MHB Calculate Square of Sum ∑1..9999

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Square Sum
AI Thread Summary
The discussion focuses on calculating the square of the sum of two series involving square roots, specifically for the range from 1 to 9999. Participants suggest simplifying the sums by using the conjugate of the denominators and explore a generalization for a function f(N) defined for N values. Numerical experimentation with small values of N suggests that f(N) approaches 1 + √2, leading to a conjecture that this holds for all N. A detailed solution is provided, confirming that f(N) equals 1 + √2, resulting in the final expression equaling 8. The conversation highlights mathematical exploration and verification of conjectures through both numerical and algebraic methods.
Albert1
Messages
1,221
Reaction score
0
$$\left(\sum_{n=1}^{9999}\frac{\sqrt{100+\sqrt{n}}}{\sqrt{100-\sqrt{n}}}+\sum_{n=1}^{9999}\frac{\sqrt{100-\sqrt{n}}}{\sqrt{100+\sqrt{n}}}\right)^2$$
 
Last edited:
Mathematics news on Phys.org
Albert said:
$(\sum_{n=1}^{9999}\dfrac{\sqrt{100+\sqrt n}}{\sqrt{100-\sqrt n}}
+\sum_{n=1}^{9999}\dfrac{\sqrt{100-\sqrt n}}{\sqrt{100+\sqrt n}})^2$

I would try to simplify these sums by multiplying the top and bottom of each by the bottom's conjugate.
 
Albert said:
$(\sum_{n=1}^{9999}\dfrac{\sqrt{100+\sqrt n}}{\sqrt{100-\sqrt n}}
+\sum_{n=1}^{9999}\dfrac{\sqrt{100-\sqrt n}}{\sqrt{100+\sqrt n}})^2$
sorry it should be:

$$\left(\frac{\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}}{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}
+\frac{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}{\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}}\right)^2$$
 
Albert said:
$$\left(\frac{\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}}{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}
+\frac{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}{\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}}\right)^2$$
8
[sp]Generalisation. This is a particular case ($N=100$) of this problem:

Find $$\biggl(f(N) + \frac1{f(N)}\biggr)^{\!2}$$, where $$f(N) = \frac{\sum\limits_{n=1}^{N^2-1}\sqrt{N+\sqrt n}}{\sum\limits_{n=1}^{N^2-1}\sqrt{N-\sqrt n}}.$$

Experimentation. To see which way the wind is blowing, I got out my trusty pocket calculator and found a numerical expression for $f(2)$: $$f(2) = \frac{\sqrt{2 + \sqrt1} + \sqrt{2 + \sqrt2} + \sqrt{2 + \sqrt3}} {\sqrt{2 - \sqrt1} + \sqrt{2 - \sqrt2} + \sqrt{2 - \sqrt3}} \approx \frac{1.732 + 1.848 + 1.932}{1.000 + 0.765 + 0.518} = \frac{5.512}{2.283} \approx 2.414.$$ I recognised that answer as being suspiciously close to $1+\sqrt2$, but I could not see either how to prove that or how to generalise it for higher values of $N$. So I did the same calculation for $N=3$, and found to my surprise that answer again looked like $1+\sqrt2$. This made me wonder what the denominator of the fraction for $f(2)$ would look like if I multiplied it by $\sqrt2$, and that led to the following table: $$ \begin{array}{r|cccl} \text{row 1} & 1.732 & 1.848 & 1.932 & \text{(numerator)} \\ \text{row 2} & 1.000 & 0.765 & 0.518 & \text{(denominator)} \\ \text{row 3} & 1.414 & 1.082 & 0.732 & \text{(row 2 times $\sqrt2$)} \\ \text{row 4} & 0.732 & 1.082 & 1.414 & \text{(row 3 reversed)} \\ \text{row 5} & 1.732 & 1.848 & 1.932 & \text{(row 2 plus row 4)} \end{array}$$ Row 5 is the same as row 1! That gave me the idea for proving that $f(N) = 1 + \sqrt2$ for all $N$.

