text:
# Pang Yan Han's blog

## Vim abbreviations for a specific file

20 Feb 2018, by Pang Yan Han

Disclaimer: Opinions expressed on this blog are solely my own and do not express the views or opinions of my employer(s), past or present.

Recently I came across Bram Moolenaar’s article Vim: Seven habits of effective text editing and learnt about vim abbreviations. Essentially, vim abbreviations allow you to specify shortform words that vim will automatically expand to their long form. For instance, with the following in your vimrc:

ab dev developer

Whenever you type the word dev followed by a space / enter / punctuation mark, vim will automatically expand it to developer. I am not entirely certain of the rule that vim uses to expand the abbreviation, but it seems to me that vim does it on a word boundary basis. Hence you do not have to be afraid that you cannot type devops for fear that it will be expanded to developerops.

This is pretty useful and can save a lot of typing. There are other pieces of wisdom in the article and I highly encourage you to read it.

## How to have file specific vim abbreviations

Which brings us to the meat of this post. How do we have abbreviations that apply to a specific file? Some Googling led to this Unix & Linux Stack Exchange answer and here’s some example code on how to do it:

au BufRead,BufNewFile /path/to/file call YourAbbrevFunction()
function YourAbbrevFunction()
ab dev developers
" define your other abbreviations here
endfunction

### Instructions

1. Copy and paste the above code to your vimrc file
2. Define all your desired abbreviations in the YourAbbrevFunction vim function
3. Change the /path/to/file to the file you want the abbreviations to apply to
4. If you have an existing vim session editing the /path/to/file, you will need to quit and start it again for the abbreviations to apply

Enjoy!

## References

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text:
# How do you write an equation in point slope form that passes through Point: (7, 4); Slope: 6?

In general, given a point $\left(\hat{x} , \hat{y}\right)$ and a slope $m$
$\left(y - \hat{y}\right) = m \left(x - \hat{x}\right)$
Using the point $\left(7 , 4\right)$ and the slope $6$
$\left(y - 4\right) = 6 \left(x - 7\right)$

format: plain


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# Cell segmentation119

Posted by Steve Eddins,

Blog reader Ramiro Massol asked for advice on segmenting his cell images, so I gave it a try. I'm not a microscopy expert, though, and I invite readers who have better suggestions than mine to add your comments below.

Let's take a look first to see what we have. I'm going to work with a cropped version of the original so that the images aren't too big for the layout of this blog.

Note: you can download the functions imcredit and imoverlay from MATLAB Central.

I = imread('http://blogs.mathworks.com/images/steve/60/nuclei.png');
I_cropped = I(400:900, 465:965);
imshow(I_cropped)
imcredit('Image courtesy of Dr. Ramiro Massol')

Strictly speaking, contrast adjustment isn't usually necessary for segmentation, but it can help the algorithm developer see and understand the image data better. This is a fairly low-contrast image, so I thought it might help. You can adjust the display contrast interactively with imtool, or you can use an automatic method such as adapthisteq. adapthisteq implements a technique called contrast-limited adaptive histogram equalization, or CLAHE. (I always thought "CLAHE" sounded like it must be some Klingon delicacy.)

I_eq = adapthisteq(I_cropped);
imshow(I_eq)

So what happens if we just apply a threshold now?

bw = im2bw(I_eq, graythresh(I_eq));
imshow(bw)

Let's clean that up and then overlay the perimeter on the original image.

bw2 = imfill(bw,'holes');
bw3 = imopen(bw2, ones(5,5));
bw4 = bwareaopen(bw3, 40);
bw4_perim = bwperim(bw4);
overlay1 = imoverlay(I_eq, bw4_perim, [.3 1 .3]);
imshow(overlay1)

Now, I'm not familiar with these cell images, so I don't know exactly what I'm looking at. I assume some of these blobs need more help to be separated properly. One possible approach is called marker-based watershed segmentation. There's a demo of this idea on The MathWorks web site.

With this method, you have to find a way to "mark" at least a partial group of connected pixels inside each object to be segmented. You also have to mark the background.

Let's try to use the bright objects, which I assume are nuclei. The extended maxima operator can be used to identify groups of pixels that are significantly higher than their immediate surrounding.

mask_em = imextendedmax(I_eq, 30);
imshow(mask_em)

Let's clean that up and then overlay it.

mask_em = imclose(mask_em, ones(5,5));
overlay2 = imoverlay(I_eq, bw4_perim | mask_em, [.3 1 .3]);
imshow(overlay2)

Next step: complement the image so that the peaks become valleys. We do this because we are about to apply the watershed transform, which identifies low points, not high points.

I_eq_c = imcomplement(I_eq);

Next: modify the image so that the background pixels and the extended maxima pixels are forced to be the only local minima in the image.

I_mod = imimposemin(I_eq_c, ~bw4 | mask_em);

Now compute the watershed transform.

L = watershed(I_mod);
imshow(label2rgb(L))

I don't know if this is a good segmentation result or not, but I hope some of the methods I've shown will give Dr. Massol some ideas to try.

Get the MATLAB code

Published with MATLAB® 7.2

kruthika replied on : 1 of 119

hy…im sorry ….amnt here to give a comment…i am working on splittin and merging algorithms..i find there is some error with the splitmerge.m function given in ‘Digital Image Processing using MATLAB’…Could u please help me…i ve just learnt the basics of image segmentation….

thnk u…

Steve replied on : 2 of 119

Kruthika, Rafael Gonzalez knows more about the splitmerge function than I do. You can find contact information for Professor Gonzalez at http://www.imageprocessingplace.com.

ishraq replied on : 3 of 119

hi everbody,,
i have a qustion about coins segmentation ,,
i didn’t find a good way to segment the coins when they are
connected (the edges of coins are connected).

any one can help me please???!?!?!

ishraq replied on : 6 of 119

Dear Mr Steve ,
Thanks alot for your fast replaying ,
but I still have difficulties in using watershed because
im beginner in using MATLAB,

I like to add 2 new urls for coin images with overlapping between the edges

my problem is about how i can segment these coins in a good way so i can use “bwlabel” and “regionprops”
to recognize the coin types .

i hope from you to help me to find a way to segment these coins because you are expert person .

thanks you alot
ishraq

Steve replied on : 7 of 119

Ishraq – I’m sorry, but I can’t work this problem out for you, or help you learn MATLAB. Go through the Getting Started in MATLAB guide to get some MATLAB knowledge. Then look at the Image Processing Toolbox Users Guide and the product demos on our web site, including the one I already linked to. Look at the material in the documentation and demos on thresholding, morphological operators such as erosion and opening, and watershed segmentation.

Wilson replied on : 8 of 119

Hi there,
I have a query on how to use Matlab for Vehicle Plate segmentation?
I read from the papers that one solution is to gray scale the picture. But how does the system be so clever to segment out the unwanted parts and leave the vehicle plate untouched?
Hope to hear from you guys soon.
Cheers..

subrajeet mohapatra replied on : 9 of 119

Hi,
I have read your book but there is nothing about adaptive histogram equalization.Can u give me some ideas or the algorithm for adaptive histogram equalization.So that i can go to CLAHE.

Steve replied on : 10 of 119

Subrajeet – There is an implementation of CLAHE in the Image Processing Toolbox. See the adapthisteq function.

subrajeet mohapatra replied on : 11 of 119

can i get the complete algorithm

Steve replied on : 12 of 119

Subrajeet – The complete algorithm is summarized at the top of the M-file, plus you can always look at the code. The M-file also contains this reference: Karel Zuiderveld, “Contrast Limited Adaptive Histogram Equalization”, Graphics Gems IV, p. 474-485, code: p. 479-484

subrajeet mohapatra replied on : 13 of 119

Hi,
is there any built in function to measure signal to noise ratio for images ,My friend is telling there is Peak signal to noise ratio (PSNR) im image processing toolbox

Steve replied on : 14 of 119

Subrajeet – There’s no PSNR function in the Image Processing Toolbox, but it’s very easy to implement it directly from the formula.

Fabiano Malhard replied on : 15 of 119

Hi steve,

I am really new at MatLab and I am trying to work with images, reducing its size and later resizing them using one of the commons methods … but i would like to compare the resulting images with the original ones …

Can you help with the SNR function, or there is any other better aproach ???

Thanks

Steve replied on : 16 of 119

Fabiano – There are several different definitions of SNR used for image comparisons, so the details vary depending on the definition you are interested in. But the heart of the computation, typically something like a mean-squared error, is straightforward in MATLAB. If you are comparing A and B, it might look something like this:

mse = mean( (A(:) - B(:)).^2 );

rosani replied on : 17 of 119

Steve,
how to compute a confusion matrix to measure accuracy between two image(rgb,in pixel) from different source.

Steve replied on : 18 of 119

Rosani—I’m confused; what’s a confusion matrix?

bob gold replied on : 19 of 119

Hi my name is Bob Gold a pharmacist in Indiana. I
am interested in developing a method to determine the
number of tablets in a bottle for the purpose of
improving compliance. I am not an expert on cameras .

Do you have any ideas.
Thanks

Rob replied on : 20 of 119

Bob Gold,

I think first of all you’ll need some really wide, flat bottles so the pills are no more than 1 deep in the image. (sorry for the sarcasm, couldn’t resist)

Other than that, you could take a picture of pills in a regular bottle and compute an *estimate* of how many pills are packed in there. Depending on the pill shape, there are many standard packing computations, such as hexagonal-close-packed, body-centered-cubic, etc. Think lattice structures, but all the ones I’ve studied are for spheres. The main failing here is that you won’t get the real number, which it sounds like you need for compliance.

Hope this helps,
Rob

Steve replied on : 21 of 119

Bob—I don’t anything to concrete to suggest beyond what Rob said. It sounds like it might not be feasible except under very particular circumstances.

Rob—Thanks for jumping in.

Steve replied on : 22 of 119

Bob—OK, I do have one thought. I’d be more tempted to use a precision scale instead of a camera and image processing to measure the number of pills. But this kind of quality assurance application is a bit beyond my expertise.

Adnan Butt replied on : 23 of 119

Hello Sir,
I am working in MATLAB image processing and one thing I do have explored that most of the commands extracting quantitaive measures work only on 2D matrices there is not much material on 3D image processing in MATLAB?
Can you guide me in 3d Image processing in MATLAB?
Regards

Steve replied on : 24 of 119

Adnan—There are many functions in MATLAB and the Image Processing Toolbox that work on three-dimensional arrays, including filtering, transforms, morphology, and region measurements. Can you be more specific about what you are looking for?

Ben Taylor replied on : 25 of 119

Improvement: Blobs/particles touching the sides of the images will mess up your size measurements and shape characterization because they are not showing the whole particle. To remove all particles or cells touching the sides I wrote the following simple script that removes any particle touching the border of the image.

%%%%////
function L=borderstrip(L)
display(‘starting border strip’)
border=[L(end,:) L(1,:) L(:,end)’ L(:,1)’]; %these are all the pixels for the borders
background=mode(mode(L)); %this finds the most common pixel for the background which is usually 2

count=0;
for i=1:length(border)
if border(1,i)==2
border(1,i)=0;
else
count=count+1;
blacklist(count)=border(1,i);
end
end
blacklist=unique(blacklist); % this strips out all of the copied black listed pixels to be removed
height=size(L,1);width=size(L,2);
for j=1:length(blacklist)
for h=1:height
for w=1:width
if L(h,w)==blacklist(j)
L(h,w)=background;
end
end
end
end
display(‘ending border strip’)

Steve replied on : 26 of 119

Ben—You can use imclearborder to remove objects touching the border.

