Comparing Two Sums of Radicals in Two Ways
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- Опубликовано: 18 сен 2024
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6:39 It would be incorrect to write f(x) is increasing for (-infinity, 102) because f(x) is only defined for the domain x belongs to (0,204) not from (-infinity, infinity)
That's right!
Elevo al quadrato 101+103+2√101√103 vs 99+105+2√99√105...semplifico √101√03 vs √99√105..elevo al quadrato 101*103 vs 99*105...102^2-1 vs (102-3)(102+3)=102^2-9...quindi √101+√103 Is greater
I did a variation on method 2: A=√(102-1) + √(102+1) so A²=102-1+102+1+2√(102-1)(102+1)=2*102+2√(102²-1²). Do the same with B and you end up with B²=2*102+2√(102²-3²). Since 3>1 B²B
Before video:
2+5 = 3+4
put both into roots.
root 4 + root 25 = root 9 + root 16
4+25 > 9 + 16
Therefore, the right side is more efficient use of the rooted numbers
Therefore, the closer together the roots are the bigger it is.
Second way to solve:
Taken to logic extreme,
root of (205) + root of (0) < root 101 + root 103, we can just clearly intuit that the right half is 20 ish, but the left is less than 20^2 (400). Gives the same general principle.
Edit: 204, not 205. Oops.
Secondly, i loved the slow methodical calculus interpretation. My solutions are very impatient compared to that, great stuff.
Good thinking!
Assume f(x)=sqrt(x) with derivative f'(x)=(1/2)*(1/sqrt(x))
Notice that f satisfies the conditions of the Mean Value Theorem in [99,101] and [103,105]
therefore there'll be x1E(99,101) and x2E(103,105) such that, after some simplifications and application of MVT, f'(x1)=(sqrt(101)-sqrt(99))/2 and f'(x2)=(sqrt(105)-sqrt(103))/2
Notice that f' is strictly decreasing in its domain and also x1 sqrt(101)-sqrt(99) > sqrt(105)-sqrt(103) => sqrt(101) + sqrt(103) > sqrt(105) + sqrt(99)
Wow!
At the end we have to compare 101*103 and 99*105 --->
99*105 = (101-2)*(103+2) = 101*103 +101*2 - 2*103 -2*2 = 101*103 - 2*(103-101) -4
99*105 = 101*103 - 8 --->sqrt(101) + sqrt(103) > sqrt(99) + sqrt(105)
A really nice way of approaching it is by taking the numbers sqrt(105) -sqrt(103) and sqrt(101)-sqrt(99) and rearanging them to 2 /((sqrt(101)+sqrt(99) ) and 2/((sqrt(103) +sqrt(105))...the last number clearly is shorter than first one since it has a larger denominator and it allowes us to realize that sqrt(105)-sqrt(103) is shorter than sqrt(101) -sqrt(99) and we get the answer!
I like this one.
Straight off the bat, you notice the issue, the "difference of 2". So the question is, which is more important, going from 101 to 99, or 103 to 105?
If you picture the square root function (or think about its inverse and work from there), you might notice that it's a concave function, and this is enough to tell you that the increase from sqrt(103) to sqrt(105) is less than the decrease from sqrt(101) to sqrt(99).
So this convinces you that the first one is greater, but isn't too formal.
For formality, you could take a calculus approach, thinking about the function sqrt(x) - sqrt(x-2) and you can take the derivative, note that it's positive, confirming that it's an increasing function.
I'm sure there's an analytical approach too, which is more rigorous still, but I've never been brilliant with analysis.
Edit: Ah, method 2 is so slick! I'm a tad gutted I didn't even consider it.
I did something like method 1, with an argument, that sqrt(x) does slow down while increasing. So if it was linear, they would be the same. Below that the first would be greater.
But the second one is cooler!
I used calculus:
d/dx(sqrt(x))=1/(2sqrt(x))
Since x is in the denominator, I know that the function f(x)=sqrt(x) increases more slowly as the x values increase.
This indicates that (sqrt(101)-sqrt(99))>(sqrt(105)-sqrt(103)). Therefore, (sqrt(101)+sqrt(103))>(sqrt(99)+sqrt(105)).
Enjoyed both methods! My calc isn't good enough to have found the first one, but I feel like I could conceivably have found the second one.
You got this!
At a glance I can say that √101 + √103 is greater. It because of how the function √x behaves.
