POTD
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They are equivalent. :confused:
"Throughout human history, we have been dependent on machines to survive. Fate, it seems, is not without a sense of irony. " - Morpheus "Real men use mspaint for writing code and notepad for designing graphics." - Anna-Jayne Metcalfe
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I see my mistake, ln(1)!=1 :(( No wait I don't...
"Throughout human history, we have been dependent on machines to survive. Fate, it seems, is not without a sense of irony. " - Morpheus "Real men use mspaint for writing code and notepad for designing graphics." - Anna-Jayne Metcalfe
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I see my mistake, ln(1)!=1 :(( No wait I don't...
"Throughout human history, we have been dependent on machines to survive. Fate, it seems, is not without a sense of irony. " - Morpheus "Real men use mspaint for writing code and notepad for designing graphics." - Anna-Jayne Metcalfe
ln(1)=0 Also if you've drawn the graph of the two funtions with each other, you will solve it for well.
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ln(1)=0 Also if you've drawn the graph of the two funtions with each other, you will solve it for well.
Picture is not proof.
"Throughout human history, we have been dependent on machines to survive. Fate, it seems, is not without a sense of irony. " - Morpheus "Real men use mspaint for writing code and notepad for designing graphics." - Anna-Jayne Metcalfe
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hmm, I'll still use 1<=ln(x)+1/x for x in (1,e), function ln(x) + 1/x is raising as well, because it's derivation, (ln(x)+1/x)' = (1/x -1/(x^2)) = (x-1)/(x^2) > 0 for x in (1,e).
"Throughout human history, we have been dependent on machines to survive. Fate, it seems, is not without a sense of irony. " - Morpheus "Real men use mspaint for writing code and notepad for designing graphics." - Anna-Jayne Metcalfe
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hmm, I'll still use 1<=ln(x)+1/x for x in (1,e), function ln(x) + 1/x is raising as well, because it's derivation, (ln(x)+1/x)' = (1/x -1/(x^2)) = (x-1)/(x^2) > 0 for x in (1,e).
"Throughout human history, we have been dependent on machines to survive. Fate, it seems, is not without a sense of irony. " - Morpheus "Real men use mspaint for writing code and notepad for designing graphics." - Anna-Jayne Metcalfe
dnh wrote:
I'll still use 1<=ln(x)+1/x
You use it as if it's true, I want you to prove it, not use it to prove itself.
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dnh wrote:
I'll still use 1<=ln(x)+1/x
You use it as if it's true, I want you to prove it, not use it to prove itself.
No I don't!
"Throughout human history, we have been dependent on machines to survive. Fate, it seems, is not without a sense of irony. " - Morpheus "Real men use mspaint for writing code and notepad for designing graphics." - Anna-Jayne Metcalfe
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dnh wrote:
I'll still use 1<=ln(x)+1/x
You use it as if it's true, I want you to prove it, not use it to prove itself.
1<=ln(x)+1/x is equivalent to (x-1)/x <= ln x. I am proving 1<=ln(x)+1/x... And my proof is that for x = 1, ln(x) + 1/x = 1. for x>1, ln(x) + 1/x > 1 because it's raising function, as I proved by showing it's derivation is > 0 on interval (1,e). for x>e, I proved it before.
"Throughout human history, we have been dependent on machines to survive. Fate, it seems, is not without a sense of irony. " - Morpheus "Real men use mspaint for writing code and notepad for designing graphics." - Anna-Jayne Metcalfe
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This is a pure math question, but its quite good.
Prove that for all x>=1
(x-1)/x <= ln x
Easy - use the following series expansion[^] ln = (x-1)/x + 1/2 ((x-1)/x)^2 + 1/3 ((x-1)/x)^3 + ... which is valid for x > 1/2 when x > 1 the second and subsequent terms are positive. For x = 0 all terms are zero QED
Graham My signature is not black, just a very, very dark blue
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Easy - use the following series expansion[^] ln = (x-1)/x + 1/2 ((x-1)/x)^2 + 1/3 ((x-1)/x)^3 + ... which is valid for x > 1/2 when x > 1 the second and subsequent terms are positive. For x = 0 all terms are zero QED
Graham My signature is not black, just a very, very dark blue
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Easy - use the following series expansion[^] ln = (x-1)/x + 1/2 ((x-1)/x)^2 + 1/3 ((x-1)/x)^3 + ... which is valid for x > 1/2 when x > 1 the second and subsequent terms are positive. For x = 0 all terms are zero QED
Graham My signature is not black, just a very, very dark blue
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Easy - use the following series expansion[^] ln = (x-1)/x + 1/2 ((x-1)/x)^2 + 1/3 ((x-1)/x)^3 + ... which is valid for x > 1/2 when x > 1 the second and subsequent terms are positive. For x = 0 all terms are zero QED
Graham My signature is not black, just a very, very dark blue
Graham Shanks wrote:
when x > 1 the second and subsequent terms are positive.