Saad Khan replied on : 27 of 119

Hi Steve,
Do you know of any matlab implementation of 3D deblurring (deconvolution)? The matlab image processing toolbox contains four deblurring functions but all of these work with images (2D data).
Thanks!

Steve replied on : 28 of 119

Saad—You can use the deconvolution routines in the Image Processing Toolbox. They are not limited to two dimensions.

VEN replied on : 29 of 119

Respected Sir,
I am Working in Matlab with Some Videos to Identify Moving
Object in that Video. I am Working in Matlab7 in which Video
Processing is not There. Now I am able to Identify the
moving objects Partially. Could you please Suggest me to Segment only that Moving objects In that Video. And also I want to Count the Number of segmented Object. Please help me in this matter.

Thanking You,

maged replied on : 30 of 119

i want functions that make segmantation of nodules in image by matlab
thanks

Shalin replied on : 31 of 119

3D deconvolution:

Hi Steve! Great page… Continuing on Saad’s query, could you help me with pointers on how to represent 3D point spread functions(in my case for optical microscope) and how to employ blind deconvolution.

Thanks
Shalin

Steve replied on : 32 of 119

Shalin—A 3D point spread function is represented as a 3D array containing the impulse response of the blur operator. There is an Image Processing Toolbox function for performing blind deconvolution. Did you have a specific question about it?

Stalin Mohapatra replied on : 33 of 119

Can you plz give me a complete MATLAB code on MEAN Filtering of Noisy Images?
It would be kind enough if you send it to my mail. I’ll be ever grateful to u.

Steve replied on : 34 of 119

Stalin—It sounds like you might just want to use imfilter (or the MATLAB function conv2) with a constant filter, such as ones(5,5)/25.

Adnan replied on : 35 of 119

Hi Steve, I posted the same query a couple of days ago but I cant see my Post now. Any way, There is a camera on road fixed in one position. I want to count the vehicles passed by the camera. But the problem is the segmentation i.e. to ignore any thing except vehicle. Kindly suggest me some proper algorithm for this problem.
Thanking in Anticipation
Regards

Eric replied on : 36 of 119

Hi Steve, I am trying to create circular and elliptical images as quality assurance tests for my image segmentation code. I can do a simple generate xy coordinates from parametric equations command, but I get a real blotchy image. I’ve figured out that it is because I’m not implementing Bresenham style line algorithums. Does matlab have built in Bresenham line drawing algorithums that can be augmented eaisly to draw nice smooth circles and ellipses in pixelated images?

Steve replied on : 37 of 119

Eric—As far as I understand, Bresenham drawing algorithms are for drawing shapes blazingly fast, with an absolute minimum of arithmetic, in low-level code. I’m not sure such a thing is really needed in MATLAB, where we get to do all the math we want. For example, the file toolbox/images/images/private/intline.m draws the same line between two pixels that a Bresenham line-drawing routine but, it uses some simple math to do it. So … what do you mean by “blotchy”?

I would probably “draw” a circle on an image using meshgrid and some sort of distance logic, like this:

[x,y] = meshgrid(linspace(-1,1,200));
bw = hypot(x,y) < = 0.25;

Wander replied on : 38 of 119

Dear Sir,
I want to find out the circularity and cell count of cell images after employing Marker Controlled Watershed transform. (Images are similar to those taken as illustration in this web-page). Can you kindly explain to me how to go ahead in this direction.

Steve replied on : 39 of 119

Wander—Use bwlabel and regionprops.

walid jerbi replied on : 40 of 119

Hi Steve,
I need to merge two regions in a labeled image. I want to know if there is a function that allow mes to do this?
Regards

Steve replied on : 41 of 119

Walid—Can you provide a more specific, detailed description of the operation you want to perform?

sherly replied on : 42 of 119

I am working with counting of cells in an image. I have segmented the cells but i need your help to count the number of cells

deepak replied on : 43 of 119

count the no: of pixels in the each segmented part

Steve replied on : 44 of 119

Sherly—If you have already segmented the cells, then use bwlabel to count them.

arindam replied on : 45 of 119

i am working on segmentation of myocardium and (general countour tracking) of heart from cardiac images . most articles seem to refer to using HMRF(hidden markov fields) and Expectation Maximisation namely

http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?arnumber=906424

the images i am working on are available from
specifically akld-mrirg-cines-truefisp-long.zip
the first set 0f 32 images.(long 1)
any guidance in the right direction will be appreciated.

Administrator replied on : 46 of 119

Arindam—I am not familiar with cardiac image segmentation.

Asanka replied on : 47 of 119

Is there any way to eliminate closely spaced markers of the wathershed segmentation inorder to avoid over segmentation?

Any suggestions to avoid over segmentation?

Steve replied on : 48 of 119

Asanka—If you are generating your markers using the minima of something like a gradient image, you might try using something like imextendedmin in order to filter out shallow minima.

TMan replied on : 49 of 119

I’m trying to calculate the number distinct objects in a huge (about 6000×6000) binary image B.

[ignore count] = bwlabel(B);

won’t work because of the double precision ‘ignore’-image not fitting my memory.

count = sum(sum(bwmorph(B,’shrink’,Inf)))

is, on the other hand, way too slow.

Anyone having efficient solutions to this problem?

silvia replied on : 50 of 119

hi steve,

I want to measure the area of cell nucleus. Could you explain to me the steps to do that?

Steve replied on : 51 of 119

TMan—You can start with bwlabel and modify it to suit your needs. Look in the file for the line:

numComponents = length(sizes);

That’s the answer you are looking for, and that line occurs before the memory-consuming output label matrix is constructed.

Steve replied on : 52 of 119

Silvia—Once you have the nuclei segmented, you can use bwlabel and regionprops to measure the area. I can’t help you much with the segmentation problem. Image segmentation usually requires some custom algorithm development that depends on the specific characteristics of your data set. Consult texts on microscopy and image processing for some general methods that you might be able to adapt to your data.

JP replied on : 53 of 119

When I copy this image, save it as a PNG, and try to use it, I get an error. Here is my code:

——————–
%test work with imfill
I_cropped = I;
imshow(I_eq)
——————–

All I’m doing is reading in the image, which I copied from the webpage. THen I get this error:

——————-
Error in ==> imfill_test at 4

??? Function ADAPTHISTEQ expected its first input, I, to be two-dimensional.
—————–

I can use adaphisteq if I read in a GIF or TIF with [X,map]=imread… and then use ind2gray, but the example image on this blog will not work for me at all, evne when saved as a TIF, using the same code that reads another TIF. I’m using R2006b. There must be something I’m missing about the definition of the image or something.

JP replied on : 54 of 119

Ok, if anybody else had the same problem as me, just make sure you grab only the top 2-D layer of the PNG after you read it in. On my computer it came in as MxNx3 and made adapthisteq choke. When I said I_cropped=I(:,:,1); it picked up a 2-D matrix and was happy with that input.

Steve replied on : 55 of 119

JP—The second paragraph of this post has a link to the original nuclei.png image for use with this image. The images you see on this web page are automatically captured screen shots from MATLAB, and they are stored in a color format even if the originals are grayscale.

I have modified the first line of code (the call to imread) to show how to read the correct original image directly, using its URL. I hope this will help others avoid similar confusion.

JP replied on : 56 of 119

Thanks, Steve! Pesky color images that aren’t color…

I will keep working to adapt the method to my images. In a nutshell, I have two images of a fuel spray taken with different filters. Each image is zeroed by a dark image, so it is nearly zero outside of the spray itself. The images are stored as matrices and then divided pixel by pixel. Within the spray itself, the resulting data is fine. But outside of the spray, you wind up dividing small noise by small noise and you get pure trash outside of the spray. I’m trying to use the same techniques to outline my spray and make a mask so I can discard everything outside the spray. I’m getting there, but I keep getting tripped up in the difference between matrices, greyscale images and color images. I’ll get it soon. Thanks!

Administrator replied on : 57 of 119

JP—Sounds like an interesting application.

Rudi replied on : 58 of 119

Steve:

Can I use this concept to segment human heads in a crowd scene? But the occlusion problem could be a significant issue for this purpose.

Thanks

Steve replied on : 59 of 119

Rudi—The concept of marker-controlled watershed segmentation is quite general, and might apply to your problem. I would expect the details to be very difficult to work out, however.

Chris replied on : 60 of 119

hi steve,

thanks for this suggestion of the cell segmentation. i have got a question: have you ever implement the pyramid linking algorithm from J. Burt? i am on a similar problem like this cell segmentation. but we’ve got colonies and sometimes the edges are blurred and sometimes they are sharp! if you want i can send you some pictures of these colonies!?!

Thanks
Chris

Steve replied on : 61 of 119

Chris—No, I’m not familiar with that method.

Rupesh Tatiya replied on : 62 of 119

Steve,

I have a similar image segmentation problem at hand where we have the RGB color image of human face and the objective is to segment the eyebrow from the rest of the face. I have tried different algorithms given in various tutorials but they do not work with good enough. So, can you help me in this matter?

Thanks and Regards,

Rupesh Tatiya

Steve replied on : 63 of 119

Rupesh—No, I generally do not have the time available to provide custom algorithm development advice for particular datasets.

Shannon replied on : 64 of 119

i am not sure if this has already been addressed/answered. I used code like yours, and incorporated what you did for segmentation.

I need to create code to count the specific cells once thresholded. like that produced a single number of the cell count, like 18.

i think i need a for loop, but i’m not really strong in matlab, so i was wondering if you knew how to count pixels in order to cell count or what… maybe had sample code or language to steer me in the right direction? THANK YOU!

shannon

Pradeep replied on : 65 of 119

Hi Steve
I am planning to do something on cell segmentation. But I am not an expert. I need your help. I need in areas of calculating more accurate cell density by keeping in mind: Partial Cell, Non-cell area and overlap cell. One more area i want to focus on is auto threshold if you have any ideas regarding this plz let me knw

Thanks

Steve replied on : 66 of 119

Shannon—Try this:

[L, num_cells] = bwlabel(segmented_cell_image);

Steve replied on : 67 of 119

Pradeep—Are your cells reasonably circular? If so, you might consider trying a circular Hough transform. There are many different variations on the literature. There’s at least one available on the MATLAB Central File Exchange.

Pradeep replied on : 68 of 119

thnx steve
i have one more question. why we only use black and white image for cell segmentation. like if we have any color image we first convert it to a grayscale.

Steve replied on : 69 of 119

Pradeep—It is convenient to use a binary (black and white) image to represent the outcome of the fundamental segmentation idea: Choosing which pixels belong to objects of interest, and which pixels belong to the background.

Sam replied on : 70 of 119

Hi ,
How can I identify objects with intermediate intensities in a image(there are bright objects, medium bright objects and the background), when the region around the brighter objects(glow around the bright objects) are of similar intensities to the medium intensity objects and needs to be excluded !

