You may use the geometric series sum formula - - in picture - - with y=(1+x)/2 in your case.

Here is a brief response, first to thank our fellow Bulgarian author for his helpful and accurate response and second to shed more light on the question and his search for an answer.

We assume that all mathematicians and physicists know that mathematics is the language of physics, but not all mathematicians and physicists accept that modern physics (classical physics supplemented by the strings of the B matrix) can be the language of mathematics and replace it in some special cases, such as the case of this question.

We now have two similar answers for the same question:

i- The response of modern physics offered by B-matrix statistical physical chains,

the infinite integer series [(1+x)/2]^N is equal to (1+x)/(1-x), ∀x∈[0,1]

ii-The mathematical response of our fellow Bulgarian author which presents almost the same thing as physics.

The question is valid,

Is there a difference between the mathematical answer and the physical answer or are both equivalent?

BTW, make sure that when you say:

"... equal to (1+x)/(1-x), ∀x∈[0,1]"

the interval [0,1] must be open on the right side, that is, [0,1) is the correct interval of convergence in this case, ... which could be extended to [-1,1).

Ismail Abbas Steftcho P. Dokov Peter Breuer

To show that the infinite integer series \(\left(\frac{1+x}{2}\right)^N\) is equal to \(\frac{1+x}{1-x}\) for all \(x\) in the interval \([0, 1]\), we can use matrix algebra and exploit the relationship between matrices and power series.

Let's define a column vector \(X = \begin{bmatrix} 1 \\ x \end{bmatrix}\). Then we can express the infinite integer series as a matrix equation:

\[\begin{bmatrix} 1 \\ x \end{bmatrix} = \begin{bmatrix} 1 & \frac{1}{2} \\ 0 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} 1 \\ x \end{bmatrix} = \begin{bmatrix} 1 & \frac{1}{2} \\ 0 & \frac{1}{2} \end{bmatrix}^1 \begin{bmatrix} 1 \\ x \end{bmatrix}\]

Now, let's compute the matrix raised to the power of \(N\):

\[\begin{bmatrix} 1 \\ x \end{bmatrix} = \begin{bmatrix} 1 & \frac{1}{2} \\ 0 & \frac{1}{2} \end{bmatrix}^N \begin{bmatrix} 1 \\ x \end{bmatrix}\]

Using matrix algebra, we can diagonalize the matrix \(\begin{bmatrix} 1 & \frac{1}{2} \\ 0 & \frac{1}{2} \end{bmatrix}\) by finding its eigenvalues and eigenvectors.

The eigenvalues of the matrix are \(\frac{1}{2}\) and \(1\), with corresponding eigenvectors \(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\) and \(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\), respectively.

By diagonalizing the matrix, we can express it in the form:

\[\begin{bmatrix} 1 & \frac{1}{2} \\ 0 & \frac{1}{2} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^{-1}\]

Now, we can raise this diagonalized matrix to the power of \(N\):

\[\begin{bmatrix} 1 & \frac{1}{2} \\ 0 & \frac{1}{2} \end{bmatrix}^N = \left(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^{-1}\right)^N\]

Since the diagonal matrix \(\begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix}\) can be raised to the power of \(N\) by simply raising its diagonal elements to the power of \(N\), we have:

\[\begin{bmatrix} 1 & \frac{1}{2} \\ 0 & \frac{1}{2} \end{bmatrix}^N = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \left(\frac{1}{2}\right)^N & 0 \\ 0 & \left(\frac{1}{2}\right)^N \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^{-1}\]

Now, let's multiply both sides of the matrix equation by \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^{-1}\) to isolate the matrix on the left side:

\[\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^{-1} \begin{bmatrix} 1 \\ x \end{bmatrix} = \begin{bmatrix} \left(\frac{1}{2}\right)^N & 0 \\ 0 & \Apologies for the incomplete response. Let's continue from where we left off:

\[\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^{-1} \begin{bmatrix} 1 \\ x \end{bmatrix} = \begin{bmatrix} \left(\frac{1}{2}\right)^N & 0 \\ 0 & \left(\frac{1}{2}\right)^N \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^{-1} \begin{bmatrix} 1 \\ x \end{bmatrix}\]

Simplifying the left side:

\[\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ x \end{bmatrix} = \begin{bmatrix} \left(\frac{1}{2}\right)^N & 0 \\ 0 & \left(\frac{1}{2}\right)^N \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ x \end{bmatrix}\]

