I am looking how can we define SVD of 2D function f(x+u,y+v)?

Anyone can pls help me to direct to it. thank you

]]>why is math so hard and challenging i know at my age i should have it by now but it is math it should be easy.

If people do not understand mathematics is simple, it is because they do not see how complicated life is.

You are welcome to join the forum

]]>In his proof he proves that HLKF is a parallelogram. You can use that.

(I do hope Im not bothering you in any way)

I enjoy helping people on MIF. I'm reasonably good at geometry but I've never gone through all the propositions like you have. That makes it harder because I have to look them up and work through several just to understand what has already been proved. My version of Java blocks the diagrams from that site as it says it is 'untrusted'. But I have a pdf with the complete Elements which uses the same words and diagrams so I can get what I need from there.

Bob

]]>There are a number of ways of defining a parallelogram:

All definitions require 4 sides but then:

opposite sides equal in length

OR

opposite sides parallel

OR

one pair of opposite sides equal and parallel

OR EVEN

diagonals bisect each other

I expect there are more. You can quickly show these are equivalent by congruent triangles.

Bob

]]>hi ducmod

I'll be happy to provide further help. You'll need to understand the formula for variance and co-variance. The book I have used to teach from is called Advanced Level Statistics by A Francis. ISBN 0-85950-451-4 My book was printed in 1986 so it may not be available now.

Post further questions when you are ready.

Bob

Bob, thank you very much for your help! Trust me, I will be back Very soon. It's challenging for me to learn math on my own, but it's the only option I have now.

Have a grate one!

Bob

]]>B

]]>Bob

]]>Welcome to the forum!

I guess you're referring to the puzzle found here:

What 5-digit number has the following features:

If we put the numeral 1 at the beginning, we get a number three times smaller than if we put the numeral 1 at the end of the number.

The answer and the solution method (algebraic) are given here.

Here's another approach, one that doesn't use an algebraic equation:

Let the 5-digit number be *abcde*, where each letter represents one of the 5 digits.

From this, the number *beginning* with 1 is 1*abcde*, and the number *ending* with 1 is *abcde*1.*Note: The fact that the number's letters are different from each other doesn't necessarily mean that the digits are all different.*

Work backwards, digit by digit, to find which digit values result in the following:

1*abcde*

x3

--------*abcde*1

=====

Start by asking this question to find *e* in 1*abcde* (and hence in *abcde*1):

The product of which digit when multiplied by 3 ends in 1?

Answer: 7 (ie, 7x3=21)...so *e*=7 (carry 2...ie, the product's tens unit)

To find *d, c, b and a*, ask the same question as for *e*, but this time there's a 'carry' to add. eg, to find *d*:

The product of which digit when multiplied by 3, plus the carry 2, ends in 7?

Answer: 5 (ie, (5x3)+2=17)...so *d*=5 (carry 1)

So far we have this partial solution:

1*abc*57

x3

--------*abc*571

=====

...and the rest I'll leave for you to solve.

]]>Bob

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