Tilman Sauter replied on : 71 of 119

Dear Sir,

I have developed a 3D image deconvolution code using the approximate method described in “Digital Image Processing, Castleman”. It uses the out of focus microscope images and the out of focus PSF images to determine the ‘noise’, which is then subtracted from the in-focus micrsocope image.

Now I would like to use for example the deconvlucy() function provided by Matlab to do the same thing in order to compare both.

Is it possible to work with out of focus planes in any of matlabs deconvolution functions to achieve 3D image deconvolution?

Tilman

Steve replied on : 72 of 119

Tilman—I don’t have the Castleman reference handy. Can elaborate on what you mean by “work with out of focus planes”?

Tilman Sauter replied on : 73 of 119

Dear Sir,

Thank you for your fast reply. The code I implemented is based on a simple equation:
in-focus-object = in-focus-image – sum(lower-out-of-focus-image convoluted with lower-out-of-focus-PSF + higher-out-of-focus-image convoluted with higher out-of-focus-PSF).
The sum sign defines the number of adjacent planes used, image the obtained image from the microscope and object the deconvoluted image.

The important thing is, that I can use information from the adjacent planes to improve the in-focus-object, whereas with deconvlucy() I can only use the information of the in-focus-image and in-focus-PSF.

Do you think it is possible to use deconvlucy() or another deconvolution function to achieve a comparable result?

With best regards
Tilman

Steve replied on : 74 of 119

Tilman—No, none of the deconvolution methods in the Image Processing Toolbox implement that technique. You would need to modify them yourself.

Steve replied on : 75 of 119

Sam—It’s hard to say without seeing a sample. Sounds like you have a difficult segmentation problem. If the object boundaries are reasonably well-defined, it might be helpful to use the gradient magnitude image.

Jon Hauris replied on : 76 of 119

Steve, I am interested in texture processing. Is there a way to search your excellent archives to see if you have covered this area and where it is located. Thank you, Jon

Steve replied on : 77 of 119

Jon—There is a MATLAB Central search box at the top of the page. You can limit the scope of the search to MATLAB Central blogs. Or you can click on the “Blog archive” link in the side panel. I haven’t written that much about texture analysis, though. You might want to try the texture analysis section in the toolbox documentation.

Giles Kingsley replied on : 78 of 119

Steve,

I have an image of a metal plate from a time-lapse series. There is a grid in the background, which I want to remove. Does this sound like a morphological operation? Throw me a bone here. Edge detection just enhances the grid, and thresholding removes area(my output) from the plate. Any ideas?
-Giles

Steve replied on : 79 of 119

Giles—An opening (IMOPEN) might work. How thick is the grid, and how thick is the object(s) of interest?

IB replied on : 80 of 119

I get the following error when I follow the commands here
“Undefined function or method ‘imoverlay’ for input arguments of type ‘uint8′.”
How could I overcome this error?

Steve replied on : 81 of 119

IB—The third paragraph in this blog post tells you how to get the function imoverlay.

IB replied on : 82 of 119

Steve do you mean the cell segmentation blog? I did not found it in the cell segmentation blog

Steve replied on : 83 of 119

IB—In this blog post, the very one you are commenting on, the third paragraph says “Note: you can download the functions imcredit and imoverlay from MATLAB Central.” The links are provided there.

azhagumani replied on : 84 of 119

i want to expand a region and find out what are all label values it reached.

Steve replied on : 85 of 119

Azhagumani—Form a binary image whose foreground is the region you are interested in, and then dilate it. Then use the dilated image as a logical mask into the original labeled image; that will pull out all the labels adjacent to the original labeled region.

huda replied on : 86 of 119

Hye.. I’m new to this image processing things.. I was told to make an image segmentation of a picture.. The image consists of 3 shapes that overlapping each other.. It was a grayscale image but with 3 different tones of gray colour.. The task was to convert to binary format first and then segment the shape individually.. If during the segmentation, the shape missing some portion, I have to complete the missing portion.. So, can you tell me how to do it in Matlab? And I wonder do I have to set a threshold value?

Steve replied on : 87 of 119

Huda—It’s difficult in general to reconstruct overlapping shapes in a segmentation problem. It’s easier for some specific shapes, though. For example, if your shapes are circles, you could use a circular Hough transform. Hough-like algorithms can often detect a partially occluded shape. The Image Processing Toolbox does not have a circular Hough algorithm, but there’s at least one available on the MATLAB Central File Exchange.

huda replied on : 88 of 119

The shapes are circle, square and rectangle.. Can i used the Hough transform for these shapes? And for the segmentation? How to do it?

Steve replied on : 89 of 119

Huda—The problem sounds kind of contrived. I don’t have any particular suggestions.

Chris Soton replied on : 90 of 119

Hi Steve,

I am MATLAB beginner and I am using a similar image to this example. I have converted my nuclei image to binary and used bwlabel to count the nuclei. However I would like to measure the distance between cell nuclei from a central point on each nuclei (centroid), but I am struggling with the code. Is there any coding help or pointers you could give me with this please?

Thanks,

Chris

Steve replied on : 91 of 119

Chris—You can pass the output of bwlabel to regionprops to get the centroid of each labeled object.

ragu replied on : 92 of 119

hi, steve
i want to develop a general algorithm for image resulting from watershed to display how many pixel occupied in each segment and average intensity of each segment .so that i need variables to sum up the the pixel and average intensity. for limited segment i can develop with specific variables but for generalize i need to get variables which should generate depends on how many segments from watershed output.

Chris Soton replied on : 93 of 119

Hi Steve,

Thanks for the reply, I am sorry but I think I was not clear. I have obtained the centroids of my cells and am struggling to get the distance between the centriods. I would like the code to automatically get the distance between all centriods.

I have tried imdistline but this seems to be a manual code and my attempts to pass the centriod data to this code have failed. Please could you give me any sample code or pointers?

Thanks,

Chris

Steve replied on : 94 of 119

Ragu—You can pass the output from watershed to the function regionprops in order to calculate the desired measurements for each labeled region.

Steve replied on : 95 of 119

Chris—Try something like this:

dist = hypot(centroid1 - centroid2);

Stefan replied on : 96 of 119

Hello Steve!

First of all I’d like to say that I very appreciate your Help here in this Blog. You’re amazing!
I have a similar problem like those discussed here. I have an Image where you can find lots of small cruxes (+), which are all different orientated. I would like to get a contour on every single crux, like you did it on the cells on top of this site. My problem is that i cant isolate the cruxes so i always get connected regions.

The image is here: stud.unileoben.ac.at/~m0335168/RC7.jpg

Thanks a lot!

Steve replied on : 97 of 119

Stefan—I’m sorry, but I don’t have specific suggestions for you.

Chris Soton replied on : 98 of 119

Thanks Steve! Your code combined with another of your examples has helped me greatly.

Ragu replied on : 99 of 119

hi steve
right now i dont have imoverlay function in 6.5.1 version

can you give any function to do the same

or how to develop algorithm

Steve replied on : 100 of 119

kavitha replied on : 101 of 119

Hi Steve.
Its a great page. I got answres for a lot of questions. and in my project i have two images for a same object. I ve segmented those two objects. but i need to calculate the area difference between those objects. how can i do this. pls help me.

Steve replied on : 102 of 119

Kavitha—Use bwlabel and regionprops.

kavitha replied on : 103 of 119

k. steve…. thank u for ur reply. i refered the page which u wrote abt bwlabel.Ur work and discussion are very useful to students like us.. Those examples helped me a lot…

Ragu replied on : 104 of 119

hi steve thanks for the reply

now i have two questions .

1.how many ways are there to make markers.

2.i need to make marker on my gray image before apply to watershed in order to segment the defects(black in color).

where should i make marker either on black parts or gray parts.

Steve replied on : 105 of 119

Ragu—Take a look at the marker-controlled watershed segmentation demo in the Image Processing Toolbox (you can also find it on our web site). You might also be interested in my watershed article in News and Notes from a few years ago.

ragu replied on : 106 of 119

in water marking
fore ground marking is based on the imregionalmin or imregionalmax followed by imclose ,imfill, bwareaopen.

in back ground marking i have little confussion of super imposing fore ground and back ground marker images on gradiant magnitude image rather than original gray image.

Steve replied on : 107 of 119

Ragu—I’m not sure exactly what your question is. Whether to use the original image, the gradient magnitude image, or some other derived image depends on your data set and your chosen method for extracting markers.

ragu replied on : 108 of 119

hi steve thanks for the question,
i have made foreground markers and background markers as per my region of interest.

now should i impose these on original image or gradiant magnitude image.

in example of matlab watermarking it mentioned magnitude gradient.
why?

i belive that imposing markers on original gray image is logically good. if it is not .why?

ragu replied on : 109 of 119

im sorry i forgot to mention that ,my ultimate aim to get segmentation of defects on gray image which is in dark portion ,by applying watershed transform.

for this i need markers to reduce oversegmentation, .

Steve replied on : 110 of 119

Ragu—In my comment #105, I gave a link to my News & Notes article on the watershed transform, which explains the basic concepts of using the transform for segmentation. I encourage you to take a look at it. Here’s an important sentence: “The key behind using the watershed transform for segmentation is this: Change your image into another image whose catchment basins are the objects you want to identify.” Often, the original grayscale image does not have the property that its catchment basins correspond to the objects you are trying to segment. Markers, minima imposition, computing gradient magnitudes, etc., are all techniques for making that happen.

Pratyusha replied on : 111 of 119

HI Steve,
I have a similar problem with segmentation of tumors in an Ultrasound image.
I use a .jpg image file and edge detection techniques, but most of the times it does not work, as the background of an US image is not that clear, i.e., there is a very smooth gradient. I could send/upload some of the files (Images) so you can have a look at them.
Please suggest a suitable edge detection technique for these kind of images.

Thank you!

Steve replied on : 112 of 119

Pratyusha—I don’t have any suggestions for you.

Pratyusha replied on : 113 of 119

Thank you for letting me know. I will try to use a kalman filter for it, or color code the image to highlight the tumor areas.

Francesco replied on : 114 of 119

Goodmorning,
in my engineering degree study i have to make a microstructural characterization of open cell aluminium foams.
After i have done a watershed transform on my 2D slice i have to do a masking of the complement of the binary starting image and the watershed one to reduce the oversegmentation but i don’t know how to make this.

Steve replied on : 115 of 119

Francesco—This blog post has some example code. You might also want to look at the watershed segmentation demo in the Image Processing Toolbox.

Francesco replied on : 116 of 119

thresh=40;
BW=(I>40);
D=bwdist(I1,’euclidean’);
D1=imcomplement(D1);
H=imhmin(D1,13,6);
L=watershed(H);
d=imcomplement(BW)&L;
Now i have reconstructed my open cell foam but if i want to make a geometrical analysis it seems to be that the image has only one region instead of hundreds,because matlab doesn’t recognize that there are many regions.
How i can make to solve this problem??