Right
Graham Shanks wrote:
For x = 0 all terms are zero
For x = 1, not 0, because x cannot be zero.
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dnh wrote:
1 <= ln(x) + 1/x
That doesn't mean anything, you must prove the whole side : (x-1)/x <= ln x
Yes it does.
-- Raaaaaaaaaaaaaaaaaaaaa!
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1<=ln(x)+1/x is equivalent to (x-1)/x <= ln x. I am proving 1<=ln(x)+1/x... And my proof is that for x = 1, ln(x) + 1/x = 1. for x>1, ln(x) + 1/x > 1 because it's raising function, as I proved by showing it's derivation is > 0 on interval (1,e). for x>e, I proved it before.
"Throughout human history, we have been dependent on machines to survive. Fate, it seems, is not without a sense of irony. " - Morpheus "Real men use mspaint for writing code and notepad for designing graphics." - Anna-Jayne Metcalfe
You are correct. I don't know what he's been smoking. :confused:
-- Raaaaaaaaaaaaaaaaaaaaa!
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This is a pure math question, but its quite good.
Prove that for all x>=1
(x-1)/x <= ln x
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Why, when it is far simpler to do what dnh did? He basically said that:
(x-1)/x <= ln x is equivalent to 1<=ln(x)+1/x
Since we know the properties of ln and ^-1, it's easy to show (by induction) that for all x >= 1, that ln(x) + 1/x >= 1. Series are creepy. :~
-- Raaaaaaaaaaaaaaaaaaaaa!
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Why, when it is far simpler to do what dnh did? He basically said that:
(x-1)/x <= ln x is equivalent to 1<=ln(x)+1/x
Since we know the properties of ln and ^-1, it's easy to show (by induction) that for all x >= 1, that ln(x) + 1/x >= 1. Series are creepy. :~
-- Raaaaaaaaaaaaaaaaaaaaa!
Can't use mathematical induction. That only works on the natural numbers (one of the steps is to prove that if it works on x then it works for x+1). The real numbers are not countable and therefore induction will not work. New POTD: prove that the real numbers are uncountable POTD #2: prove that the rational numbers are countable POTD #3: prove that there are exactly the same number of rational numbers as there are positive integers
Graham My signature is not black, just a very, very dark blue
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Can't use mathematical induction. That only works on the natural numbers (one of the steps is to prove that if it works on x then it works for x+1). The real numbers are not countable and therefore induction will not work. New POTD: prove that the real numbers are uncountable POTD #2: prove that the rational numbers are countable POTD #3: prove that there are exactly the same number of rational numbers as there are positive integers
Graham My signature is not black, just a very, very dark blue
Graham Shanks wrote:
Can't use mathematical induction.
That's reason why I asked if x was real. But you can easily show that f(x) > f(y) <=> x>y with first derivation of f.
"Throughout human history, we have been dependent on machines to survive. Fate, it seems, is not without a sense of irony. " - Morpheus "Real men use mspaint for writing code and notepad for designing graphics." - Anna-Jayne Metcalfe
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Can't use mathematical induction. That only works on the natural numbers (one of the steps is to prove that if it works on x then it works for x+1). The real numbers are not countable and therefore induction will not work. New POTD: prove that the real numbers are uncountable POTD #2: prove that the rational numbers are countable POTD #3: prove that there are exactly the same number of rational numbers as there are positive integers
Graham My signature is not black, just a very, very dark blue
Graham Shanks wrote:
That only works on the natural numbers (one of the steps is to prove that if it works on x then it works for x+1). The real numbers are not countable and therefore induction will not work.
Maybe I used the wrong term (didn't know induction was reserved for natural numbers only). If you can show that f(x + dx) > f(x) where dx >, that equation which dnh originally proposed, must be equally valid as any series... right? :~ We know that ln(x+dx) > ln(x) for all x > 1, and we also know that 1/x > 0 for all x > 1. The initial number x = 1 yields ln(1) + 1/1 = 1. I just don't see why you have to bring in fancy pants series to solve this problem. That seems to me like using the sledgehammer just to hit tiny nails...
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