Solution. For $1\leqslant n< N$, $$\begin{aligned} 2\bigl(N - \sqrt{N^2-n}\bigr) &= 2N - 2\sqrt{N^2-n} \\ &= (N + \sqrt n) - 2 \sqrt{(N+\sqrt n)(N - \sqrt n)} + (N - \sqrt n) \\ &= \bigl(\sqrt{N + \sqrt n} - \sqrt{N - \sqrt n}\bigr)^2. \end{aligned}$$ Take the positive square root of both sides, to get $\sqrt2\sqrt{N - \sqrt{N^2-n}} = \sqrt{N + \sqrt n} - \sqrt{N - \sqrt n}$ and therefore $\sqrt{N - \sqrt n} + \sqrt2\sqrt{N - \sqrt{N^2-n}} = \sqrt{N + \sqrt n}.$ Sum that from $n=1$ to $N^2-1$: $$ \sum_{n=1}^{N^2-1}\sqrt{N - \sqrt n}\: + \sum_{n=1}^{N^2-1}\sqrt2\sqrt{N - \sqrt{N^2-n}} = \sum_{n=1}^{N^2-1}\sqrt{N + \sqrt n}.$$ Now reverse the order of summation for the second sum on the left side, by replacing $n$ by $N^2-n$: $$ \sum_{n=1}^{N^2-1}\sqrt{N - \sqrt n}\: + \sum_{n=1}^{N^2-1}\sqrt2\sqrt{N - \sqrt n} = \sum_{n=1}^{N^2-1}\sqrt{N + \sqrt n}.$$ The left side is then $$\bigl(1+\sqrt2\bigr)\sqrt{N - \sqrt n}$$, from which it follows that $f(N) = 1+\sqrt2.$

Conclusion. $$\biggl(f(N) + \frac1{f(N)}\biggr)^{\!2} = \bigl((1+\sqrt2) + (1+\sqrt2)^{-1}\bigr)^2 =\bigl((1+\sqrt2) + (\sqrt2 - 1)\bigr)^2 = \bigl(2\sqrt2\bigr)^2 = 8.$$[/sp]
 
Opalg's has given one so detail solution and my hat is off to him!

My solution that is essentially quite the same as Opalg's method:

If we let

$a=\sqrt{100+\sqrt n}-\sqrt{100-\sqrt n}$

Then squaring, rearranging and taking square root of it we have

$a^2=100+\sqrt n+100-\sqrt n-2(\sqrt{100+\sqrt n})(\sqrt{100-\sqrt n})$

$a^2=200--2\sqrt{100^2-n})$

$a^2=2(100-\sqrt{100^2-n})$

$a=\sqrt{2}\sqrt{(100-\sqrt{100^2-n})}$

$\therefore \sqrt{2}\sqrt{(100-\sqrt{100^2-n})}=\sqrt{100+\sqrt n}-\sqrt{100-\sqrt n}$

Notice that

$\begin{align*}\displaystyle \sum\limits_{n=1}^{9999}\sqrt{2}\sqrt{(100-\sqrt{100^2-n})}&=\sqrt{2}\sum\limits_{n=1}^{9999}\sqrt{(100-\sqrt{100^2-n})}\\&=\sqrt{2}\left(\sqrt{100-\sqrt{9999}}+\sqrt{100-\sqrt{9998}}+\cdots +\sqrt{100-\sqrt{2}}+\sqrt{100-\sqrt{1}}\right)\\&=\sqrt{2}\left(\sqrt{100-\sqrt{1}}+\sqrt{100-\sqrt{2}}+\cdots +\sqrt{100-\sqrt{9998}}+\sqrt{100-\sqrt{9999}}\right)\\&=\sqrt{2}\sum\limits_{n=1}^{9999}\sqrt{(100-\sqrt{n})}\end{align*}$

So we have
$\displaystyle \sum\limits_{n=1}^{9999} \sqrt{2}\sqrt{(100-\sqrt{100^2-n})}= \sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}- \sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}$

$\displaystyle\sqrt{2}\sum\limits_{n=1}^{9999}\sqrt{(100-\sqrt{n})}=\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}- \sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}$

$\displaystyle\sum\limits_{n=1}^{9999}\sqrt{(100-\sqrt{n})}(\sqrt{2}+1)=\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}$

Replacing this into the first fraction inside the square of the expression gives

$\begin{align*}\displaystyle \frac{\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}}{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}&=\frac{\sum\limits_{n=1}^{9999}\sqrt{(100-\sqrt{n})}(\sqrt{2}+1)}{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}\\&=\sqrt{2}+1\end{align*}$

Hence,

$\begin{align*}\displaystyle \frac{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}{\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}}&=\dfrac{1}{\sqrt{2}+1}\\&=\sqrt{2}-1 \end{align*}$

Last, we see that

$\displaystyle \left(\frac{\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}}{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}
+\frac{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}{\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}}\right)^2=(\sqrt{2}+1+\sqrt{2}-1)^2=8$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Back
Top