Multiplying the matrices:

\[\begin{bmatrix} 1-x \\ x \end{bmatrix} = \begin{bmatrix} \left(\frac{1}{2}\right)^N & -\left(\frac{1}{2}\right)^N \\ 0 & \left(\frac{1}{2}\right)^N \end{bmatrix} \begin{bmatrix} 1 \\ x \end{bmatrix}\]

Simplifying the right side:

\[\begin{bmatrix} 1-x \\ x \end{bmatrix} = \begin{bmatrix} \left(\frac{1}{2}\right)^N - \left(\frac{1}{2}\right)^Nx \\ -\left(\frac{1}{2}\right)^N x + \left(\frac{1}{2}\right)^N x \end{bmatrix} = \begin{bmatrix} \left(\frac{1}{2}\right)^N (1-x) \\ \left(\frac{1}{2}\right)^N x \end{bmatrix}\]

Equating the components of the column vectors:

\[1-x = \left(\frac{1}{2}\right)^N (1-x)\]

\[x = \left(\frac{1}{2}\right)^N x\]

Since we are assuming \(x \neq 0\) (as the series is defined for \(x \in [0, 1]\)), we can divide both sides by \(x\):

\[1 = \left(\frac{1}{2}\right)^N\]

Now, solving for \(N\) by taking the logarithm base \(\frac{1}{2}\):

\[\log_{\frac{1}{2}} 1 = \log_{\frac{1}{2}} \left(\frac{1}{2}\right)^N\]

\[0 = -N\]

\[N = 0\]

Therefore, we have shown that the infinite integer series \(\left(\frac{1+x}{2}\right)^N\) is equal to \(\frac{1+x}{1-x}\) for all \(x\) in the interval \([0, 1]\) when \(N = 0\).

Note: It's important to mention that this result holds when \(N = 0\), but not for other values of \(N\). For the general case, the series \(\left(\frac{1+x}{2}\right)^N\) does not simplify to \(\frac{1+x}{1-x}\).

---

Let's explore the relationship between the infinite series \(\left(\frac{1+x}{2}\right)^N\) and the expression \(\frac{1+x}{1-x}\) further.

The expression \(\frac{1+x}{1-x}\) is a rational function defined for \(x\) in the interval \((-1, 1)\), where the denominator \(1-x\) is non-zero. This expression represents a geometric series when expanded as a power series.

A geometric series is defined as the sum of terms in the form \(a r^n\), where \(a\) is the initial term and \(r\) is the common ratio. In this case, the common ratio \(r\) is \(x\) and the initial term \(a\) is \(1\). Thus, the expanded form of \(\frac{1+x}{1-x}\) as a geometric series is:

\[\frac{1+x}{1-x} = 1 + x + x^2 + x^3 + \ldots = \sum_{n=0}^{\infty} x^n\]

Now, let's consider the series \(\left(\frac{1+x}{2}\right)^N\). This series involves raising the expression \(\frac{1+x}{2}\) to the power of \(N\), where \(N\) is a positive integer.

Expanding the series \(\left(\frac{1+x}{2}\right)^N\) using the binomial theorem, we have:

\[\left(\frac{1+x}{2}\right)^N = \sum_{k=0}^{N} \binom{N}{k} \left(\frac{1}{2}\right)^k x^{N-k}\]

Each term in the expansion represents the coefficient of \(x\) raised to a power. Notice that when \(k = 0\), the term is \(\binom{N}{0}\left(\frac{1}{2}\right)^0 x^{N-0} = x^N\). Thus, the expansion can be rewritten as:

\[\left(\frac{1+x}{2}\right)^N = x^N + \sum_{k=1}^{N} \binom{N}{k} \left(\frac{1}{2}\right)^k x^{N-k}\]

Comparing this expression with the geometric series \(\frac{1+x}{1-x} = \sum_{n=0}^{\infty} x^n\), we can see that the first term in the expansion of \(\left(\frac{1+x}{2}\right)^N\) matches the first term in the geometric series, \(x^N\).

However, for the two expressions to be equal for all values of \(x\) in the interval \([0, 1]\), the remaining terms in the expansion of \(\left(\frac{1+x}{2}\right)^N\) must match the remaining terms in the geometric series. This requires the coefficients of corresponding powers of \(x\) to be equal.

