.1117,0][-10.8372,9.46492,0][9.1817,9.46492,0][-10.8372,-12.1117,23.5746][9.1817,-12.1117,23.5746][-10.8372,9.46492,23.5746][9.1817,9.46492,23.5746]"<FACES>"12[1.0,3.0,4.0,0.0,0.0,-1.0,2][4.0,2.0,1.0,0.0,0.0,-1.0,2][5.0,6.0,8.0,0.0,0.0,1.0,4][8.0,7.0,5.0,0.0,0.0,1.0,4][1.0,2.0,6.0,0.0,-1.0,0.0,8][6.0,5.0,1.0,0.0,-1.0,0.0,8][2.0,4.0,8.0,1.0,0.0,0.0,16][8.0,6.0,2.0,1.0,0.0,0.0,16][4.0,3.0,7.0,0.0,1.0,0.0,32][7.0,8.0,4.0,0.0,1.0,0.0,32][3.0,1.0,5.0,-1.0,0.0,0.0,64][5.0,7.0,3.0,-1.0,0.0,0.0,64]"<TEXCOORDS>"12[0,0,0][1,0,0][0,1,0][1,1,0][0,0,0][1,0,0][0,1,0][1,1,0][0,0,0][1,0,0][0,1,0][1,1,0]"<TEXFACES>"[10,12,11][11,9,10][9,10,12][12,11,9][5,6,8][8,7,5][1,2,4][4,3,1][5,6,8][8,7,5][1,2,4][4,3,1]"<\MESH>""<ENDGEO>"

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Hey, thanks for the quick reply.

Yea I know about the indexed texture stuff, I forgot to put that part of my code in there, but I do get the texture indices for the faces and then once its all loaded, get the actual texture coordinate values for each face. Thanks again, but I think i''m having a different problem

-rj

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# How do I find the standard deviation of a portfolio? [closed]

Compute the expected return $\mu_V$ and standard deviation $\sigma_V$ of a portfolio consisting of three securities with weights $\omega_1=40\%$, $\omega_2=-20\%$, $\omega_3=80\%$, given that the securities have expected reuturns $\mu_1=8\%$, $\mu_2=10\%$, $\mu_3=6\%$, standard deviations $\sigma_1=0.15$, $\sigma_2=0.05$, $\sigma_3=0.12$, and correlations $\rho_{12}=0.3$, $\rho_{23}=0$, $\rho_{31}=-0.2$.

I know how to compute the expected return of the portfolio, I got $\mu_V=0.06$, but I don't know how to calculate the standard deviation of a portfolio? What is the formula I need to use given the information? Do I need to find the variances given the standard deviations?

• @BobJansen So this is basic? – BCLC Feb 23 '16 at 18:39

You can calculate variance of a portfolio/basket by taking direct weighed averages of the components and then adding the relevant correlation terms * weights for each pair.

Can take sqrt of the expression obtained to have Standard deviation.

Exact formula for calculation goes like this :

(source: benetzkorn.com)

• The site that had your image went down. Could you please find a way to replace the image? – Mithical Mar 8 '17 at 12:46


text:
Standard:

Description:

Standard:

# Math.8.EE.4 or 8.EE.A.4

Description:

## Perform operations with numbers expressed in scientific notation, including problems where both decimal and scientific notation are used. Use scientific notation and choose units of appropriate size for measurements of very large or very small quantities (e.g., use millimeters per year for seafloor spreading). Interpret scientific notation that has been generated by technology.

8th Grade Math - Scientific Notation Lesson


text:
## Innovations in Incidence Geometry

### The universal representation group of Huybrechts's dimensional dual hyperoval

#### Abstract

A $d$-dimensional dual hyperoval can be regarded as the image $S = ρ ( Σ )$ of a full $d$-dimensional projective embedding $ρ$ of a dual circular space $Σ$. The affine expansion $Exp ( ρ )$ of $ρ$ is a semibiplane and its universal cover is the expansion of the abstract hull $ρ ˜$ of $ρ$.

