POTD
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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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This is a pure math question, but its quite good.
Prove that for all x>=1
(x-1)/x <= ln x
x is real, or integer?
"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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x is real, or integer?
"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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ok, If I do this (and I can because x>=1): (x-1)/x <= ln(x) 1-(1/x) <= ln(x) 1 <= ln(x) + 1/x for x=1 both sides are equal. Now, since ln(x) is raising function*, and 1/x >0 for all x>1, 1 < ln(x) + 1/x obviously. *in case it's called differently, I mean that ln(x) > ln(y) <=> x > y
"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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ok, If I do this (and I can because x>=1): (x-1)/x <= ln(x) 1-(1/x) <= ln(x) 1 <= ln(x) + 1/x for x=1 both sides are equal. Now, since ln(x) is raising function*, and 1/x >0 for all x>1, 1 < ln(x) + 1/x obviously. *in case it's called differently, I mean that ln(x) > ln(y) <=> x > y
"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:
1 <= ln(x) + 1/x
That doesn't mean anything, you must prove the whole side : (x-1)/x <= ln x
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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
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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ok, If I do this (and I can because x>=1): (x-1)/x <= ln(x) 1-(1/x) <= ln(x) 1 <= ln(x) + 1/x for x=1 both sides are equal. Now, since ln(x) is raising function*, and 1/x >0 for all x>1, 1 < ln(x) + 1/x obviously. *in case it's called differently, I mean that ln(x) > ln(y) <=> x > y
"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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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!