Steve replied on : 117 of 119

Francesco—It’s your algorithm that doesn’t recognize that there are many regions. It’s hard to say for sure without looking at your image, but I’m not sure I understand why all your steps make sense. You might want to take a look at my News & Notes article about watershed-based segmentation.

deXter Holland replied on : 118 of 119

http://img3.imageshack.us/my.php?image=sample01.jpg

hi all! i need to do cell detection on the image above (identify the presence of those ‘big’ cells). my main problem is the background (noise), i couldn’t remove them properly, so i couldn’t do the usual segmentation easily. for now i’m processing it on grayscale, based on the materials i’m reading.
any suggestion?

Steve replied on : 119 of 119

Dexter—You might try using bwareaopen to remove the small “noise” objects detected.

These postings are the author's and don't necessarily represent the opinions of MathWorks.


text:
# How to add color to Maths formulas which contain power exponent or fraction?

Hi!
I'd like to add highlight to my Maths formulas.
I've found that "\colorbox{aqua}{...}" works great but it doesn't like when there are exponent or fraction:

This formula with exponent doesn't work:
$\colorbox{aqua}{E = mc^2}$

Error:

KaTeX parse error: Expected 'EOF', got '^' at position 23: …ox{aqua}{E = mc^̲2}

And this formula with fraction inside neither:
$\colorbox{aqua}{U = -\frac{G m_1 m_2}{r} }$
gives the error:

KaTeX parse error: Can't use function '\frac' in text mode at position 17: …colorbox{aqua}{\̲f̲r̲a̲c̲{

How can I find a way to highlight my Maths formulas when they contain power exponent or fraction?
LaTeX expects the second argument to be text. If you write math there, you need the $…$ delimiters.


text:
In mathematics, a Diophantine equation is a polynomial equation, usually in two or more unknowns, such that only the integer solutions are sought or studied.

Diophantine problems have fewer equations than unknown variables and involve finding integers that work correctly for all equations. In more technical language, they define an algebraic curve, algebraic surface, or more general object, and ask about the lattice points on it.

The word Diophantine refers to the Hellenistic mathematician of the 3rd century, Diophantus of Alexandria, who made a study of such equations and was one of the first mathematicians to introduce symbolism into algebra. The mathematical study of Diophantine problems that Diophantus initiated is now called Diophantine analysis.

Finding the solution or solutions to a Diophantine equation is closely tied to modular arithmetic and number theory. Often, when a Diophantine equation has infinitely many solutions, the parametric form is used to express the relation between the variables of the equation.

## Hardy-Ramanujan Number

One of the examples of a 3rd degree diophantine equation.

Its popularly known as the Hardy-Ramanujan number. Well, this number has an interesting story behind it.

Godfrey Hardy was a professor of mathematics at Cambridge University. One day he went to visit a friend, the brilliant young Indian mathematician Srinivasa Ramanujan, who was ill. Both men were mathematicians and liked to think about numbers.

When Ramanujan heard that Hardy had come in a taxi he asked him what the number of the taxi was. Hardy said that it was just a boring number – 1729. Ramanujan replied that 1729 was not a boring number at all,rather it was a very interesting one. He explained that it was the smallest number that could be expressed by the sum of two cubes in two different ways.

Linear Combination

A Diophantine equation in the form $ax+by=c$ is known as a linear combination. If two relatively prime integers $a$ and $b$ are written in this form with $c=1$, the equation will have an infinite number of solutions. More generally, there will always be an infinite number of solutions when $\gcd(a,b)=1$.

If GCD(a,b) not equal to c, then there are no solutions to the equation. To see why consider the equation –$3x-9y=3(x-3y)=17$$3$ is a divisor of the LHS (also notice that  $x-3y$ must always be an integer). However, $17$ will never be a multiple of $3$, hence, no solutions exist.

Now consider the case where $c=0$. Thus, $ax=-by$. If $a$ and $b$ are relatively prime, then all solutions are obviously in the form $(bk,-ak)$ for all integers $k$. If they are not, we simply divide them by their greatest common divisor.

## Pythagorean Triples

A Pythagorean triple is a set of three integers that satisfy the Pythagorean Theorem, $a^2+b^2=c^2$. There are three main methods of finding Pythagorean triples:

### Method of Pythagoras

If $m>1$ is an odd number, then  $m, \frac {m^2 -1}{2}, \frac {m^2 + 1}{2}$ is a Pythagorean triple.

### Method of Plato

If $n>1$$2n, n^2 - 1, n^2 + 1$ is a Pythagorean triple.

### Babylonian Method

For any $m,n$$m^2 - n^2, 2mn, m^2 + n^2$ is a Pythagorean triple.

## Sum of Fourth Powers

An equation of form,  $x^4+y^4=z^2$ has no integer solutions, as follows:

We assume that the equation does have integer solutions, and consider the solution which minimizes $z$ . Let this solution be $(x_0,y_0,z_0)$. If $\gcd (x_0,y_0)\neq 1$ then their GCD $d$ must satsify $d^2|z$. The solution $\left(\frac{x_0}{d},\frac{y_0}{d},\frac{z_0}{d}\right)$ would then be a solution less than $z_0$, which contradicts our assumption. Thus, this equation has no integer solutions.

If $\gcd(x_0,y_0)=1$, we then proceed with casework, in $\mod 4$.

Note that every square, and therefore every fourth power, is either $1$ or $0\mod 4$. The proof of this is fairly simple, and you can show it yourself.

Case 1: $x_0^4 \equiv y_0^4 \equiv 1\mod 4$

This would imply $z^2 \equiv 2\mod 4$, a contradiction.

Case 2: $x_0^4 \equiv y_0^4 \equiv 0\mod 4$

This would imply $z^2 \equiv 0\mod 4$, a contradiction since we assumed $\gcd(x_0,y_0)=1$.

Case 3: $x_0^4 \equiv 1\mod 4, y_0^4 \equiv 0\mod 4$, and $z_0^2 \equiv 1\mod 4$

We also know that squares are either $-1, 0$ or $1 \mod 5$. Thus, all fourth powers are either $0$ or $1 \mod 5$.

By similar approach, we show that:

$x_0^4 \equiv 0\mod 5, y_0^4 \equiv 1\mod 5$, So $z_0^2 \equiv 1\mod 5$.

This is a contradiction, as $z_0^2 \equiv 1\mod 4$ implies $z_0$ is odd, and $z_0^2 \equiv 1\mod 5$ implies $z_0$ is even. QED

## Pell Equations

A Pell equation is a type of Diophantine equation in the form $x^2-Dy^2=\pm1$ for natural number $D$. The solutions to the Pell equation when $D$is not a perfect square are connected to the continued fraction expansion of $\sqrt D$. If $a$ is the period of the continued fraction and $C_k=P_k/Q_k$is the $k$ th convergent, all solutions to the Pell equation are in the form $(P_{ia},Q_{ia})$ for positive integer $i$.

## Methods of Solving

### Coordinate Plane

Note that any linear combination can be transformed into the linear equation $y=\frac{-b}{a}x+\frac{c}{a}$, which is just the slope-intercept equation for a line. The solutions to the diophantine equation correspond to lattice points that lie on the line. For example, consider the equation $-3x+4y=4$ or $y=\frac{3}{4}x+1$. One solution is (0,1). If you graph the line, it’s easy to see that the line intersects a lattice point as x and y increase or decrease by the same multiple of $4$ and $3$, respectively (wording?). Hence, the solutions to the equation may be written parametrically $x=4t, y=3t+1$ (if we think of $(0,1)$ as a “starting point”).

### Modular Arithmetic

Sometimes, modular arithmetic can be used to prove that no solutions to a given Diophantine equation exist. Specifically, if we show that the equation in question is never true mod $m$, for some integer $m$, then we have shown that the equation is false. However, this technique cannot be used to show that solutions to a Diophantine equation do exist.

### Induction

Sometimes, when a few solutions have been found, induction can be used to find a family of solutions. Techniques such as infinite Descent can also show that no solutions to a particular equation exist, or that no solutions outside of a particular family exist.

## General Solutions

It is natural to ask whether there is a general solution for Diophantine equations, i.e., an algorithm that will find the solutions for any given Diophantine equations. This is known as Hilbert’s tenth problem. The answer, however, is no.

## Fermat’s Last Theorem

$x^n+y^n=z^n$ is known as Fermat’s Last Theorem for the condition $n\geq3$. In the 1600s, Fermat, as he was working through a book on Diophantine Equations, wrote a comment in the margins to the effect of “I have a truly marvelous proof of this proposition which this margin is too narrow to contain.” Fermat actually made many conjectures and proposed plenty of “theorems,” but wasn’t one to write down the proofs or much other than scribbled comments. After he died, all his conjectures were re-proven (either false or true) except for Fermat’s Last Theorem. After over 350 years of failing to be proven, the theorem was finally proven by Andrew Wiles after he spent over 7 years working on the 200-page proof, and another year fixing an error in the original proof.

Besides this, you can view our video and blog collections in the Video Section & Blog Section of the website.


text:
# What is the percentage "ammonium" content in "ammonium dichromate"?

Jan 27, 2017

In ${\left(N {H}_{4}\right)}_{2} C {r}_{2} {O}_{7}$............?

#### Explanation:

The molecular mass of $\text{ammonium dichromate}$ is $252.07 \cdot g \cdot m o {l}^{-} 1$. We nominate the $N {H}_{4}^{+}$ unit as the cation. This has a mass of $18.04 \cdot g \cdot m o {l}^{-} 1$.

And thus percentage of cation by mass is:

(2xx18.04*g*mol^-1)/(252.07*g*mol^-1)xx100%~=14%.

What is the percentage by mass of metal?