In this paper we consider Huybrechts’s dual hyperoval, namely $ρ ( Σ )$ where $Σ$ is the dual of the affine space $AG ( n , 2 ) ⊂ PG ( n , 2 )$ and $ρ$ is induced by the embedding of the line grassmannian of $PG ( n , 2 )$ in $PG n + 1 2 − 1 , 2$.

It is known that the universal cover of $Exp ( ρ )$ is a truncation of a Coxeter complex of type $D 2 n$ and that, if $U ˜$ is the codomain of the abstract hull $ρ ˜$ of $ρ$, then $U ˜$ is a subgroup of the Coxeter group $D$ of type $D 2 n$, $| U ˜ | = 2 2 n − 1$ but $U ˜$ is non-commutative. This information does not explain what the structure of $U ˜$ is and how $U ˜$ is placed inside $D$. These questions will be answered in this paper.

#### Article information

Source
Innov. Incidence Geom., Volume 3, Number 1 (2006), 121-148.

Dates
Accepted: 22 May 2006
First available in Project Euclid: 28 February 2019

https://projecteuclid.org/euclid.iig/1551323244

Digital Object Identifier
doi:10.2140/iig.2006.3.121

Mathematical Reviews number (MathSciNet)
MR2267610

Zentralblatt MATH identifier
1229.51008

#### Citation

Del Fra, Alberto; Pasini, Antonio. The universal representation group of Huybrechts's dimensional dual hyperoval. Innov. Incidence Geom. 3 (2006), no. 1, 121--148. doi:10.2140/iig.2006.3.121. https://projecteuclid.org/euclid.iig/1551323244

#### References

• B. Baumeister, T. Meixner and A. Pasini, $\GF(2)$-expansions, Geom. Dedicata 67 (1997), 163–180.
• B. Baumeister and A. Pasini, On flat flag-transitive $c.c^*$-geometries, J. Algebraic Combin. 6 (1997), 2–26.
• B. Baumeister and G. Stroth, The non-canonical gluings of two affine spaces, in Groups and Geometries (L. of Martino et al. eds.), Birkhäuser, Basel (1998), 9–28.
• M. Buratti and A. Del Fra, Semi-Boolean Steiner quadruple systems and dimensional dual hyperovals, Adv. Geom., special issue in honour of Adriano Barlotti (2003), S254–S270.
• J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker and R. A. Wilson, Atlas of Finite Groups, Clarendon Press, Oxford, 1985.
• B. N. Cooperstein and J. A. Thas, On generalized $k$-arcs in $\PG(2n,q)$, Ann. Comb. 5 (2001), 141–152.
• A. Del Fra, On $d$-dimensional dual hyperovals, Geom. Dedicata 79 (2000), 157–178.
• A. Del Fra and S. Yoshiara, Dimensional dual hyperovals associated with Steiner systems, European J. Combin. 26 (2005), 173–194.
• C. Huybrechts, Dimensional dual hyperovals in projective spaces and $c.\AG^*$-geometries, Discrete Math. 255 (2002), 193–223.
• C. Huybrechts and A. Pasini, Flag-transitive extensions of dual affine spaces, Beitr\Hage zur Algebra und Geometrie, 40 (1999), 503–531.
• A. V. Ivanov and S. V. Shpectorov, Geometries of Sporadic Groups II. Representations and Amalgams, Cambridge University Press, Cambridge 2002.
• A. Pasini, Diagram Geometries, Oxford University Press, Oxford 1994.
• A. Pasini, Embeddings and expansions, Bull. Belg. Math. Soc. Simon Stevin 10 (2003), 585–626.
• A. Pasini and S. Yoshiara, On a new family of flag-transitive semibiplanes, European J. Combin. 22 (2001), 529–545.
• A. Pasini and S. Yoshiara, New distance regular graphs arising from dimensional dual hyperovals, European J. Combin. 22 (2001), 547–560.
• J. A. Thas and H. Van Maldeghem, Characterizations of quadric and Hermitian Veroneseans, J. Geom. 76 (2003), 282–293.
• J. A. Thas and H. Van Maldeghem, Characterizations of the finite quadric Veroneseans ${\cal V}^{2^n}_n$, Quart. J. Oxford 55 (2004), 99–113.
• S. Yoshiara, A family of $d$-dimensional dual hyperovals in $\PG(2d+1,2)$, European J. Combin. 20 (1999), 489–503.
• S. Yoshiara, Ambient spaces of dimensional dual arcs, J. Alg. Combin. 19 (2004), 5–23.
• S. Yoshiara, Dimensional dual arcs. A survey, to appear.


