The value of the limit
L = y → ∞ lim 7 ( 8 y + 3 ) ( y + 1 ) ( y + 2 ) ( y + 3 ) ( 2 y + 3 ) ( 2 y + 5 ) ( 4 y + 5 ) − 2 y
is of the form B A where A and B are co-prime integers.
Find the value of A + B .
Note : Do NOT use wolfram alpha.
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Heh, I tried to find some sort of relationship to AM-GM (seeing 7 terms multiplied to the one seventh power).
The slope of the asymptote of the radical
7 ( 8 y + 3 ) ( y + 1 ) ( y + 2 ) ( y + 3 ) ( 2 y + 3 ) ( 2 y + 5 ) ( 4 y + 5 )
is 2 y , because the highest order term in the radical (when expanded) is 1 2 8 y 7 . Hence, this is cancelled by the subtraction of 2 y from the radical, leaving a limit value which is the same as the y intercept of the asymptote. Let a be that y intercept. Then the coefficient of the 2 n d term of both the expanded polynomial in the radical and ( 2 y + a ) 7 has to be the same. Thus a has to be
a = 1 2 8 ( 7 ) ( 2 1 ) 1 2 8 ( 8 3 + 1 1 + 1 2 + 1 3 + 2 5 + 4 5 ) = 2 8 9 3 = B A
A weird solution ----
We need to find
L = y → ∞ lim 7 ( 8 y + 3 ) ( y + 1 ) ( y + 2 ) ( y + 3 ) ( 2 y + 3 ) ( 2 y + 5 ) ( 4 y + 5 ) − 2 y .
We can rewrite it as
L = y → ∞ lim 2 × 7 ( y + 8 3 ) ( y + 1 ) ( y + 2 ) ( y + 3 ) ( y + 2 3 ) ( y + 2 5 ) ( y + 4 5 ) − 2 y
Now, consider the terms ( y + 8 3 ) , ( y + 1 ) , ( y + 2 ) , ( y + 3 ) , ( y + 2 3 ) , ( y + 2 5 ) and ( y + 4 5 ) .
As y → ∞ , all these terms become equal.
We can check this by dividing each of the above terms by every other term and then applying y → ∞ lim to it.
Hence, we can apply the fact that for equal terms, A . M = G . M .
Doing so,
y → ∞ lim 7 ( y + 8 3 ) ( y + 1 ) ( y + 2 ) ( y + 3 ) ( y + 2 3 ) ( y + 2 5 ) ( y + 4 5 ) = y → ∞ lim 7 ( y + 8 3 ) + ( y + 1 ) + ( y + 2 ) + ( y + 3 ) + ( y + 2 3 ) + ( y + 2 5 ) + ( y + 4 5 ) .
Hence,
L = y → ∞ lim 2 ( 5 6 9 3 + y ) − 2 y
L = 2 8 9 3
∴ A + B = 1 2 1
Your argument is not mathematically correct. You have the right idea, but need to figure out how to show the limit properly.
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What's the mistake in this method??..i have also solved this question by using this method.... Please tell me
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The statement of "As y → ∞ , all these terms become equal. Hence we can apply the fact that for equal terms AM= GM." is false.
For example, we know that lim 2 y and lim 2 y also approach the same limit (namely infinity). Does this mean that
lim 2 y × 2 y = lim 2 2 y + 2 y = 4 5 y ?
This approach "happens to work" for "linear factors with the same coefficient". As such, you need a proper explanation for that fact.
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@Calvin Lin
–
Can't we say that in comparison to y other terms like 3..... can be neglected
So AM=GM.
Also in your example
Lim \frac{2×y}{y÷2} =4 not 1
Please reply and correct me if I am wrong
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@Ayush Jain – As I said
This approach "happens to work" for "linear factors with the same coefficient". As such, you need a proper explanation for that fact.
My example isn't perfectly equivalent. So, what is the mathematical reasoning for why it works in the scenario that you're thinking of? Is "limit of ratios = 1" sufficient?
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@Calvin Lin
–
Let us take x=10,000. Now 2x=20,000
Whereas x+3=10,003 now what I am trying to say through this is if we multiply something to a big number it increases or decreases appreciably but if we add a small number to it it does not change much . That is why
AM=GM method is working only where coeficient of x=1.
Calvin Lin sir please reply and help me to improve myself
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What is the mathematical reasoning for why it works in the scenario that you're thinking of? Is "limit of ratios = 1" sufficient?
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@Calvin Lin – I think necessary condition are: X should tend to infinity Coefficient of x should be 1 The term added with x should be finite. Please reply. Thanks
You can check that with the following limit(s)--
L = y → ∞ lim ( y + 1 y + 8 3 ) = 1
And similarly for every other term.
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Yes, the limit of the ratio is equal to 1 is the relevant property here. Now, how do you use this to justify the interchange of limits?
Take out factor of 2 y and use Binomial Expansion on each of the sevenths roots.
Solving the problem by using CAUCHY's INEQUALITY:
L = lim y → ∞ [ 7 ( 8 y + 3 ) ( y + 1 ) ( y + 2 ) ( y + 3 ) ( 2 y + 3 ) ( 2 y + 5 ) ( 4 y + 5 ) − 2 y ] = 2 lim y → ∞ [ 7 ( y + 3 / 8 ) ( y + 1 ) ( y + 2 ) ( y + 3 ) ( y + 3 / 2 ) ( y + 5 / 2 ) ( y + 5 / 4 ) − y ]
by Cauchy's inequality:
7 ( y + 3 / 8 ) ( y + 1 ) ( y + 2 ) ( y + 3 ) ( y + 3 / 2 ) ( y + 5 / 2 ) ( y + 5 / 4 ) ≤ ≤ 7 ( y + 3 / 8 ) + ( y + 1 ) + ( y + 2 ) + ( y + 3 ) + ( y + 3 / 2 ) + ( y + 5 / 2 ) + ( y + 5 / 4 ) = y + 5 6 9 3
where equality holds for y + 3 / 8 ≈ y + 1 ≈ y + 2 ≈ y + 3 ≈ y + 3 / 2 ≈ y + 5 / 2 ≈ y + 5 / 4 it works if y → ∞
⟹ lim y → ∞ [ 7 ( y + 3 / 8 ) ( y + 1 ) ( y + 2 ) ( y + 3 ) ( y + 3 / 2 ) ( y + 5 / 2 ) ( y + 5 / 4 ) = y + 5 6 9 3
⟹ L = 2 ( y + 5 6 9 3 − y ) = 2 8 9 3 = B A ⟹ A n s w e r i s A + B = 1 2 1
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Let f ( y ) be that seventh root. We have that f ( y ) − 2 y = f ( y ) 6 + ( 2 y ) f ( y ) 5 + ⋯ + ( 2 y ) 6 f ( y ) 7 − 1 2 8 y 7 . The top of this fraction is a polynomial of degree 6 , equal to 1 4 8 8 y 6 + c 5 y 5 + ⋯ + c 0 . Divide the top and bottom by y 6 to get ( f ( y ) / y ) 6 + 2 ( f ( y ) / y ) 5 + ⋯ + 2 6 1 4 8 8 + c 5 / y + ⋯ + c 0 / y 6 .
Since y → ∞ lim y f ( y ) = 2 , the limit we want is 2 6 + 2 6 + ⋯ 2 6 1 4 8 8 = 4 4 8 1 4 8 8 = 2 8 9 3 . So the answer is 1 2 1 .