Pls can you help with that

MathsEconomics:


Number Theory and Algebraic Geometry can do it.

MathsEconomics:


I have that book with me at home and I've read it before a good number of times.
I'll not lie to you, it's not a good book and it's best you stop wasting your time on it.
The book has little theoretical background but contains a lot of extremely difficult problems as exercises.
I suggest you stop using problem solving books for now. Use books like Chrystal's Algebra or Plane Trigonometry by Loney or Hobson (Both are equally good but Hobson is too rigorous for beginners).

The proofs presented in those Olympiad problem solving books will not really teach you how to construct your own proofs because they hide their steps and just give you the short and final answer leaving you with no hint as to how to construct your own proofs.

Also. As for the proofs you gave. I'll tell you something really important.

Just because you derived a valid inequality from a given inequality does not necessarily mean that that original inequality is valid.
For example. -4 < -3, -2 < -5 so by multiplication
8 < 15. Notice that the inequality we derived is valid but this is not a valid proof because our assertion that -2 < - 5 is false.
So regardless of how you manipulate inequalities, arriving at a true statement does not make that inequality true. That's why I dissected my proof above with induction to make it look like I'm actually deriving the result. Inequalities can be tricky.


Also. The conditions you imposed on you lemma actually invalidated it since the conditions you used do not satisfy the conditions imposed by the problem.

I initially got Topics in Algebra and Analysis because of the range of topics it covered that I wasn't seeing a lot elsewhere.

I have Chrystal's algebra already. What do you think of the book Challenge and Thrills Of Pre-College Mathematics?

gpercuxionz:
Pls can you help with that

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Onlinedet:
Pls someone should try this....

Someone should try this in what way? Is it that you cannot solve them or what or you don't want to make efforts at all??

Darivie04:

I initially got Topics in Algebra and Analysis because of the range of topics it covered that I wasn't seeing a lot elsewhere.

I have Chrystal's algebra already. What do you think of the book Challenge and Thrills Of Pre-College Mathematics?

Can you find a?
5log₄ a + 48log_a 4 = a/8

naturalwaves:

Someone should try this in what way? Is it that you cannot solve them or what or you don't want to make efforts at all??

Naah don't stress it bro..... Someone sent me the question to help him solve, but it looked strange to me (not like the normal matrix i knw).... If u can solve it, fine, if u wont, fine...... Jst allow others to do it....

Onlinedet:

Naah don't stress it bro..... Someone sent me the question to help him solve, but it looked strange to me (not like the normal matrix i knw).... If u can solve it, fine, if u wont, fine...... Jst allow others to do it....

What the hell do you mean by determinant?

Onlinedet:

Naah don't stress it bro..... Someone sent me the question to help him solve, but it looked strange to me (not like the normal matrix i knw).... If u can solve it, fine, if u wont, fine...... Jst allow others to do it....

Okay. I get you now. It is not matrix, it is differentiation bro. Although it says determinant, it is not asking for the determinant of a matrix. You need to find dee Y, dee X written as dy/dx. Let me solve it for you. BRB.

MathsEconomics:

Alright.

waiting for the solution

Onlinedet:

Naah don't stress it bro..... Someone sent me the question to help him solve, but it looked strange to me (not like the normal matrix i knw).... If u can solve it, fine, if u wont, fine...... Jst allow others to do it....

Here is the solution bro. Like I said, it is not matrix but differentiation. You learn the concept under Calculus aspect of Mathematics.

Silentscreamer:
waiting for the solution

Do you know the solution? Just asking.

Martinez39:
Do you know the solution? Just asking.

Nope. need the solution in preparation for an exam.

Silentscreamer:
Nope. need the solution in preparation for an exam.

Harmonic Analysis?
Given 6 date points, it is required to express a function in terms of sine and consines. Right?
I'm working on your problems though. The computations involved are mind boggling.

gpercuxionz:
Pls can you help with that

If you reduce the problem to a Diophantine Equation. Number Theory comes in. All you need to do next is to show that x = 2 is the only solution.

Silentscreamer:
Help me solve this.
fit the equation y=Co+C1sinpix+C2cospix to the data points (0.2,1)(0.4,2)(0.6,3)(0.8,5)(1,5)(1.2,3) using method of least squares. Co, C1 and C2 are constants.

Are you sure it's just this 6 data points that was given in the problem set?

MathsEconomics:


Harmonic Analysis?
Given 6 date points, it is required to express a function in terms of sine and consines. Right?
I'm working on your problems though. The computations involved are mind boggling.

it is not under harmonic analysis. it is under curve fittings.
and the computations involved are not mind boggling, I've actually solved a few, but this particular type is giving me trouble.

Silentscreamer:
it is not under harmonic analysis. it is under curve fittings.
and the computations involved are not mind boggling, I've actually solved a few, but this particular type is giving me trouble.