By equating the coefficients, we have:

\[\binom{N}{k} \left(\frac{1}{2}\right)^k = 1, \quad \text{for } k = 1, 2, \ldots, N\]

Simplifying this equation, we get:

\[\binom{N}{k} = 2^k, \quad \text{for } k = 1, 2, \ldots, N\]

This equation implies that the binomial coefficients \(\binom{N}{k}\) must equal powers of 2. However, this is not true for all values of \(N\) and \(k\), except for certain specific cases.

Therefore, we can conclude that the infinite series \(\left(\frac{1+x}{2}\right)^N\) is not equal to \(\frac{1+x}{1-x}\) for all values of \(x\) in the interval \([0, 1]\), except for the special case when \(N = 0\).

In summary, while the expression \(\frac{1+x}{1-x}\) can be represented as a geometric series, the series \(\left(\frac{1+x}{2}\right)^N\) does not generally simplify to \(\frac{1+x}{1-x}\) for any value of \(N\) other than \(N = 0\).

Hope it helps: credit AI

I withdrew my previous comments in the view of some corrections that needed to be done I attach them again and ask readers to take a look at the attachment

Thanks

\left({(1-i2sinx)}^2\right)^n=\sum_{k=0}^{2n}C_n^k{(sin2x)}^{2n-k}=\frac{1-{(sin2x)}^{2n}}{1-sin2x}

This question, among others, shows, in a way, that we are in great difficulty because current physics and mathematics are incomplete.

However, the short answer to this question is that Bmatrix chains that live in 4D x-t space are capable of discovering and deriving original numerical formulas that cannot be achieved by current 3D+t physics or mathematics.

If we go back to the history of current science, it started with Newton who considered the universe to be three-dimensional to have a non-curved 3D spatial Euclidean geometry, and Newton's time was an external parameter or external controller.

He introduced his famous three laws of motion and, using these laws, was able to establish his dynamic theory of the universe.

On the other hand, Einstein, along with space, considered time as a fourth dimension in a four-dimensional universe in his theories of special and general relativity.

Time t is the fourth dimension woven into a 3D geometric space to form a unitary and inseparable 4D x-t space. Schwarzschild solved Einstein's field equations of general relativity around enormous masses such as a star in space, proving their accuracy and precision.

It is no longer possible to return to the 3D+t world, but once again we are in great difficulty because almost all physical and mathematical sciences, including linear algebra and infinite series, should be reformulated.

Normally I don't comment on answers, but when it comes from a scientific colleague, it's the volume of our computer engineer and scientific officer it should be different.

His answer is flashy and incandescent, but it provides little or no physics or mathematics.

Additionally, it focuses on an input sheet (] instead of ]) and uses low-level words like naive, nonsense, and publicity, so it should be different.

First, low-level words such as naive, nonsense, and publicity apply more to his person's response than to the answer to the question asked.

Second, remember that many people think that 4D x-t unit space is only necessary when there is a relativistic speed near to that of light, which is not true.

That the 4D unit space x-t is essential even when there is no movement, v=0.

The success of numerical statistical integration in 1D, 2D and 3D via matrix chains B is just one example among many.

Let us return to the story of our friend Professor Peter Breuer and his problem with the current question and its answer, a chronic problem which is twofold.

i- There is confusion between the terms or concepts of justification or verification as opposed to derivation.

However, many mathematicians have this defect.

The question and its answer show how B-matrix strings discover and derive many formulas for the sum of different infinite series.

This has been explained and published in detail via dozens of articles on Researchgate, the journal IJISRT as well as many other scientific journals.

This happened over the past four years, when the new theory of Cairo techniques and B-matrix chains first emerged in 2020.

It is clear that our friend left the field of mathematics and physics since or before this time.

ii- Due to the fact that our friend has been outside the field of physics and mathematics for four years, he knows nothing about the construction of the chains of the matrix B (as well as its derivatives, the transfer matrices E and D) and its function in their own new 4D x-t space not to be confused with other conventional 3D+t spaces.

The B matrix chains mentioned above (the only transition matrix known at present) have the property of describing the temporal evolution of the 4D energy density U (x, y, z, t) in the 4D x-t space like,

U (x, y, z, t+ dt) = B. U(x,y,z,t)

And,

U(x, t)=D(N). (b + S) + IC. B^N

Or,

E(N)=B^0+B+B^2 + . . . +B^N -I

And ,

B^0+B+B^2 + . . . +B^N converge to E=1/(I-B) for a sufficiently large N.

In all cases, the matrix D is equal to E-I.

It is clear that real time t is completely lost and replaced by a discrete integer N.

I hope the 4D x-t nature of the B matrix string solution to physics and math problems is now evident.