text:
# How to stop a program HSERIN blocking

#### camerart

Joined Feb 25, 2013
2,113
Hi,
EDITED:
I'm using HSERIN an a program, but it blocks
How do I stop this please?
As this has been edited, some early replies, may look confusing.
Camerart.

Last edited:

#### ericgibbs

Joined Jan 29, 2010
10,846
hi C,
You could check for a character in the serial buffer, by using an ASM statement, prior to trying HSERIN.
E

#### camerart

Joined Feb 25, 2013
2,113
hi C,
You could check for a character in the serial buffer, by using an ASM statement, prior to trying HSERIN.
E
Hi E,
I can't 'do' ASM, and as I'm tidying up, I'll leave this till later in my project.
Thanks.
C.

#### ericgibbs

Joined Jan 29, 2010
10,846
hi C,
E

If PIR1.RCIF = 0 Then Goto getmsg
char = RCREG
............................

#### Ian Rogers

Joined Dec 12, 2012
807
You should use HSERGET.. This is non blocking but will return 0 if nothing has arrived...

#### camerart

Joined Feb 25, 2013
2,113
Hi E and I,
I found the [ getmsg ] programs, and now understand how to use it.

I tested HSERGET instead of HSERIN and it works fine, especially, as sometimes a message starts then gets garbled, and HSERGET READS it, so some DATA can be collected

At the moment I'm checking using the Simulator and at this point;
--------------------------------------------------------------------------------------------
Hserget char
If char = "$" Then 'start str1(0) = char str2(0) = char rxi = 1 Else Goto skip1 Endif ------------------------------------------------------------------------ if I mess up the beginning of the message which starts with [$ ] then it goes to [ skip1 ] but leaves the rest of the message blocking the [ Hardware UART sim ]
Will this be the same LIVE (I'll be LIVE testing later.)
Thanks,
C.

#### ericgibbs

Joined Jan 29, 2010
10,846
hi C,
As your serial inputs are not synchronised, recall what we had to do sometime ago.
If the RX buffer gets buffer gets data and it is not Read, then the OERR flag is set, if you do not test for this flag, when you read the buffer it could be 'old' rubbish data.
We used the following routine.

get_neo:
If RCSTA.OERR = 1 Then 'if over run error then flush RXD buffer
RCSTA.CREN = 0
RCSTA.CREN = 1
char = RCREG
char = RCREG
PIR1.RCIF = 0
Endif

sync1: 'wait for a $start of string If PIR1.RCIF = 0 Then Goto sync1 char = RCREG If char <> 0x24 Then Goto sync1 '$'
str1(1) = char
rxi = 2

#### camerart

Joined Feb 25, 2013
2,113
hi C,
As your serial inputs are not synchronised, recall what we had to do sometime ago.
If the RX buffer gets buffer gets data and it is not Read, then the OERR flag is set, if you do not test for this flag, when you read the buffer it could be 'old' rubbish data.
We used the following routine.

get_neo:
If RCSTA.OERR = 1 Then 'if over run error then flush RXD buffer
RCSTA.CREN = 0
RCSTA.CREN = 1
char = RCREG
char = RCREG
PIR1.RCIF = 0
Endif

sync1: 'wait for a $start of string If PIR1.RCIF = 0 Then Goto sync1 char = RCREG If char <> 0x24 Then Goto sync1 '$'
str1(1) = char
rxi = 2
Hi E,
We changed the [ get_neo ] to [ get_data ] where some changes were made, here is what is used at the moment:
I notice there isn't [ PIR1.RCIF ] in the later CODE.

