Solving for d in ((a+b)^n)-(a^n+b^n) - ((c+d)^n)-(c^n+d^n)

[eddybob123]

Hi, I need help solving this equation, get d on one side of the equal sign:

(((a+b)^n)-(a^n+b^n))-(((c+d)^n)-(c^n+d^n))=d
Thanks in advance. I don't know how to use the fancy code thing

 

[Vorde]

To make it easier: (a+b)^n-(a^n+b^n)-(c+d)^n-(c^n+d^n)=d

Not sure how to go about it though, I'd try isolating all the terms with d to begin with. Then maybe you could use logs to simplify it? I'll have to work though this.

 

[eddybob123]

To make it easier: (a+b)^n-(a^n+b^n)-(c+d)^n-(c^n+d^n)=d

Not sure how to go about it though, I'd try isolating all the terms with d to begin with. Then maybe you could use logs to simplify it? I'll have to work though this.

Before you start though, your (c+d)^n-(c^n+d^n) is supposed to be ((c+d)^n-(c^n+d^n))

 

[sin2beta]

A possible first step is to simplify (a+b)^n - (a^n + b^n) as (a+b)^(n-1). Likewise with the c and d terms.

 

[Vorde]

Before you start though, your (c+d)^n-(c^n+d^n) is supposed to be ((c+d)^n-(c^n+d^n))

Would it make a difference?

A possible first step is to simplify (a+b)^n - (a^n + b^n) as (a+b)^(n-1). Likewise with the c and d terms.

That doesn't simplify like that, right? Can someone reinforce that for me though?

 

[sin2beta]

Would it make a difference?

Yes, because of the subtraction before the left most parentheses.

That doesn't simplify like that, right? Can someone reinforce that for me though?

No it shouldn't. I don't know what I was doing. There will be coefficients on those two terms in the binomial theorem that need to be taken care of. Disregard my previous post.

 

[Borek]

A possible first step is to simplify (a+b)^n - (a^n + b^n) as (a+b)^(n-1)

Looks to me like (a+b)^n - (a^n + b^n) is just a constant and there is no need to simplify it - call it a' or something.

 

[Vorde]

I think n is supposed to be a variable.

 

[eddybob123]

n is an integer that is greater than 2. Actually, all of a,b,c, and d should be greater than 2. Is it really that tough of an algebra question?

 

[Matt Benesi]

Note that for the message below, I assume you meant to isolate d on one side of the equals sign, from the following equality:

(a+b)^n-(a^n+b^n)-((c+d)^n-(c^n+d^n))=0Just take the d^n out of the one side of the equation, and take the nth root of both sides. You end up with:

d=\sqrt[n]{{\left( d+c\right) }^{n}-{c}^{n}-{\left( b+a\right) }^{n}+{b}^{n}+{a}^{n}}

Of course, if you want to entirely remove d from the one side (under the radical above), you'd probably need to calculate for specific n.

For example, solved in maxima... lazily ;), with n equal to 3:

d=-\frac{\sqrt{{c}^{4}+\left( 4\,a\,{b}^{2}+4\,{a}^{2}\,b\right) \,c}+{c}^{2}}{2\,c}
...or...
d=\frac{\sqrt{{c}^{4}+\left( 4\,a\,{b}^{2}+4\,{a}^{2}\,b\right) \,c}-{c}^{2}}{2\,c}

Keep in mind that it gets crazy complicated as you increase powers of n.

 

[Nabeshin]

Indeed, this looks in general to be an (n-1)th order equation in d. No general solution in terms of radicals exists for n \geq 6.

 

[eddybob123]

Matt, I mean to solve the equation on the first post for d, with relation to the variables a,b,c, and n, not just specific values.
I might use the binomial theorem, but I don't know how. Please help me.

 

[Borek]

Matt, I mean to solve the equation on the first post for d, with relation to the variables a,b,c, and n, not just specific values.

As Nabeshin wrote there are no general solutions for finding roots of polynomials of degree higher than 5 (as long as you are looking for a way of expressing them using addition, subtraction, multiplication, division and roots of any degree; this has been proven almost 200 years ago). So it is quite likely that the problem can be solved for some small n, but there is no general solution that you are looking for.

There is always a small chance it is possible to use some fancy trick to simplify the equation, but I would not hold my breath.

 

[eddybob123]

But, from what I know, it should be possible. Even humans have dealt with higher power equations, so this shouldn't be much more different.
I think I would use the summation definition of a binomial, then subtract the first and last terms, which are a^n and b^n, and c^n and d^n. This will leave us with the summation starting at 1 and ending at n-1.

 

[micromass]

But, from what I know, it should be possible. Even humans have dealt with higher power equations, so this shouldn't be much more different.
I think I would use the summation definition of a binomial, then subtract the first and last terms, which are a^n and b^n, and c^n and d^n. This will leave us with the summation starting at 1 and ending at n-1.

No. It has been proven that most polynomial equations can not be solved. This is a polynomial equation in multiple variables even. So I wouldn't expect a neat solution.

 

[eddybob123]

Then let me ask you this question: what is the highest expected value d with relation to a,b,c, and n. Let me define this more precisely. The highest power of a binomial will determine its greatest value, so the terms ((a+b)^n)-(a^n +b^n) must be greatest possible, so a and b are fairly close together. Likewise, the c and d terms of the left side must be least possible, so c and d have to be furthest apart.

 

[haruspex]

Then let me ask you this question: what is the highest expected value d with relation to a,b,c, and n. Let me define this more precisely. The highest power of a binomial will determine its greatest value, so the terms ((a+b)^n)-(a^n +b^n) must be greatest possible, so a and b are fairly close together. Likewise, the c and d terms of the left side must be least possible, so c and d have to be furthest apart.

If you want the largest d for a given a, b, and n, then, yes, just make c as small as is allowed. If that makes it much smaller than a and b (hence much smaller than d) then you can approximate (c+d)ⁿ - (cⁿ + dⁿ) as ncd^n-1.

 

[JJacquelin]

Since d^n disappears, the remaining term of highest power is d^(n-1).
It is known that analytical solutions of polynomial equations are possible up to power 4.
So, your equation is analytically solvable up to n=5 (inclusive)
In case of n>5, you have to use numerical computation in order to obtain approximate values for the solutions.