If algebra use factorise method,

Which method do maple use to dsolve differential equation?

I am trying to perform the following integral:

Which spits the integral back out at me.

I've also tried

Which, again, spits the integral back out at me.

My last attempt was this

Which... Still spit back out the integral.

Is there something special I should be doing for functions I'm integrating with a natural log? I need to get an exact value for this, not an approximation (because I am trying to check the accuracy of an approximation with this!).

Thanks!

Hi

How call each column of matrix by loop?

Hello there, my first time posting in this forum and using Maple, so sorry if I make some mistakes.

I'm trying to read a wav audio file, but there's this weird error that I can't understand : 

with(AudioTools);

[Audio, Clip, Convolution, Create, Duration, Extract, FormatFromName, Formats, Modulate, Normalize, Play, Preview, Read, Record, Resample, Scale, ToMono, ToStereo, Write]


audio := Read("C:\\Users\\Simon\\Desktop\\Test\\440.wav");

Error, (in readbytes) WAVE Error: unexpected end of file

Do you know what it can mean ? I'm sure it's a wave file, so I don't really get it

hi,

i'am beginers in  the maple programmation, i want to solve the einstien equation in the spherical coordinate,

Dear All,

I am working on ODEs and have obtained the plot for "variable vs time". I would like to know if it is possible and how to analyze those data in the frequency domain.ODEs.mw

Thank you.

Very kind wishes,

Wang Zhe

When Maple 2017?. Are there anybody about the new features of this version?. I,m waiting for a special project.

Hello,

I am working with numerical function with two indices, such as f[i,j], where i,j run over 1 to N, where N is n integer number.

The resluts stored in vector colummn for each value of j. i.e.

for each j=1,     G1:=Vector[column](N,i-->f(i,1))

for each j=2,     G2:=Vector[column](N,i-->f(i,2))

and so on.

The procedure is working for the first value j=1.

for the second value the following message appear

{--> enter Terminate, args = 
{--> enter Terminate, args = 
<-- exit Terminate (now in Terminate) = }
<-- exit Terminate (now at top level) = }
 

Hi, I try this numerical integral with:
int((1/2)*exp(-(1/2)*z^2)*Jo*sqrt(2)*(1-tanh(sqrt(q)*z/T)^2)/(sqrt(Pi)*T), z = -infinity .. infinity, numeric)
But does not work.
I have also tried it without success with:

evalf(Int((1/2)*exp(-(1/2)*z^2)*Jo*sqrt(2)*(1-tanh(sqrt(q)*z/T)^2)/(sqrt(Pi)*T), z = -infinity .. infinity, method = _d01amc, methodoptions = [maxintervals = 1000]))
How can i do?
Regards.

I have a nested for loop that iterates through a range of values for x and y coordinates to create a 3d surface for illustration of my research. after the x loop there is a y loop, and inside of that y loop is a series of commands to find some eigenvalues of a matrix (which become the z coordinates) and sort them into already open files. This isn't bad when the precision i require is more than .02, but some of my matrices require up to 0.005 or less. The latter precision costs hours of computation time on just one processor. However my laptop has an i7, so I want to see if i can get the for loop to send its next iteration to the next processor in line while it has the previous ones still calculating. Have any tips?

I have to crypt and decrypt with vigenere.

(procedures need lists)

"In a Caesar cipher, each letter of the alphabet is shifted along some number of places; for example, in a Caesar cipher of shift 3, A would become D, B would become E, Y would become B and so on. The Vigenère cipher consists of several Caesar ciphers in sequence with different shift values.

To encrypt, a table of alphabets can be used, termed a tabula recta, Vigenère square, or Vigenère table. It consists of the alphabet written out 26 times in different rows, each alphabet shifted cyclically to the left compared to the previous alphabet, corresponding to the 26 possible Caesar ciphers. At different points in the encryption process, the cipher uses a different alphabet from one of the rows. The alphabet used at each point depends on a repeating keyword.[citation needed]

For example, suppose that the plaintext to be encrypted is:

ATTACKATDAWN
The person sending the message chooses a keyword and repeats it until it matches the length of the plaintext, for example, the keyword "LEMON":

LEMON
Each row starts with a key letter. The remainder of the row holds the letters A to Z (in shifted order). Although there are 26 key rows shown, you will only use as many keys (different alphabets) as there are unique letters in the key string, here just 5 keys, {L, E, M, O, N}. For successive letters of the message, we are going to take successive letters of the key string, and encipher each message letter using its corresponding key row. Choose the next letter of the key, go along that row to find the column heading that matches the message character; the letter at the intersection of [key-row, msg-col] is the enciphered letter.

