# logaritims and matrix mathematics

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I am a year 10 student. This is not acctually homework, but I want to know how to do logaritims and matrix mathematics. can someone try and explain this to me?

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I am a year 10 student. This is not acctually homework, but I want to know how to do logaritims and matrix mathematics. can someone try and explain this to me?

You are asking for a whole course! Post some specific problems that you have and we may be able to help you.

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Thanks. I think I can understand the Matrix mathematics now, but I'm not so sure on the lorgarithms. maybe I'll just ask my Maths teacher about those...

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Thanks. I think I can understand the Matrix mathematics now, but I'm not so sure on the lorgarithms. maybe I'll just ask my Maths teacher about those...

Well... do you understand inverse functions?

I.e. $f^{-1}(f(x)) = x$

Like $\arcsin{(\sin{x})} = x$

or $\sqrt{x^2}=x$

will logarithms are the inverse functions of exponentiation.

$\log_n{n^x} = n^{\log_n{x}} = x$.

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Then there's the exponent of a matrix http://www.sosmath.com/matrix/expo/expo.html.

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Well... do you understand inverse functions?

I.e. $f^{-1}(f(x)) = x$

Like $\arcsin{(\sin{x})} = x$

or $\sqrt{x^2}=x$

will logarithms are the inverse functions of exponentiation.

$\log_n{n^x} = n^{\log_n{x}} = x$.

$f^{-1}(f(x)) = x$ does that mean thae same as $f^{-1}(fx)$? I understand that well enough.

I haven't come across arcsin yet, can you explain that?

$\sqrt{x^2}=x$ I understand too.

$\log_n{n^x} = n^{\log_n{x}} = x$? Can you go over that again, explaining it a bit more, please? I don't really understand that.

Booker, I didn't find that website very helpful, it was a bit too confusing.

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For trigonometric functions there seems to be the convention of calling the inverse of a function "arc". Specifically, arcsin is just the inverse of sin.

You can simply look at wiki for the basic properties of logs. As Bignose says, they are probably best thought of as inverses to exponents.

If you want to take the exponential of a matrix* (or some other non-commuting objects ) you use the formula $e^{A} = 1 + A + \frac{1}{2!}A^{2} + \frac{1}{3!}A^{3} \cdots$.

Now, the only slight complication is the fact that it is no so straightforward to solve $Z = \log (e^{A}e^{B})$. More specifically, it is not simply $A+B$. The solution is known as the Baker-Campbell-Hausdorff formula. Explicitly, the first few terms are

$Z = A+B +\frac{1}{2}[A,B] + More \: commutators$

______________________________________________________

*Note that for two matrices the commutator, $[A,B]= AB-BA \neq 0$ in general.

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$\log_n{n^x} = n^{\log_n{x}} = x$? Can you go over that again, explaining it a bit more, please? I don't really understand that.

well, you follow what $n^x$ means, right?

Let's apply some other function to $n^x$, let's just call it G.

$G(n^x)$

Now, G is a special function that is defined to be the inverse of $n^x$ so that $G(n^x)=x$.

The way people normally write out that function $G()$ is $\log_n ()$ so that

$\log_n{n^x} = n^{\log_n{x}} = x$

Exponentiation and log are inverses of each other. Log is the operation you do to perform the inverse of exponentiation.

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• 2 weeks later...

Some Common Logarithm Rules to note:

$\log_a(x)+\log_a(y) = \log_a(xy)$

$\log_a(x)-\log_a(y) = \log_a\left(\frac{x}{y}\right)$

$\log_a(x^y) = y\log_a(x)$

$\log_a(x) = \frac{\log_y(x)}{\log_y(a)}$

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okay, I understand logaritims better now.

Another thing that I didn't quite get from the wikipedia matrix page is kronecker product, it uses two 2-by-2 matricies, and that makes its explanation a bit weird. can someone please give me a better example, and also tell me how to do that with *normal* (what they teach you in lower high school) numbers?

I'm selecting advanced mathematics for one of my senior high school subjects next year... I should do well according to my teachers...

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Try understanding this:

let us take a simple 2*2 matrix.

[1 2] [3 4]

[5 6] [7 8]

Now according to the kronecker product we have to multiply:

{For first coioum of new matrix}

1)first element of first matrix with the first element of second matrix then the first element of first element with the second element of second matrixi.i.e;1*3 then 1*4.

2)second element of first matrix with the frist element of second matrix then the second element of first element with the second element of second matrix.i.e; 2*3 and 2*4.

{for second coloum of the new matrix}

3)now the first element of the frist matrix with the second element in the 1st row of second matrix and the first element of the frist matrix with the second element in the 2nd row of second matrix.

4)now the second element of the frist matrix with the second element in the 1st row of second matrix and the second element of the frist matrix with the second element in the 2nd row of second matrix.

repeat the above 4 steps for the elements of the rest elements of the frist matrix.

Remember that kronecker product of 2*2 matrix will have 4 rows and 4 coloums.

kronecker product for the above example is:

[1.3 1.4 2.3 2.4]

[1.7 1.8 2.7 2.8]

[5.3 5.4 6.3 6.4]

[5.7 5.8 6.7 6.8]

which on further solving gives:

[03 04 06 08]

[07 08 14 16]

[15 20 18 24]

[35 40 42 48]

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• 2 weeks later...

Ok, I understand that much, that's explained on Wiki, but what if we have a 3 by 3 matix and a 2 by 2 matrix? what then?

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