text:
# [NTG-context] Margins around figures

Kumar Appaiah akumar at ee.iitm.ac.in
Fri Sep 7 19:46:50 CEST 2007

```On Fri, Sep 07, 2007 at 12:31:26PM +0200, Wolfgang Schuster wrote:
> He can set the offset in ConTeXt and not in MetaPost, e.g.
>
> \offset[leftoffset=...]{\externalfigure[fugurename]}
>
> The \offset command is described in the details manual, he can also
> look at the definition in core-box.

Well, that seems to be a nice option, but it fills the offset region
in the background colour instead of white. I'll try to figure out if
there's some workaround for that. Also, I'll try Oliver's solution,
though it seems to generate full pages and not individual images.

But, is there still an elegant way to have more margin in MetaPost
images?

Thanks!

Kumar
--
Kumar Appaiah,
458, Jamuna Hostel,


text:
# Connection of “modern” runtimes and number of steps on a Turing machine

Why an evaluation of Turing machine efficiency is equal to the algorithm which is implemented by this machine and vise versa? For example, we can say that efficiency of merge sorting algorithm is O(nlog(n)) and therefore Turing machine which implements this algorithm will do O(nlog(n)) steps. Why is this conclusion correct? And vise versa: if there's no deterministic Turing machine which takes less than exponential number of steps to solve some problem, we aren't able to find efficient algorithm solving this problem.

However, not all problems can be solved exactly as fast on a turing machine. For example, consider the problem PALINDROME to determine if the input is a palindrome. On a modern computer there is an obvious $O(n)$ algorithm, where you just read the front and back character, and if they match recurse on the inner substring. This problem is known to require $\Omega(n^2)$ time for a classical turing machine to solve. This means turing machines are decidedly "weaker" by some amount compared to an ideal modern computer.
• As I said, a turing machine can simulate a modern machine with polynomial slowdown -- thus, if a modern machine can solve a problem in, say, $O(n^3)$, then a turing machine can solve it within $O(n^5)$ or so. In other words, multiplying two polynomials always gives you a polynomial, so in terms of polytime vs not-polytime we may as well treat TMs and modern machines as the same thing even though they are not exactly the same. – Kurt Mueller Aug 11 '14 at 13:45


text:
# Chapter 10 - Exponents and Radicals - 10.4 Dividing Radical Expressions - 10.4 Exercise Set - Page 654: 61

$\dfrac{5}{\sqrt{55}}$

#### Work Step by Step

Rationalizing the numerator, the given expression, $\sqrt{\dfrac{5}{11}} ,$ is equivalent to \begin{array}{l} \sqrt{\dfrac{5}{11}\cdot\dfrac{5}{5}} \\\\= \sqrt{\dfrac{(5)^2}{55}} \\\\= \dfrac{\sqrt{(5)^2}}{\sqrt{55}} \\\\= \dfrac{5}{\sqrt{55}} .\end{array}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
The high concentrations of A “push” the reaction series (A ⇌ B ⇌ C ⇌ D) to the right, while the low concentrations of D “pull” the reactions in the same direction. Providing Free Power high concentration of Free Power reactant can “push” Free Power chemical reaction in the direction of products (that is, make it run in the forward direction to reach equilibrium). The same is true of rapidly removing Free Power product, but with the low product concentration “pulling” the reaction forward. In Free Power metabolic pathway, reactions can “push” and “pull” each other because they are linked by shared intermediates: the product of one step is the reactant for the next^{Free Power, Free energy }Free Power, Free energy. “Think of Two Powerful Magnets. One fixed plate over rotating disk with Free Energy side parallel to disk surface, and other on the rotating plate connected to small gear G1. If the magnet over gear G1’s north side is parallel to that of which is over Rotating disk then they both will repel each other. Now the magnet over the left disk will try to rotate the disk below in (think) clock-wise direction. Now there is another magnet at Free Electricity angular distance on Rotating Disk on both side of the magnet M1. Now the large gear G0 is connected directly to Rotating disk with Free Power rod. So after repulsion if Rotating-Disk rotates it will rotate the gear G0 which is connected to gear G1. So the magnet over G1 rotate in the direction perpendicular to that of fixed-disk surface. Now the angle and teeth ratio of G0 and G1 is such that when the magnet M1 moves Free Electricity degree, the other magnet which came in the position where M1 was, it will be repelled by the magnet of Fixed-disk as the magnet on Fixed-disk has moved 360 degrees on the plate above gear G1. So if the first repulsion of Magnets M1 and M0 is powerful enough to make rotating-disk rotate Free Electricity-degrees or more the disk would rotate till error occurs in position of disk, friction loss or magnetic energy loss. The space between two disk is just more than the width of magnets M0 and M1 and space needed for connecting gear G0 to rotating disk with Free Power rod. Now I’ve not tested with actual objects. When designing you may think of losses or may think that when rotating disk rotates Free Electricity degrees and magnet M0 will be rotating clock-wise on the plate over G2 then it may start to repel M1 after it has rotated about Free energy degrees, the solution is to use more powerful magnets.
Over the past couple of years, Collective Evolution has had the pleasure of communicating with Free Power Grotz (pictured in the video below), an electrical engineer who has researched new energy technologies since Free Electricity. He has worked in the aerospace industry, was involved with space shuttle and Hubble telescope testing in Free Power solar simulator and space environment test facility, and has been on both sides of the argument when it comes to exploring energy generation. He has been involved in exploring oil and gas and geothermal resources, as well as coal, natural gas, and nuclear power-plants. He is very passionate about new energy generation, and recognizes that the time to make the transition is now.
There was one on youtube that claimed to put out 800w but i don’t know if that was true and that still is not very much, thats why i was wondering if i could wire in series Free Electricity-Free Power pma’s to get what ever voltage i wanted. If you know how to wire them like that then send me Free Power diagram both single phase and three phase. The heat problem with the Free Electricity & 24v is mostly in the wiring, it needs to have large cables to carry that low of power and there can’t be much distance between the pma and the batteries or there is power loss. Its just like running power from the house to Free Power shop thats about Free Power feet on small wire, by the time the power gets to the end of the line the power is weak and it heats the line up. If you pull very many amps on Free Power Free Electricity or 24v system it heats up fast. Also, i don’t know the metric system. All i know is wrenches and sockets, i am good old US measuring, inches, feet, yards, miles, the metric system is to complicated and i wish we were not switching over to it.

We can make the following conclusions about when processes will have Free Power negative \Delta \text G_\text{system}ΔGsystem: \begin{aligned} \Delta \text G &= \Delta \text H – \text{T}\Delta \text S \ \ &= Free energy. 01 \dfrac{\text{kJ}}{\text{mol-rxn}}-(Free energy \, \cancel{\text K})(0. 022\, \dfrac{\text{kJ}}{\text{mol-rxn}\cdot \cancel{\text K})} \ \ &= Free energy. 01\, \dfrac{\text{kJ}}{\text{mol-rxn}}-Free energy. Free Power\, \dfrac{\text{kJ}}{\text{mol-rxn}}\ \ &= -0. Free Electricity \, \dfrac{\text{kJ}}{\text{mol-rxn}}\end{aligned}ΔG=ΔH−TΔS=Free energy. 01mol-rxnkJ−(293K)(0. 022mol-rxn⋅K)kJ=Free energy. 01mol-rxnkJ−Free energy. 45mol-rxnkJ=−0. 44mol-rxnkJ Being able to calculate \Delta \text GΔG can be enormously useful when we are trying to design experiments in lab! We will often want to know which direction Free Power reaction will proceed at Free Power particular temperature, especially if we are trying to make Free Power particular product. Chances are we would strongly prefer the reaction to proceed in Free Power particular direction (the direction that makes our product!), but it’s hard to argue with Free Power positive \Delta \text GΔG! Our bodies are constantly active. Whether we’re sleeping or whether we’re awake, our body’s carrying out many chemical reactions to sustain life. Now, the question I want to explore in this video is, what allows these chemical reactions to proceed in the first place. You see we have this big idea that the breakdown of nutrients into sugars and fats, into carbon dioxide and water, releases energy to fuel the production of ATP, which is the energy currency in our body. Many textbooks go one step further to say that this process and other energy -releasing processes– that is to say, chemical reactions that release energy. Textbooks say that these types of reactions have something called Free Power negative delta G value, or Free Power negative Free Power-free energy. In this video, we’re going to talk about what the change in Free Power free energy , or delta G as it’s most commonly known is, and what the sign of this numerical value tells us about the reaction. Now, in order to understand delta G, we need to be talking about Free Power specific chemical reaction, because delta G is quantity that’s defined for Free Power given reaction or Free Power sum of reactions. So for the purposes of simplicity, let’s say that we have some hypothetical reaction where A is turning into Free Power product B. Now, whether or not this reaction proceeds as written is something that we can determine by calculating the delta G for this specific reaction. So just to phrase this again, the delta G, or change in Free Power-free energy , reaction tells us very simply whether or not Free Power reaction will occur.
The song’s original score designates the duet partners as “wolf” and “mouse, ” and genders are unspecified. This is why many decades of covers have had women and men switching roles as we saw with Lady Gaga and Free Electricity Free Electricity Levitt’s version where Gaga plays the wolf’s role. Free Energy, even Miss Piggy of the Muppets played the wolf as she pursued ballet dancer Free Energy NureyeFree Power
But I will send you the plan for it whenever you are ready. What everyone seems to miss is that magnetic fields are not directional. Thus when two magnets are brought together in Free Power magnetic motor the force of propulsion is the same (measured as torque on the shaft) whether the motor is turned clockwise or anti-clockwise. Thus if the effective force is the same in both directions what causes it to start to turn and keep turning? (Hint – nothing!) Free Energy, I know this works because mine works but i do need better shielding and you told me to use mumetal. What is this and where do you get it from? Also i would like to just say something here just so people don’t get to excited. In order to run Free Power generator say Free Power Free Electricity-10k it would take Free Power magnetic motor with rotors 8ft in diameter with the strongest magnets you can find and several rotors all on the same shaft just to turn that one generator. Thats alot of money in magnets. One example of the power it takes is this.
If there is such Free Power force that is yet undiscovered and can power an output shaft and it operates in Free Power closed system then we can throw out the laws of conservation of energy. I won’t hold my breath. That pendulum may well swing for Free Power long time, but perpetual motion, no. The movement of the earth causes it to swing. Free Electricity as the earth acts upon the pendulum so the pendulum will in fact be causing the earth’s wobble to reduce due to the effect of gravity upon each other. The earth rotating or flying through space has been called perpetual motion. Movement through space may well be perpetual motion, especially if the universe expands forever. But no laws are being bent or broken. Context is what it is all about. Mr. Free Electricity, again I think the problem you are having is semantics. “Perpetual- continuing or enduring forever; everlasting. ” The modern terms being used now are “self-sustaining or sustainable. ” Even if Mr. Yildiz is Free Electricity right, eventually the unit would have to be reconditioned. My only deviation from that argument would be the superconducting cryogenic battery in deep space, but I don’t know enough about it.
