" then place the approximation (expanded expression) into the preceding equation; T^2+exp(-T)+T^3+...=0, and solve the simplified equation to get the exact T "

Of course, it will NOT be the exact T value of the original equation but it might be close enough for your needs. One way to evaluate the error would be to approximate again but with higher precision and see if the value you get converges...

Of course, if you know the solution is close to 5, you could solve it numerically by Newton method starting from this good approximate value! Any math computation software will do it (Matlab, Maple, Mathematica, etc.)

The equation I wrote is only an example for 'another typical equation' and T =5 is the solution to that 'another typical equation', not the solution to the equation I wrote. The equation I wrote, was only for exemplifying the problem. I am to algebraically find T, but Taylor's expansion is infinite series (not closed-form), hence I cannot find an exact numerical solution for T, by plugging the infinite Taylor series into the equation, without introducing truncation error by cutting the series at a special power.

Thanks Bruno, I see and got what you advised.

My last comment was an answer for Dr. Farid.

Thank you Peter. I will have a look over the book. Yeah, I missed analysis for algebra. You are correct; ''that's analysis, not algebra''.

Knowing that T ≈ 5, and looking for an expansion to be plugged into the equation, the bad point is that, I want to find T in terms of another parameter, analytically, not numerically. For example, assume we are given, a+T+T^2+exp(-T)=0, and then truncating the Maclaurin series of;

exp(-T)=1-T+T^2/2, then plugging this into the preceding equation gives:

a+T+T^2+(1-T+T^2/2)=0,

and this is the quadratic equation that yields the analytic expression for T in terms of the unknown variable; a. This is the goal of the question, I've queried in this post.

It seems, that's analysis coupled with algebra!.

I emphasize for the second time: e-t = 1/2 + (1-t)2/2 is only an example, and not the equation I am looking to solve. My equation is -(a+b+2c+bT+2cT+cT^2)exp(-T)+d=1.

This is to be solved to get T in terms of a, b, c, d. That's it.

I recommended that you should try Newton method using matlab or mathematica. It's easy to implement and always promising.

Thank you Jamilu and all others. I now have my question work out.

Thank you, Peter. As already I said, the question is work out.

For your notice, a,b,c,d are parameters (constants) of another typical problem (fluid physics problem), casted as the follow-up equation, to which I am looking for a soltion as a question in this post. It means a,b,c,d are constants, while T is to be evaluated in terms of a,b,c,d, where satisfying: (a+b+2c+bT+2cT+cT^2)exp(-T)+d=1.

Thank you Peter, for your serious pursuit of the question. You've been resourceful. Bests. Saeb.

Dear Peter and all others. I finally recap my question. Consider a quadratic equation; ax^2+bx+c=0. The solution to this quadratic equation is simply:

x=(-b ± sqrt(b^2-4ac))/(2a).

Now add an exponential term to the former quadratic equation, it means: exp(x)+ax^2+bx+c=0.

Similar to the quadratic case, the quest of a solution x to this equation is sought and solicited.

I guess it is not analytically approachable. Thank you all. Bests. Saeb.

Dear Saeb,

I read the nice contributions of the followers about your question, I am sure my answer is not the best, but I have a little bit to add just to share you the presented ideas.

In general, there are algebraic equations ( as polynomials) and transcendental equations( that involves logarithms, trigonometric, exponential, etc.)

In most cases, exact solutions ( roots, or zeros of F(x) = 0) do not exist, so one of the major topics in numerical analysis is to find solutions.

The idea is like this, consider any smooth curve, a differentiable function of one variable, y = f(x). At any point A(a,f(a)) belongs to the curve, we can determine the tangent line at that point.

( y - f(a) = m(x-a) ) this line can replace the function f(x), it is called linearization of f(x) in the neighborhood of A.

In general and for more precision f(x) is replaced by its power series up to some order.

In the equation T^3 + T^2 +e^(-T)+ ....= 0

we have a problem with e^(-T)

The Macalurine series of e^(-T) = 1 + (-T) +(-T)^2 /2! + (-T)^3 /3! ....

( the expansion in the neighborhood of 0 ) ,

For T = 1, this means the approximate value of e^(-1) =1 -1+1/2 -1/6...

(1/e)=approx.= 0.36788..

1-1+(1/2)-(1/6)= 0.33333 (using 4 terms)

1-1+(1/2)-(1/6)+(1/(24))= 0.375 (using 5 terms)

1-1+(1/2)-(1/6)+(1/(24))-(1/(120))= 0.36667 ( using 6 terms)

more terms give best approximation and converge to real value of (1/e)

Indeed , the power series 1 + (-T) +(-T)^2 /2! + (-T)^3 /3! .... can replace e^(-T), with some minor error ( which can be controlled by adding more terms),

So, we can replace e^(-T) in the equation T^3 + T^2 +e^(-T)+ ....=0

we obtain: T^3 + T^2 +(1-T)+....=0

or T^3 + T^2 +(1 + (-T) +(-T)^2 /2! )+...

you can replace it by ( n terms) according to your need up to precision.

In the latest post, you mentioned the equation

(a+b+2c+bT+2cT+cT^2)exp(-T)+d=1.

Your goal is find T= T(a,b,c,d)

It is a transcendental equation, and exact solutions cease to exist, then the approximation of exp(-T) is a must. Next, after replacing exp(-T) by its approximation, the new equation is easier the original one and give a good approximation to the required solution.

[Try to substitute 4 distinct approximations of e^(-T) at T0

T0 + eps1, T0 + eps2, T0 + eps3, where eps(i), i=1,2,3 are very small numbers, then solve the linear system for a, b ,c and d

i.e, you can approximate, a, b ,c and d at T = T0 ]

This based on the nature of your mathematical model.

This simple idea has many versions and can be generalized to functions of several variables.

Best regards

@Issam. Thank you friend. Just before your answer, I added an update and recap statement of my question. I think as you mentioned in your kind answer, the equation is transcendental and ceases to yield even an exact numerical solution, how about an analytic solution ! (it would not have to high probability any analytic solution). Bests, Saeb.

@ Saeb,

e^(-T) is analytic everywhere in all cases.

I didn't say the solution doesn't exist; I said exact solutions cease to exist.

Approximations work almost all the time under certain conditions.

Regards

Numerical solution methods such as Newton method will give you a solution. The accuracy of this solution can be up to 10 to power - 7 or more.

@Abdul Saif.

Thank you. But I'm looking for analytical solution.

The 'menu' in the Answer states (quote): Identify yourself as an expert by providing high-quality answers. My (this) answer to 'A philosophy to validate analytical approximation to solve a mathematical algebraic equation?' is certainly not expert or Mathematical. However, I would like to point out (and emphasise)that all the 'posts'/answers to the question are very informative, instructive and well thought out (even the single line/sentence answers and/or responses). This is one of the best discussion topics that I have seen for a long time and MOST CERTAINLY keeps the (or a) mathematical mind interested: Even those who enjoy so called 'recreational mathematics' at different levels of simplicity or complexity.  

Thank you, Mr. Chemaly. Regards. AmirAhmadi.