Is there a reason for the second [ char = RCREG ] marked with XXXXXXXXXXXXXX
C.
Code:
------------------------------------------------------------------------------------
get_data:
'Await GPS RXD -----------------GPS RMC ----------------------
''''''PASTE: $GNRMC,123519,A,4807.038,N,01131.000,E,022.4,084.4,230394,003.1,W*6A? '''OR PASTE:$REMOTE,1100,1200,1300,1400,1500,1600,?

If RCSTA.OERR = 1 Then  'if over run error then flush RXD buffer
RCSTA.CREN = 0
char = RCREG
char = RCREG  'XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
Hserout "RX Err!", CrLf
Goto skip1
Endif
'Hserin char
'HSERGET TEST
Hserget char

C

#### Attachments

• 67.6 KB Views: 2
Last edited:

#### camerart

Joined Feb 25, 2013
2,113
Hi,
I've tried the simulator where #15 is true.

LIVE tests:
Using HSERGET does keep the program flowing allowing the other modules (Compass etc) are serviced, but doesn't appear to process the CHARS so I think HSERIN has to be used. Any comments?

Using HSERIN, as shown in the simulator the UART blocks.

IMPORTANT NOTE:
In the program I use a DATA SWITCH to switch between the GPS and 'BASE/REMOTE', each INPUT switches to the other as each is READ
C

Last edited:

#### camerart

Joined Feb 25, 2013
2,113
Hi,
Reading #14 again, Am I correct that [ On UART RX ] would need a TIMER? If so then I don't have it set, would it be better if I did?.

Here is the bottom of the MAIN LOOP, showing the GOTO and the SUBROUTINE.
C.

#### Attachments

• 3.6 KB Views: 11

#### ericgibbs

Joined Jan 29, 2010
10,846
hi C,
You are Exiting a GOTO call to subroutine with a RETURN.
Its the Program thats hanging not the UART
E

get_data:
If RCSTA.OERR = 1 Then 'if over run error then flush RXD buffer
RCSTA.CREN = 0
char = RCREG '1
char = RCREG '2
PIR1.RCIF = 0 'clear the RCIF flag, which had been set by the earlier err data.
Hserout "RX Err!", CrLf
Goto skip1 ' ?????????????????????????????????
Endif
......................................................................
skip1:

For x = 0 To 79
str1(x) = 0
Next x

For x = 0 To 79
str2(x) = 0
Next x

Return '???????????????????????????????

#### camerart

Joined Feb 25, 2013
2,113
hi C,
You are Exiting a GOTO call to subroutine with a RETURN.
Its the Program thats hanging not the UART
E

get_data:
If RCSTA.OERR = 1 Then 'if over run error then flush RXD buffer
RCSTA.CREN = 0
char = RCREG '1
char = RCREG '2
PIR1.RCIF = 0 'clear the RCIF flag, which had been set by the earlier err data.
Hserout "RX Err!", CrLf
Goto skip1 ' ?????????????????????????????????
Endif
......................................................................
skip1:

For x = 0 To 79
str1(x) = 0
Next x

For x = 0 To 79
str2(x) = 0
Next x

Return '???????????????????????????????
Hi E,
The RETURN is returning to the earlier [ Gosub get_data ] See #17
C.

Last edited:

#### camerart

Joined Feb 25, 2013
2,113
Hi,
As each test using the fantastic simulator takes quite a while, I quite often loose track of what I'm testing (Even with notes) So I tend to test LIVE for this.

Today I woke up with a FLOWCHART in my head, but while I can 'see' it, I can't tell if it would work as a program.

Using the example in #17, I understand how each element works (Almost!) Here is what I want to try, and I think I can write it.
-------------------------------------------------------------------------------------------------
Use the [ If RCSTA.OERR = 1 Then ] section at the beginning.