It all smells of scam. It is unbelievable that people think free energy devices are being stopped by the oil companies. Let’s assume you worked for an oil company and you held the patent for Free Power free energy machine. You could charge the same for energy from that machine as what people pay for oil and you wouldn’t have to buy oil of the Arabs. Thus your profit margin would go through the roof. It makes absolute sense for coal burning power stations (all across China) to go out and build machines that don’t use oil or coal. wow if Free Energy E. , Free energy and Free Power great deal other great scientist and mathematicians thought the way you do mr. Free Electricity the world would still be in the stone age. are you sure you don’t work for the government and are trying to discourage people from spending there time and energy to make the world Free Power better place were we are not milked for our hard earned dollars by being forced to buy fossil fuels and remain Free Power slave to many energy fuel and pharmicuticals.
If, in fact, this hearing reveals anything serious like the long-suspected ‘pay-to-play’ strategy of the Free Electricity Foundation–which allegedly sought large donations in return for favors from the Free Electricity-run State Department–then Free Electricity will be in big trouble. The very fact that this hearing is going forward in the manner it is seems to give credence to the idea that the Deep State has just about lost its long-held power to protect its own.
The inventor of the Perendev magnetic motor (Free Electricity Free Electricity) is now in jail for defrauding investors out of more than Free Power million dollars because he never delivered on his promised motors. Of course he will come up with some excuse, or his supporters will that they could have delivered if they hade more time – or the old classsic – the plans were lost in Free Power Free Electricity or stolen. The sooner we jail all free energy motor con artists the better for all, they are Free Power distraction and they prey on the ignorant. To create Free Power water molecule X energy was released. Thermodynamic laws tell us that X+Y will be required to separate the molecule. Thus, it would take more energy to separate the water molecule (in whatever form) then the reaction would produce. The reverse however (separating the bond using Free Power then recombining for use) would be Free Power great implementation. But that is the bases on the hydrogen fuel cell. Someone already has that one. Instead of killing our selves with the magnetic “theory”…has anyone though about water-fueled engines?.. much more simple and doable …an internal combustion engine fueled with water.. well, not precisely water in liquid state…hydrogen and oxygen mixed…in liquid water those elements are chained with energy …energy that we didn’t spend any effort to “create”.. (nature did the job for us).. and its contained in the molecular union.. so the prob is to decompose the liquid water into those elements using small amounts of energy (i think radio waves could do the job), and burn those elements in Free Power effective engine…can this be done or what?…any guru can help?… Magnets are not the source of the energy.
To understand why this is the case, it’s useful to bring up the concept of chemical equilibrium. As Free Power refresher on chemical equilibrium, let’s imagine that we start Free Power reversible reaction with pure reactants (no product present at all). At first, the forward reaction will proceed rapidly, as there are lots of reactants that can be converted into products. The reverse reaction, in contrast, will not take place at all, as there are no products to turn back into reactants. As product accumulates, however, the reverse reaction will begin to happen more and more often. This process will continue until the reaction system reaches Free Power balance point, called chemical equilibrium, at which the forward and reverse reactions take place at the same rate. At this point, both reactions continue to occur, but the overall concentrations of products and reactants no longer change. Each reaction has its own unique, characteristic ratio of products to reactants at equilibrium. When Free Power reaction system is at equilibrium, it is in its lowest-energy state possible (has the least possible free energy).
Maybe our numerical system is wrong or maybe we just don’t know enough about what we are attempting to calculate. Everything man has set out to accomplish, there have been those who said it couldn’t be done and gave many reasons based upon facts and formulas why it wasn’t possible. Needless to say, none of the ‘nay sayers’ accomplished any of them. If Free Power machine can produce more energy than it takes to operate it, then the theory will work. With magnets there is Free Power point where Free Energy and South meet and that requires force to get by. Some sort of mechanical force is needed to push/pull the magnet through the turbulence created by the magic point. Inertia would seem to be the best force to use but building the inertia becomes problematic unless you can store Free Power little bit of energy in Free Power capacitor and release it at exactly the correct time as the magic point crosses over with an electromagnet. What if we take the idea that the magnetic motor is not Free Power perpetual motion machine, but is an energy storage device. Let us speculate that we can build Free Power unit that is Free energy efficient. Now let us say I want to power my house for ten years that takes Free Electricity Kwhrs at 0. Free Energy /Kwhr. So it takes Free energy Kwhrs to make this machine. If we do this in Free Power place that produces electricity at 0. 03 per Kwhr, we save money.
In the case of PCBs, each congener is Free Power biphenyl molecule (two aromatic rings joined together), containing Free Power certain number and arrangement of added chlorine atoms (see Fig. Free Electricity. Free Electricity). Historically, there were many commercially marketed products (e. g. , Aroclor) containing varying mixtures of PCB congeners.) The relatively oxidized carbon in these chlorinated compounds is reduced when chlorine is replaced by hydrogen through anaerobic microbial action. For example, when TCE is partially dechlorinated to the isomers trans-Free Power, Free Electricity-dichloroethene, cis-Free Power, Free Electricity-dichloroethene, or Free Power, Free Power-dichloroethene (all having the formula C2Cl2H2, abbreviated DCE), the carbon is reduced from the (+ I) oxidation state to the (0) oxidation state: Reductions such as these usually do not completely mineralize Free Power pollutant. Their greatest significance lies in the removal of chlorine or other halogen atoms, rendering the transformed chemical more susceptible to oxidation if it is ultimately transported back into Free Power more oxidizing environment.
Remember, when it comes to getting offended, we all decide what offends us and how to get offended by what someone says. TRUE empowerment means you have the control within yourself. We don’t have to allow things to offend us simply because someone says something, and this also doesn’t mean everyone is going to be mean to us all the time, this is an unsubstantiated fear.
Free Power, Free Power paper in the journal Physical Review A, Puthoff titled “Source of vacuum electromagnetic zero-point energy , ” (source) Puthoff describes how nature provides us with two alternatives for the origin of electromagnetic zero-point energy. One of them is generation by the quantum fluctuation motion of charged particles that constitute matter. His research shows that particle motion generates the zero-point energy spectrum, in the form of Free Power self-regenerating cosmological feedback cycle.
Remember, when it comes to getting offended, we all decide what offends us and how to get offended by what someone says. TRUE empowerment means you have the control within yourself. We don’t have to allow things to offend us simply because someone says something, and this also doesn’t mean everyone is going to be mean to us all the time, this is an unsubstantiated fear.
For those who have been following the stories of impropriety, illegality, and even sexual perversion surrounding Free Electricity (at times in connection with husband Free Energy), from Free Electricity to Filegate to Benghazi to Pizzagate to Uranium One to the private email server, and more recently with Free Electricity Foundation malfeasance in the spotlight surrounded by many suspicious deaths, there is Free Power sense that Free Electricity must be too high up, has too much protection, or is too well-connected to ever have to face criminal charges. Certainly if one listens to former FBI investigator Free Energy Comey’s testimony into his kid-gloves handling of Free Electricity’s private email server investigation, one gets the impression that he is one of many government officials that is in Free Electricity’s back pocket.
Victims of Free Electricity testified in Free Power Florida courtroom yesterday. Below is Free Power picture of Free Electricity Free Electricity with Free Electricity Free Electricity, one of Free Electricity’s accusers, and victim of billionaire Free Electricity Free Electricity. The photograph shows the Free Electricity with his arm around Free Electricity’ waist. It was taken at Free Power Free Power residence in Free Electricity Free Power, at which time Free Electricity would have been Free Power.

#### If power flows from the output shaft where does it flow in? Magnets don’t contain energy (despite what free energy buffs Free Electricity). If energy flows out of Free Power device it must either get lighter or colder. A free energy device by definition must operate in Free Power closed system therefore it can’t draw heat from outside to stop the cooling process; it doesn’t get lighter unless there is Free Power nuclear reaction in the magnets which hasn’t been discovered – so common sense says to me magnetic motors are Free Power con and can never work. Science is not wrong. It is not Free Power single entity. Free Electricity or findings can be wrong. Errors or corrections occur at the individual level. Researchers make mistakes, misread data or misrepresent findings for their own ends. Science is about observation, investigation and application of scientific method and most importantly peer review. Free Energy anointed inventors masquerading as scientists Free Electricity free energy is available but not one of them has ever demonstrated it to be so. Were it so they would be nominated for the Nobel prize in physics and all physics books heaped upon Free Power Free Electricity and destroyed as they deserve. But this isn’t going to happen. Always try to remember.

This expression has commonly been interpreted to mean that work is extracted from the internal energy U while TS represents energy not available to perform work. However, this is incorrect. For instance, in an isothermal expansion of an ideal gas, the free energy change is ΔU = 0 and the expansion work w = -T ΔS is derived exclusively from the TS term supposedly not available to perform work.


text:
# Dealing with errors¶

Attention

Finnish university students are encouraged to use the CSC Notebooks platform.

Others can follow the lesson and fill in their student notebooks using Binder.

## Interpreting error messages¶

So far in the course we have encountered a number of different types of error messages in Python, but have not really discussed how to understand what the computer is trying to tell you when you get an error message. We’ll do that below. For most Python errors you will see and exception raised when the error is encountered, providing some insight into what went wrong and where to look to fix it.

Let’s imagine you’ve written the code below called to convert wind speeds from km/hr to m/s and you’re dying to figure out how windy it is in Halifax, Nova Scotia, Canada where they report wind speeds in km/hr.

Unfortunately, when you run your script you observe the following:

wind_speed_km = 50
wind_speed_ms = wind_speed_km * 1000 / 3600

print(f"A wind speed of {wind_speed_km} km/hr is {wind_speed_ms} m/s.)

  File "/tmp/ipykernel_454/2231829726.py", line 4
print(f"A wind speed of {wind_speed_km} km/hr is {wind_speed_ms} m/s.)