Though the calculations were very tedious and I postponed over and over again but I have solved your question. I used a technique in linear algebra. If you are interested, I could explain it to you.
C₀ = 2.43078894348,
C₁ = 0.88872620907,
C₂ = - 2.42743109456

Below is a picture of the resulting graph with points A(0.2, 1), B(0.4, 1), C(0.8, 5), D(1, 5), E(1.2, 3), F(0.6, 3)

MathsEconomics:


If you reduce the problem to a Diophantine Equation. Number Theory comes in. All you need to do next is to show that x = 2 is the only solution.

f(x) = 2^x + 3^x is a one to one function, it is smooth and continuous in it's domain and it's domain is (-∞,∞). It's range is (0, ∞), therefore the graph of g(x) = 13 will intersect the graph f(x) at one and only one point. Hence the equation 2^x + 3^x = 13 will have only one.

If x = 2 is a solution to 2^x + 3^x = 13 then it is the only solution.

Cc. gpercuxionz

Martinez39:
Though the calculations were very tedious and I postponed over and over again but I have solved your question. I used a technique in linear algebra. If you are interested, I could explain it to you.
C₀ = 2.43078894348,
C₁ = 0.88872620907,
C₂ = - 2.42743109456

Below is a picture of the resulting graph with points A(0.2, 1), B(0.4, 1), C(0.8, 5), D(π, 5), E(1.2, 3), F(0.6, 3)

pls do (the explanation). thanks.

Silentscreamer:
pls do (the explanation). thanks.

I will write and send the solution later, at least tomorrow morning. How soon do you need it?

Martinez39:
I will write and send the solution later, at least tomorrow morning. How soon do you need it?

tomorrow is OK if u can.

Silentscreamer:
tomorrow is OK if u can.

In the picture below, form the matrices X, Y and C as shown.

Using the curve fitting application in linear algebra,

C = (X^TX)^-1X^TY

What it means is that you multiple the transpose of X, X^T , on the right by X to form a new matrix, say B, such that
B = X^TX

Then, in a similar fashion, you find the matrix product
A = X^TY

Finally, to get C, Multiple the B^-1, on the right by A, so
C = B^-1A
N.B: you would have to do the workings on your own to verify my answer. I can show my workings but the computation is very very very very tedious, much and long. I only have them on a rough sheet. I was about abandoning it until you said you needed the solution for an exam.

Darivie04:
Nice little problem.

How do you and MathsEconomics compose stuff using that white background? Where can I get it?

Martinez39:

In the picture below, form the matrices X, Y and C as shown.

Using the curve fitting application in linear algebra,


What it means is that you multiple the transpose of X, X^T , on the right by X to form a new matrix, say B, such that
B = X^TX

Then, in a similar fashion, you find the matrix product
A = X^TY

Finally, to get C, Multiple the B^-1, on the right by A, so
C = B^-1A
N.B: you would have to do the workings on your own to verify my answer. I can show my workings but the computation is very very very very tedious, much and long. I only have them on a rough sheet. I was about abandoning it until you said you needed the solution for an exam.

Thanks a lot.

Martinez39:

In the picture below, form the matrices X, Y and C as shown.

Using the curve fitting application in linear algebra,


What it means is that you multiple the transpose of X, X^T , on the right by X to form a new matrix, say B, such that
B = X^TX

Then, in a similar fashion, you find the matrix product
A = X^TY

Finally, to get C, Multiple the B^-1, on the right by A, so
C = B^-1A
N.B: you would have to do the workings on your own to verify my answer. I can show my workings but the computation is very very very very tedious, much and long. I only have them on a rough sheet. I was about abandoning it until you said you needed the solution for an exam.

The computation involved is truly tedious. The question is under curve fittings in Engineering Analysis. The ones I've solved before involve polynomial, exponential and logarithmic equations and are pretty straightforward. This particular one involving trigonometric equation is the one I couldn't tackle. Thanks for helping out.

Silentscreamer:
The computation involved is truly tedious. The question is under curve fittings in Engineering Analysis. The ones I've solved before involve polynomial, exponential and logarithmic equations and are pretty straightforward. This particular one involving trigonometric equation is the one I couldn't tackle. Thanks for helping out.

This technique would also work for polynomial functions. In such case, the question would be straight forward.

Martinez39:
How do you and MathsEconomics compose stuff using that white background? Where can I get it?

Download MathMagic Lite from the Playstore and type with it.

Martinez39:

In the picture below, form the matrices X, Y and C as shown.

Using the curve fitting application in linear algebra,


What it means is that you multiple the transpose of X, X^T , on the right by X to form a new matrix, say B, such that
B = X^TX

Then, in a similar fashion, you find the matrix product
A = X^TY

Finally, to get C, Multiple the B^-1, on the right by A, so
C = B^-1A
N.B: you would have to do the workings on your own to verify my answer. I can show my workings but the computation is very very very very tedious, much and long. I only have them on a rough sheet. I was about abandoning it until you said you needed the solution for an exam.

Your handwriting is beautiful and neat.

naturalwaves:

Download MathMagic Lite from the Playstore and type with it.

Thanks.