[ If char = "\$" Then ] start LOOP of this until>
[ If str1(1) = "R" Then ] section, then look for that sentence [ ? ]
OR
[ If str1(5) = "C" or "R" Then ] section, then look for that sentence [ ? ]

(Now that sentence is in STR1 or 2) switch [ DATASWITCH ] to look for the alternate type of sentence, while processing this sentence. (This would hopefully save going though all of the approx 10x NMEA sentences when checking the GPS)

If either type of sentence are not RECEIVED then use an INTERRUPT to switch the DATASWITCH until one or the other sentence is RECEIVED.

I hope this is clear, I'll keep checking, while I attempt to write this, if someone would check the flow please?

EDIT: MARKED BOLD
C.
----------------------------------------------------------------------------------------------------

Last edited:


text:
# Nice way to find the integral

1. Sep 22, 2008

### roeb

1. The problem statement, all variables and given/known data
$$\int_0^{\inf} \frac{x^3}{e^x - 1}dx$$

2. Relevant equations

3. The attempt at a solution
I am having a hard time with this one... Not sure what to do

I did the easier integral $$\int \frac{1}{e^x - 1} dx= ln(e^x - 1) - x$$

I then proceeded to attempt integration by parts, but I can't seem to come up with a nice way to find the integral.

$$u = 1/(e^x - 1).................du = \frac{-e^x}{(e^x-1)^2 }dx$$
$$dv = x^3 dx........v = x^4/4$$
But as you can see that seems to complicate the integral so I gave up there...

I also tried by saying:
$$u = x^3......du = 3x^2$$
$$dv = 1/(e^x - 1)..............v = ln(e^x - 1) - x$$

But this is even worse....

Integration by parts for this integral doesn't seem to work. I am thinking since the integral goes from 0 to inf that there may be some other trick that I am not aware of. Does anyone know what to do for this integral?

2. Sep 22, 2008

### cellotim

Re: Integration

The antiderivative can't be expressed in closed form. Try expanding $$\frac{1}{e^x - 1}$$ into a geometric series.

3. Sep 22, 2008

### roeb

Re: Integration

Hmm interesting, thanks. Let me see if I understand correctly,

e^x = 1 + x + ....

$$\int_0^{\inf} \frac{x^3}{e^x - 1}dx = \int_0^{\inf} \frac{x^3}{1 + x - 1}dx = \int_0^{\inf} x^2 dx = \inf ?$$

I don't believe I've ever done an integral by using a series substitution like that, but I was under the impression that the answer shouldn't be infinity? Am I doing it incorrectly yet?

4. Sep 22, 2008

### Dick

Re: Integration

No, I don't think that's what roeb meant. Use the geometric series expansion 1/(1-e^(-x))=1+e^(-x)+e^(-2x)+... That means 1/(e^x-1)=e^(-x)+e^(-2x)+... This isn't an elementary integral. Now look up the definition of the gamma function. The terms in your expansion look like x^3*exp(-nx). By a change of variable you can put that into the form of the gamma function and get that it's gamma(4)*1/n^4. So you've got gamma(4)*(1/1^4+1/2^4+1/3^4+...). Now look up the definition of the zeta function.

Last edited: Sep 22, 2008
5. Sep 23, 2008

### Dick

Re: Integration

Zero? That doesn't look right, the integrand is non-negative. Your initial series expansion is divergent and the final series doesn't alternate. Oh, and when I said "I don't think that's what roeb meant" in post 4, I meant "I don't think that's what cellotim meant". The series expansion is good idea. I don't think you've got the parts following that right yet.

Last edited: Sep 23, 2008
6. Sep 23, 2008

### Almanzo

Re: Integration

This is an improper integral, because the integrand x3/(ex-1) is not defined at the edge of the integration domain, x=0. e0 - 1 = 1 - 1 = 0. 03/0 = 0/0 is undefined. So one must first ascertain whether it converges.