^
SyntaxError: EOL while scanning string literal

Let’s break this example down and see what the error message says.

A SyntaxError, annotated

As you can see, there is quite a bit of useful information here. We have some information about script that was run to execute this code cell and which line of the cell was a problem. We also have the type of error, a SyntaxError in this case, as well as where it occurred on the line, and a bit more information about its meaning. The location on the line won’t always be correct, but Python makes its best guess for where you should look to solve the problem. Clearly, this is handy information.

Let’s consider another example, where you have fixed the SyntaxError above and now have made a function for calculating a wind speeds in m/s.

When you run this script you encounter a new and bigger error message:

def convert_wind_speed(speed):
return speed * 1000 / 3600

wind_speed_km = '30'
wind_speed_ms = convert_wind_speed(wind_speed_km)

print(f"A wind speed of {wind_speed_km} km/hr is {wind_speed_ms} m/s.")

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
/tmp/ipykernel_454/2952723812.py in <module>
3
4 wind_speed_km = '30'
----> 5 wind_speed_ms = convert_wind_speed(wind_speed_km)
6
7 print(f"A wind speed of {wind_speed_km} km/hr is {wind_speed_ms} m/s.")

/tmp/ipykernel_454/2952723812.py in convert_wind_speed(speed)
1 def convert_wind_speed(speed):
----> 2     return speed * 1000 / 3600
3
4 wind_speed_km = '30'
5 wind_speed_ms = convert_wind_speed(wind_speed_km)

TypeError: unsupported operand type(s) for /: 'str' and 'int'

In this case we see a TypeError that is part of a traceback, where the problem in the code arises from something other than on the line where the code was run. In this case, we have a TypeError where we try to divide a character string by a number, something Python cannot do. Hence, the TypeError indicating the data types are not compatible. That error, however, does not occur when the code is run until the point where the function is used. Thus, we see the traceback showing that not only does the error occur when the function is called, but also that the problem is in the function definition.

The traceback above may look a bit scarier, but if you take your time and read through what is output, you will again find that the information is helpful in finding the problem in your code. After all, the purpose of the error message is to help the user find a problem :).

### Common errors and exceptions¶

Now that we have some idea of how to read an error message, let’s have a look at a few different types of common Python exceptions that are displayed for different program errors.

#### IndexErrors¶

An IndexError occurs when you attempt to reference a value with an index outside the range of values. We can easily produce an IndexError by trying to access the value at index 5 in the following list of cities: cities = ['Paris', 'Berlin', 'London'].

cities = ["Paris", "Berlin", "London"]

cities[5]

---------------------------------------------------------------------------
IndexError                                Traceback (most recent call last)
/tmp/ipykernel_454/3963876545.py in <module>
----> 1 cities[5]

IndexError: list index out of range

Here we get the rather clear error message that the index used for the list cities is out of the range of index values for that list.

#### NameErrors¶

A NameError occurs when you reference a variable that has not been defined. We can produce a NameError by trying station_id = stations[1].

station_id = stations[1]

---------------------------------------------------------------------------
NameError                                 Traceback (most recent call last)
/tmp/ipykernel_454/2799791259.py in <module>
----> 1 station_id = stations[1]

NameError: name 'stations' is not defined

In this instance we receive a NameError because the list stations has not been defined, and we’re thus not able to access a value in that list.

#### IndentationErrors¶

An IndentationError is raised whenever a code block is expected to be indented and is either not, or is indented inconsistently. Let’s consider two examples below.

for city in cities:
city = city + " is a city in Europe"
print(city)

  File "<tokenize>", line 3
print(city)
^
IndentationError: unindent does not match any outer indentation level

for city in cities:
city = city + " is a city in Europe"
print(city)

  File "/tmp/ipykernel_454/1297196788.py", line 2
city = city + " is a city in Europe"
^
IndentationError: expected an indented block

In both of the examples above, an IndentationError is raised. In the first case, the indentation level is inconsistent. In case two, indentation is expected for the code below a for statement.

#### TypeErrors¶

A TypeError is raised whenever two incompatible data types are used together. For example, if we try to divide a character string by a number or add a number to a boolean variable, a TypeError will be raised.

cities[0] / 5

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
/tmp/ipykernel_454/2693491430.py in <module>
----> 1 cities[0] / 5

TypeError: unsupported operand type(s) for /: 'str' and 'int'

In this case, the TypeError is because it is not possible to divide a character string by a number.

#### Other kinds of errors¶

There are certainly other kinds of errors and exceptions in Python, but this list comprises those you’re most likely to encounter. As you can see, knowing the name of each error can be helpful in trying to figure out what has gone wrong, and knowing what these common error types mean will save you time trying to fix your programs.


text:
# find the square root of 144 by prime factorization method

c
Comments Off on find the square root of 144 by prime factorization method
November 29th, 2020

Ask your question. Then, 72 ÷ 2 = 36. It can be observed that the prime factors in the prime factorisation of a square number, occur twice the number of times, it occurs in the number itself. 1 2 = 1, the square root of 1 is 1. But, it can only be used if a number is a perfect square. Square Root by Prime Factorization Example Problems. 1. Therefore, if we take the square root on both the sides, we get; √144 = √(2x2x2x2x3x3) What is the prime factorization of 144 [SOLVED] The Prime Factorization of 144 Answer: 2 4 • 3 2 144 is not a prime number. (ii) Inside the square root, for every two same numbers multiplied, one number can be taken out of the square root. Prime Factorization of 144. We'll start by dividing the number, i.e., 144 by the smallest integer, which is 2. 3. 1764. (ii) Make the pair of similar factors such that the both factors in each pair are equal. 441. Is 144 an odd number? Ex 6.3, 4 Find the square roots of the following numbers by the Prime Factorization Method. (i) 729We use prime factorization to find square root.Thus, 729 = 3 × 3 × 3 × 3 × 3 × 3Square root of 729 = 3 × 3 × 3 = 9 × 3 = 27 Ex 6.3, 4 Find the square roots of t Related Links : Is 144 an even number? Step IV: Take one factor from each pair. (vii) 5776 Rough 144 × 4 = 576 145 × 5 = 725 146 × 6 = 876 Therefore, √5776 = 76 Ex 6.4, 1 Find the square root of each of the following numbers by Division method. 49. 72 ÷ 2 = 36. 165. 9 ÷ 3 = 3 (ii) Make the pair of similar factors such that the both factors in … In prime factorisation, we have to find the prime factors in the given number. Prime Factorisation. mmmmcl1975 mmmmcl1975 29.06.2018 Math Secondary School Find the square root of 169 by repeated subtraction method 2 See answers arshbbcommander arshbbcommander 169 - 1 =168 168 - 3 = 165 165-5 = 160 160-7=153 153-9=144 144-11=133 133-13=120 120-15= 105 … Join now. 3. 2. Thew following steps will be useful to find square root of a number by prime factorization. This method is easy to apply, as we have learned about prime factors in our earlier classes. For example 12 and 144, 12 = 2 × 2 × 3. What are the factors of 144? Prime Factorization of 144 by division method is the first way to calculate a number's factors. 144 = 2 × 2 × 3 × 2 × 2 × 3. (viii) 7921 Rough 167 × 7 = 1169 168 × 8 = 1344 169 × 9 = 1521 Therefore, √7921 = 89 Ex 6.4, 1 Find the square root of each of the following numbers by Division method. Step VI: The product obtained in step V is the required square root. Log in. Log in. HOPE , IT HELPS U....!!! Square root through prime factorisation - law To find the square root of the given number through prime factorization method we follows the following steps: (i) First we divide the given number in to its prime factor. 7 $1764 = 2\times 2\times 3\times 3\times 7\times 7$ Step 2: Identical factors are paired. Find the square root of 27225 by prime factorization 2 See answers gaurav2013c gaurav2013c 3 | 27225 3 | 9075 5 | 3025 5 | 605 11 | 121 11 | 11 27225 = 3 × 3 × 5 × 5 × 11 × 11 => sqrt ( 27225) = 3 × 5 × 11 => sqrt (27225) = 165 Brainly User Brainly User HENCE , THE SQUARE ROOT OF THIS NUMBER BY P.F is. 2. Click hereto get an answer to your question  Find the square root of 144 by prime fractions method. 36 ÷ 2 = 18. (iii) Take one factor … Step V: Find the product of factors obtained in step IV. Square root through prime factorisation - law To find the square root of the given number through prime factorization method we follows the following steps: (i) First we divide the given number in to its prime factor. 36 ÷ 2 = 18. Step 1: The first step is to divide the number 144 with the smallest prime factor, say 2. Join now. Step 2: Again divide 72 by 2 and the process goes on. 882. Is 144 a prime number? 144 ÷ 2 = 72. Please write the result of L.C.M (prime factorization) also. 147. For example, let us find the prime … (i) Decompose the number inside the square root into prime factors. 18 ÷ 2 = 9. So continue with the next prime factor, say 3. Prime Factorization Calculator. 18 ÷ 2 = 9. Is 144 a perfect square? Solution: Step 1: The given number is resolved into its prime factors. 7. Find the square root of 1764 using the prime factorization method. Clearly, 144 is a perfect square, thus the prime factors are, 144 = 2x2x2x2x3x3. Find the square root of 169 by repeated subtraction method - 4440731 1. Step 3: You will get a fractional number if you divide 9 by 2. Is 144 a composite number? Let us consider the prime factors of a number and its square. The process will start with, 144 ÷ 2 = 72. 1. New questions in Math. 4 2 = 16, square root of 16 is 4 and so on. Now let us know how to calculate the prime factors of a number.


text:
# nLab homotopy groups of spheres

### Context

#### Stable Homotopy theory

stable homotopy theory

Introduction

# Homotopy groups of spheres

My initial inclination was to call this book The Music of the Spheres, but I was dissuaded from doing so by my diligent publisher, who is ever mindful of the sensibilities of librarians. (D. Ravenel 86, preface)

With all due respect to anyone interested in them, the stable homotopy groups of spheres are a mess. (J. F. Adams 74, p 204)

## Idea

The homotopy groups of spheres $\pi_{n+k}(S^n)$ are the homotopy classes of maps $S^{n+k} \longrightarrow S^n$

$\pi_{n+k}(S^n) \coloneqq [S^{n+k}, S^n] \,.$

For fixed $k$, the colimit over $n$ with respect to the suspension homomorphism

$\pi_{n+k}(S^n) \longrightarrow \pi_{n+k+1}(S^{n+1})$

over all $\pi_{n+k}(S^n)$ (called the $k$-stem) is called the stable homotopy groups of spheres (also: the “stable $k$-stem”)

$\pi_k^S = \coloneqq \underset{\longrightarrow}{\lim}_n \pi_{n+k}(S^n) \,.$

In fact, by the Freudenthal suspension theorem, the value of the $\pi_{n+k}(S^n)$ stabilizes for $n \gt k+1$ (depend only on $k$ in this range), whence the name.

The stable homotopy groups of sphere are equivalently the homotopy groups of a spectrum for the sphere spectrum $\mathbb{S}$

$\pi_k^S = \pi_k(\mathbb{S}) \,.$

The stable homotopy groups of spheres are notorious for their immense computational richness. Many of the tools of algebraic topology and stable homotopy theory were devised to compute more and more of the stable stems. This notably include the Adams spectral sequence, the Adams-Novikov spectral sequence.

## Tables

The first few stable homotopy groups of the sphere spectrum $\mathbb{S}$ are

$k =$0123456789101112131415$\cdots$
$\pi_k(\mathbb{S}) =$$\mathbb{Z}$$\mathbb{Z}/2$$\mathbb{Z}/2$$\mathbb{Z}/{24}$$0$$0$$\mathbb{Z}/2$$\mathbb{Z}/{240}$$(\mathbb{Z}/2)^2$$(\mathbb{Z}/2)^3$$\mathbb{Z}/6$$\mathbb{Z}/{504}$$0$$\mathbb{Z}/3$$(\mathbb{Z}/2)^2$$\mathbb{Z}/{480} \oplus \mathbb{Z}/2$$\cdots$