It would also be true if we were dealing with logarithms.
log 101 + log 103 would also come out to be greater than log 99 + log 105
When It comes to math, "at a glance" means nothing. A well known math problem is to prove that there's no natural number in ]0,1[...would be the answer "at a glance"? Btw, being able to prove It is like finaly starting to get some wide understanding of what math is. It's Just a reflection, I don't mean to bother you.😂😂
Before I watch the video: square both expressions. Squaring a positive number >= 1 is a monotone operation, so any less than relationship is preserved when squaring. When you square you will see that what matters is the product of the two numbers, because the 204's cancel. I can't do this in my head.
I thought:
f(x)=sqrt(x) has the second derivative f''(x)=0.
Threrefore:
sqrt(101)-sqrt(99)>sqrt(105)-sqrt(103).
And we are done!
Too synthetic? Have I neglected anything?
Thanks for any feedback!
Let x = 100 so that A = √(100 +1) + √(100 + 3) and B = √(100 - 1) + (100 + 5) . . . square and compare.
No need to square the two remaining radicals.
I use basic logic. A square has the greatest area for any quadrilateral perimeter. Hence, the closer the sides are to equal, the greater the area. So, group set (101,103) is larger than (95,105).
However, what you wrote is not a proof.
احسنت وبارك الله فيك يا استاذنا
شكرا جزيلا لك!
*@ SyberMath* -- Here is a third method that is relatively very efficient.
(Use the square root symbol instead to streamline it to improve it.)
Let n = the center number of 102. Then, n - 3 = 99, n + 3 = 105, etc.
The relation becomes:
sqrt(n - 1) + sqrt(n + 1) versus sqrt(n - 3) + sqrt(n + 3)
Square each side:
n - 1 + n + 1 + 2sqrt(n - 1)*sqrt(n + 1) versus n - 3 + n + 3 + 2sqrt(n - 3)*sqrt(n + 3)
Combine like terms and then divide each side by 2:
sqrt(n - 1)*sqrt(n + 1) versus sqrt(n - 3)*sqrt(n + 3)
Square both sides:
n^2 - 1 versus n^2 - 9
Subtract n^2 from each side:
-1 > - 9
Therefore, sqrt(101) + sqrt(103) > sqrt(99) + sqrt(105).
Nice!
Let the symbols "" Be an unknown ordering, and let ">¿
You can also write "vs." between expressions to stand for "versus."
Nice!
Thanks!
101·103=(102+1)(102-1)=102^2-1
99·105=(102+3)(102-3)=102^2-9
Another method:
Denote:
x = √101 + √103 , y = √99 + √105
Therefore,
x - y = (√101 - √99) + (√103 - √105) = 2 / (√101 + √99) - 2 / (√103 + √105)
Therefore,
x - y = 2 / (√101 + √99) (√103 + √105) ( (√103 + √105) - (√101 + √99) )
Therefore,
x - y = A B
where:
A = 2 / ( (√101 + √99) (√103 + √105) ) > 0
B = (√103 + √105) - (√101 + √99) = (√103 - √99) + (√105 - √101)
However, √103 - √99 > 0 and √105 - √101 > 0 ⟹ B > 0
Therefore, x - y = AB > 0 ⟹ x > y
Therefore, √101 + √103 > √99 + √105
Second method is better than first
Taylor's expansion solves it easier. The 1st order : 0.5*(1+3) = 0.5*(-1+5). equal! The 2nd order : -(1/8)(1^2+3^2) > -(1/8)(-1^2+5^2). Therefore, sqrt(101)+sqrt(103) is larger.
-1^2 is not the square of -1. That is equal to -1. (-1)^2 is the square of -1, which equals 1.
2nd method I use. 😋😋😋😋😋😋
Great 👍
√101-√99 vs √105-√103 or 2/(√101+√99 ) vs 2/( √105+√103 ) 1 min....
Nice. That would be the third method 😍
However, you would need to fill in all of the connecting steps to give justifications and also one or two more steps after that final parenthesis in your line. What you wrote is incomplete. It is an outline.
@@robertveith6383That's why I said 1 min. My outline is just few seconds.
√101+√103 ∨ √99 +√105
square
101+103+2√(101•103) ∨
99+105+2√(99•105)
204+2√(101•103) ∨ 204+2√(99•105)
2√(101•103) ∨ 2√(99•105)
√(101•103) ∨ √(99•105)
√10403 ∨ √10395
10403 ∨ 10395
The left side is greater.
So
(√101+√103) >
(√99 +√105)