To find a primitive function, one might substitute ex = u, or x = elog(u). Then x3/(ex-1) = {elog(u)}3/(u-1), and dx = (1/u)du. Integrating by parts, and remembering that 1/{u(u-1)} = 1/(u-1) - 1/u would yield some nice expressions, but also the nasty elog(u+1)*d{elog(u)}. I don't know what to do with that, but perhaps someone else does.

7. Sep 23, 2008

### Dick

Re: Integration

Near x=0 it behaves like x^2, roeb's power series expansion shows that clearly enough. x=0 is a removable singularity and nothing to worry about. There is no elementary primitive function. As I've said, you need a zeta function.

8. Sep 24, 2008

### Almanzo

Re: Integration

Last night, I have thought about it some more.

x3/(ex-1) = (x3/ex)*1/(1-e-x)
x3/(ex-1) = (x3/ex)*(1 + e-x + e-2x + e-3x + e-4x + ...)
x3/(ex-1) = x3 * (e-x + e-2x + e-3x + e-4x + ...)

Let Gn(x) = (-1/n)*(x3 + 3x2/n + 6x/n2 + 6/n3)*e-nx
gn(x) = dGn(x)/dx = (x3 + 3x2/n + 6x/n2 +6/n3)*e-nx - (1/n)*(3x2 + 6x/n + 6/n2)*e-nx
gn(x) = x3 * e-nx

Therefore, if dF(x)/dx = x3/(ex-1)
F(x) = G1(x) + G2(x) + G3(x) + G4(x) + ...

Each of the G's is zero at x=0 and approaches zero if x increases without bound.
But is the same now true for F? This must still be proved.

If it is true for F, then the integral comes to 0 - 0 = 0.

Last edited: Sep 24, 2008
9. Sep 24, 2008

### Dick

Re: Integration

Gn(0) isn't 0. It's 6/n^4. (Not -6/n^4). x^3/(exp(x)-1) IS POSITIVE for x>0. How can you even think to claim the integral is zero? It's the sum over n of 6/n^4. It's 6*zeta(4).

10. Sep 24, 2008

### Dick

Re: Integration

Ooops. Sorry. You are correct. Gn(0)=(-6/n^4). It's the integral of the nth term that is +6/n^4.

11. Sep 24, 2008

### dirk_mec1

Re: Integration

$$\int_{0}^{\infty} \frac{x^3}{e^x-1}\ \mbox{d}x = \int_0^{\infty} \frac{e^{-x}\ x^3}{1-e^{-x}}\ \mbox{d}x = \int_0^{\infty} \sum_{n=0}^{\infty} e^{-(n+1)x}\ x^3\ \mbox{d}x = \int_0^{\infty} \sum_{n=0}^{\infty} \frac{6}{(n+1)^3}\ e^{-(n+1)x}\ \mbox{d}x= \sum_{n=0}^{\infty} \frac{6}{(n+1)^4} = \frac{6 \pi^4}{90}$$

12. Sep 25, 2008

### Almanzo

Re: Integration

Oh, yes. Gn(0) = (-1/n)*(6/n3)*e0=(-6/n4). Not zero at all.
But Gn(x) does tend to zero if x increases without bound.
So, the integral of gn(x) from 0 to infinity would be (+6/n4).

Strangely, because someone had earlier suggested zero as the answer, I accepted a result of zero without carefully checking my own algebra. Sorry.

And, if the integral converges on both ends, the integral of f(x) from 0 to infinity comes to:
6 * (1 + 1/16 + 1/81 + 1/256 + ....). Exactly the sum in the previous post.

(By the way, I wonder how to produce integral and sum signs in my posts.)

Last edited: Sep 25, 2008
13. Sep 25, 2008

### cellotim

Re: Integration

Use the tex, /tex commands each in brackets to do latex.

14. Sep 25, 2008

### Dick

Re: Integration

If you click on the symbols in dirk_mec's post, it should pop up a window showing you how the tex symbols were entered. And yes, you've got the integral. The sum of inverse powers is called the zeta function. (As I've said in every post).