The following tables show the p-primary decomposition of these and the following stable homotopy groups.

The horizontal index is the degree $n$ of the stable homotopy group $\pi_n$. The appearance of a string of $k$ connected dots vertically above index $n$ means that there is a direct summand primary group of order $p^k$. The bottom rows in each case are given by the image of the J-homomorphism.

See at fundamental theorem of finitely generated abelian grouops – Graphical representation for details on the notation used in these table, and see example 1 below for illustration.

(the following graphics are taken from Hatcher, based on (Ravenel 86)

$p = 2$-primary component

$p = 3$-primary component

$p = 5$-primary component

###### Example

The finite abelian group $\pi_3(\mathbb{S}) \simeq \mathbb{Z}_{24}$ decomposes into primary groups as $\simeq \mathbb{Z}_8 \oplus \mathbb{Z}_3$. Here $8 = 2^3$ corresponds to the three dots above $n = 3$ in the first table, and $3 = 3^1$ to the single dot over $n = 3$ in the second.

The finite abelian group $\pi_7(\mathbb{S}) \simeq \mathbb{Z}_{240}$ decomposes into primary groups as $\simeq \mathbb{Z}_{16} \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_5$. Here $16 = 2^4$ corresponds to the four dots above $n = 7$ in the first table, and $3 = 3^1$ to the single dot over $n = 7$ in the second and $5 = 5^1$ to the single dot over $n = 7$ in the third table.

## Properties

### Serre finiteness theorem

###### Theorem

(Serre finiteness theorem)

The homotopy group $\pi_{n+k}(S^k)$ is a finite group except

1. for $n = 0$ in which case $\pi_k(S^k) = \mathbb{Z}$;

2. $k = 2m$ and $n = 2m -1$ in which case

$\pi_{4m - 1}(S^{2m}) \simeq \mathbb{Z} \oplus F_m$

for $F_m$ a finite group.

(Serre 53)

### Relation to the framed bordism ring

By Thom's theorem, for any (B,f)-structure $\mathcal{B}$, there is an isomorphism (of commutative rings)

$\Omega^{\mathcal{B}}_\bullet \overset{\simeq}{\longrightarrow} \pi_\bullet(M\mathcal{B})$

from the cobordism ring of manifolds with stable normal $\mathcal{B}$-structure to the homotopy groups of the universal $\mathcal{B}$-Thom spectrum.

Now for $\mathcal{B} = Fr$ framing structure, then

$M Fr \simeq \mathbb{S}$

is equivalently the sphere spectrum. Hence in this case Thom's theorem states that there is an isomorphism

$\Omega^{fr}_\bullet \overset{\simeq}{\longrightarrow} \pi_\bullet(\mathbb{S})$

between the framed cobordism ring and the stable homotopy groups of spheres.

For discussion of computation of $\pi_\bullet(\mathbb{S})$ this way, see for instance (Wang-Xu 10, section 2) and (Putnam).

For instance

• $\Omega^{fr}_0 = \mathbb{Z}$ because there are two $k$-framings on a single point, corresponding to $\pi_0(O(k)) \simeq \mathbb{Z}_2$, the negative of a point with one framing is the point with the other framing, and so under disjoint union, the framed points form the group of integers;

• $\Omega^{fr}_1 = \mathbb{Z}_2$ because the only compact connected 1-manifold is the circle, there are two framings on the circle, corresponding to $\pi_1(O(k)) \simeq \mathbb{Z}_2$ and they are their own negatives.

The following characterizes the image of the J-homomorphism

$J \;\colon\; \pi_\bullet(O) \longrightarrow \pi_\bullet(\mathbb{S})$

from the homotopy groups of the stable orthogonal group to the stable homotopy groups of spheres. This was first conjectured in (Adams 66) (since called the Adams conjecture) and then proven in (Quillen 71).

###### Remark

By the discussion at orthogonal group – homotopy groups we have that the homotopy groups of the stable orthogonal group are

$n\;mod\; 8$01234567
$\pi_n(O)$$\mathbb{Z}_2$$\mathbb{Z}_2$0$\mathbb{Z}$000$\mathbb{Z}$

Because all groups appearing here and in the following are cyclic groups, we instead write down the order

$n\;mod\; 8$01234567
${\vert\pi_n(O)\vert}$221$\infty$111$\infty$
###### Theorem

The stable homotopy groups of spheres $\pi_n(\mathbb{S})$ are the direct sum of the (cyclic) image of the J-homomorphism, and the kernel of the Adams e-invariant.

Moreover,

• for $n = 0 \;mod \;$ and $n = 1 \;mod \; 8$ and $n$ positive the J-homomorphism is injective, hence its image is $\mathbb{Z}_2$,

• for $n = 3\; mod\; 8$ and $n = 7 \; mod \; 8$ hence for $n = 4 k -1$, the order of the image is equal to the denominator of $B_{2k}/4k$, where $B_{2k}$ is the Bernoulli number

• for all other cases the image is necessarily zero.

$n$012345678910111213141516
homotopy groups of stable orthogonal group$\pi_n(O)$$\mathbb{Z}_2$$\mathbb{Z}_2$0$\mathbb{Z}$000$\mathbb{Z}$$\mathbb{Z}_2$$\mathbb{Z}_2$0$\mathbb{Z}$000$\mathbb{Z}$$\mathbb{Z}_2$
stable homotopy groups of spheres$\pi_n(\mathbb{S})$$\mathbb{Z}$$\mathbb{Z}_2$$\mathbb{Z}_2$$\mathbb{Z}_{24}$00$\mathbb{Z}_2$$\mathbb{Z}_{240}$$\mathbb{Z}_2 \oplus \mathbb{Z}_2$$\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$$\mathbb{Z}_6$$\mathbb{Z}_{504}$0$\mathbb{Z}_3$$\mathbb{Z}_2 \oplus \mathbb{Z}_2$$\mathbb{Z}_{480} \oplus \mathbb{Z}_2$$\mathbb{Z}_2 \oplus \mathbb{Z}_2$
image of J-homomorphism$im(\pi_n(J))$0$\mathbb{Z}_2$0$\mathbb{Z}_{24}$000$\mathbb{Z}_{240}$$\mathbb{Z}_2$$\mathbb{Z}_2$0$\mathbb{Z}_{504}$000$\mathbb{Z}_{480}$$\mathbb{Z}_2$

## References

### General

Introductions and surveys include

A tabulation of stable homotopy groups of spheres is in

Original articles on basic properties include

• Jean-Pierre Serre Groupes d’homotopie et classes de groupes abelien, Ann. of Math. 58 (1953), 258–294.

### Image of the J-homomorphism

Discussion of the image of the J-homomorphism is due to

• John Adams, On the groups $J(X)$ IV, Topology 5: 21,(1966) Correction, Topology 7 (3): 331 (1968)

• Daniel Quillen, The Adams conjecture, Topology. an International Journal of Mathematics 10: 67–80 (1971)

### Formalization in homotopy type theory

For formalization in homotopy type theory see at

Last revised on June 19, 2018 at 17:29:28. See the history of this page for a list of all contributions to it.


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Homework Help: Infinite set

1. Sep 20, 2011

burak100

infinite set

1. The problem statement, all variables and given/known data
$f: \mathbb{R} \rightarrow \mathbb{Q}$,

show that there is a $q \in \mathbb{Q}$ st. $f^{-1}(q)$ is infinite set in $\mathbb{R}$.

2. Relevant equations

3. The attempt at a solution

how can we show that is true?

2. Sep 20, 2011

Dick

Re: infinite set

Do you know R is uncountable? That would be a clue.

3. Sep 20, 2011

burak100

Re: infinite set

infinite set means ; we can't find 1-1 correspondence between the set {1,...,n} and $f^{-1}(q)$ , here $f^{-1}(q)$ has n elements.
So, can we say $f^{-1}(q)$ is uncountable then we can't find 1-1 correspondence between the set {1,...,n}....????

4. Sep 20, 2011

Dick

Re: infinite set

Try to get the idea of your proof before you start writing symbols. Assume the opposite, that all of the f^(-1)(q) are finite. And you didn't answer my question. Do you know R is uncountable, and do you know what that means?

5. Sep 20, 2011

burak100

Re: infinite set

yeah R is uncountable but I can't find a relation to this question..

6. Sep 20, 2011

Dick

Re: infinite set

Every element if R is in one of the f^(-1)(q) sets, right? Suppose they are all finite?

7. Sep 20, 2011

burak100

Re: infinite set

I didn't understand clearly but ;

for all $q \in \mathbb{Q}$, we suppose all the sets, $f^{-1}(q)$ are finite. That means, every element of $\mathbb{R}$ is one of these sets, $f^{-1}(q)$ ...right?

8. Sep 20, 2011

Dick

Re: infinite set

Every element of R is in one of the f^(-1)(q) whether those sets are finite or not, just because f:R->Q. Pick an element x of R, for which q in Q is x in f^(-1)(q)??

9. Sep 20, 2011

burak100

Re: infinite set

I think that is right because just now we said
" every element of R is in one of the set f^{-1}(q) "

10. Sep 20, 2011

Dick

Re: infinite set

Yes, but can you tell me why?? Just because we said it doesn't prove it. You are going to need this as part of your proof.

Last edited: Sep 20, 2011
11. Sep 21, 2011

burak100

Re: infinite set

Ok. I think; we have $\mathbb{R}= \bigcup\limits_{q \in \mathbb{Q}} f^{-1}(q)$, and we suppose that for all q in Q , $f^{-1}(q)$ finite, then union would be countable but R is uncountable so contradiction... right?

12. Sep 21, 2011

Dick

Re: infinite set

Right, if you are clear on why the union of all the f^(-1)(q) is R.


text:
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Matrices

# Let $A\;=\begin{bmatrix}2 & 4\\3 & 2\end{bmatrix}, B\;=\begin{bmatrix}1 & 3\\-2 & 5\end{bmatrix}, C\;=\begin{bmatrix}-2 & 5\\3 & 4\end{bmatrix}$. Find $3A-C$

$\begin{array}{1 1} \begin{bmatrix}8 & 7\\6 & 0\end{bmatrix} \\ \begin{bmatrix}8 & -7\\6 & 2\end{bmatrix} \\ \begin{bmatrix}8 & 7\\6 & 2\end{bmatrix} \\ \begin{bmatrix}8 & 7\\6 & 2\end{bmatrix} \end{array}$

Can you answer this question?

Toolbox:
• The scalar multiplication $cA$ of a matrix $A$ and a number $c$ (also called a scalar in the parlance of abstract algebra) is given by multiplying every entry of $A$ by $c$.
• The difference $A-B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A - B)_{i,j} = A_{i,j} - B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
Given $A\;=\begin{bmatrix}2 & 4\\3 & 2\end{bmatrix}$, $3A = \begin{bmatrix}6 & 12\\9 & 6\end{bmatrix}$.
Therefore $3A-C = \begin{bmatrix}6 & 12\\9 & 6\end{bmatrix} - \begin{bmatrix}-2 & 5\\3 & 4\end{bmatrix} = \begin{bmatrix}6 - (-2) & 12-5\\9-3 & 6-4\end{bmatrix} = \begin{bmatrix}8 & 7\\6 & 2\end{bmatrix}$
answered Feb 27, 2013


text:
ï»¿ Answer the following briefly : (a)Â How does a muscle shorten during its contraction and lengthen during its relaxation? (b)Â What biological functions are served by the skeletal system ?(c)Â Why a red muscle fibre can work for a prolonged period while a white musule fibre suffers from fatigue after a shorter work ? Â  from Biology Locomotion and Movement Class 11 Assam Board

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Class 10 Class 12

(a)Â How does a muscle shorten during its contraction and lengthen during its relaxation?

(b)Â What biological functions are served by the skeletal system ?
(c)Â Why a red muscle fibre can work for a prolonged period while a white musule fibre suffers from fatigue after a shorter work ?

Â

(a) Shortening of muscle is due to the sliding of actin filament over myosin filament. According to sliding filament theory, the cross bridges take energy from ATP molecules and get rotated inward. Thus sarcomeres contract as Z-lines come closer. Due to contraction of sarcomeres finally whole muscle fibre contracts. During relaxation link between cross bridges and actin filaments get broken and cross bridge rotate outward. Thus sarcomeres expand as Z-lines get apart resulting in increase in the length of the mucle fibre.

(b)Â Functions of skeletal system :

1.Â It provides mechanical support.

2.Â It protects the essential organs as heart, brain, spinal cord etc.

3.Â It plays role in breathing.

4.Â It helps in locomotion.

5.Â Â It checks entry of food into glottis at the time of swallowing of food.

(c)Â Red muscle fibres contain myoglobin which stores oxyen as oxymyoglobin and it also has more number of mitochondria. Due to sufficient supply of oxygen little lactic acid is formed. These muscles contract slowly and remain in the contracted state for a long time. TDue to more energy and less lactic acid these muscles do not fatigue in less time.Â

White muscle fibres have no myoglobin. So, lactic acid is formed in them. Moreover, their contraction is fast, thus they get fatigued.

Â

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The thoracic vertebrae, ribs and sternum together form the rib cage.Â
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Gout is the inflammation of joints due to accumulation of uric acid crystals.Â
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Total number of pair of ribs?

12
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how many pairs of floating ribs are present in humns?

2
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8th, 9th and 10th pairs of ribs are called as false ribs or vertebrondral ribs.
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# Where do I put this?

Where do I put this?

I'm pretty sure I need to use Macworks for the Lisa, but can't find anywhere to put the disk.