For example, the first letter of the plaintext, A, is paired with L, the first letter of the key. So use row L and column A of the Vigenère square, namely L. Similarly, for the second letter of the plaintext, the second letter of the key is used; the letter at row E and column T is X. The rest of the plaintext is enciphered in a similar fashion:

Plaintext:	ATTACKATDAWN
Key:	LEMON
Ciphertext:	LXFOPVEFRNHR
Decryption is performed by going to the row in the table corresponding to the key, finding the position of the ciphertext letter in this row, and then using the column's label as the plaintext. For example, in row L (from LEMON), the ciphertext L appears in column A, which is the first plaintext letter. Next we go to row E (from LEMON), locate the ciphertext X which is found in column T, thus T is the second plaintext letter."

I think that it can be done with a for loop but I do not know where to start.

Thanks in advance!

My professor insists for a homework problem that we write down a number answer to this one problem which boils down to...

I'm aware that's incredibly small. So much so that when trying to get a value to display using the command

at its max, it's still 0.00000000000000..0 as an answer.

I might just give up and write down the exponential form. But I was just curious if there is a way to get a number in scientific notation.

Hi everybody, 

Only weights of type "numeric" can be used in a weighted graph (package GraphTheory) .
Is it possible to bypass this limitation and affect an infinite weight to an edge ?
If not possible, does it exist some "highest numeric value" in Maple I could use instead ?
(the trick of a very high value doesn't suit me)

Subsidiary question : is it possible to change the printing format for those weights ?
For instance   GraphTheory:-SetEdgeWeight(MyGraph, MyEdge, 123.4) plots the value 123 on edfe MyEdge

Thanks in advance

Hey there!

I wonder how i can solve the following using maple.

I got an equation containting 2 variables; x and T.

I want to plot x versus T, with the range of T from 300 to 800.

How can I do this? 

This is the formula I am talking about:

Thanks in advance!

Kind regards,

Geard

I have the following expression:

expand((((o * (((h * h - h * v) * (a * h - a * v) * h * h * h * h * a * a * h - h * h * h * a * a * h * o * (h * h - h * v) * (a * h - a * v)) * a * a * h * h * h * h * h - h * h * h * a * a * h * h * h * h * h * a * a * h * (a * h - a * o) * (h * h - h * v) * (h - v)) * h * a * a * h * h * h * h * a * a * h * a * a * h * h * h * h * h - h * h * h * a * a * h * h * h * h * h * a * a * h * a * a * h * h * h * h * h * ((v * (a * h - a * v) * h * h * h * a * a * h - h * a * a * h * (h * h - h * v) * v * (a * h - a * v)) * a * a * h * h * h * h * h - h * a * a * h * h * h * h * a * a * h * (a * h - a * o) * (v * h * h * h - h * (h * h - h * v) * v))) * h * a * a * h * h * a * a * h * a * a * h * h - h * h * h * a * a * h * h * h * h * h * a * a * h * a * a * h * h * h * h * h * h * a * a * h * h * h * h * a * a * h * a * a * h * h * h * h * h * o * (((a * h - a * v) * h * a * a * h - a * a * h * o * (a * h - a * v)) * a * a * h * h - a * a * h * h * a * a * h * (a * h - a * o) * (h - v))) * h * h * a * a * h * h * h * h * a * a * h * a * a * h * h * h * h * h - h * h * h * a * a * h * h * h * h * h * a * a * h * a * a * h * h * h * h * h * h * a * a * h * h * h * h * a * a * h * a * a * h * h * h * h * h * h * a * a * h * h * a * a * h * a * a * h * h * (o * (v * (a * h - a * v) * h * h * h * a * a * h - h * a * a * h * (h * h - h * v) * v * (a * h - a * v)) * a * a * h * h * h * h * h - h * a * a * h * h * h * h * a * a * h * (a * h - a * o) * (v * v * h * h * h - h * (h * h - h * v) * v * v))) );

When I tun it through the Maple GUI I get this result:

-2*a^23*h^40*o*v^2+2*a^23*h^40*v^3

This is also what Mathematica gives me. However, if I run it through the Maple command line I get this result:

-a^77*h^97*o*v^2+a^77*h^97*v^3-a^23*h^40*o*v^2+a^23*h^40*v^3

Why are the results different?