Prob. 5, Sec. 6.2, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: How to show this function is strictly decreasing using derivativeProb. 2, Sec. 3.3, in INTRO. TO REAL ANALYSIS by Bartle & Sherbert: What is the limit of this sequence?Prob. 6, Sec. 3.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: How to find this limit?Prob. 7, Sec. 3.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: Is this sequence convergent?Prob. 10, Sec. 3.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: Is this sequence convergent?Prob. 14, Sec. 3.4, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: The supremum of a bounded sequence not in the rangeProb. 5 (e), Sec. 4.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: How to find $lim_xto 0- fracsqrtx+1x$?Prob. 12, Sec. 5.2, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: Continuity of additive functionsProb. 14, Sec. 5.4, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: Every continuous periodic function is bounded and uniformly continuousDefinition 5.6.6 in Bartle & Sherbert's INTRO TO REAL ANALYSIS: How to show that this notion is well-defined?Prob. 17, Sec. 6.1, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: Straddle Lemma
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Prob. 5, Sec. 6.2, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: How to show this function is strictly decreasing using derivative
Prob. 2, Sec. 3.3, in INTRO. TO REAL ANALYSIS by Bartle & Sherbert: What is the limit of this sequence?Prob. 6, Sec. 3.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: How to find this limit?Prob. 7, Sec. 3.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: Is this sequence convergent?Prob. 10, Sec. 3.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: Is this sequence convergent?Prob. 14, Sec. 3.4, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: The supremum of a bounded sequence not in the rangeProb. 5 (e), Sec. 4.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: How to find $lim_xto 0- fracsqrtx+1x$?Prob. 12, Sec. 5.2, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: Continuity of additive functionsProb. 14, Sec. 5.4, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: Every continuous periodic function is bounded and uniformly continuousDefinition 5.6.6 in Bartle & Sherbert's INTRO TO REAL ANALYSIS: How to show that this notion is well-defined?Prob. 17, Sec. 6.1, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: Straddle Lemma
$begingroup$
Here is Prob. 5, Sec. 6.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Let $a > b > 0$ and let $n in mathbbN$ satisfy $n geq 2$. Prove that $a^1/n - b^1/n < (a-b)^1/n$. [Hint: Show that $f(x) colon= x^1/n - (x-1)^1/n$ is decreasing for $x geq 1$, and evaluate $f$ at $1$ and $a/b$.]
My Attempt:
We find that for $x > 1$,
$$
beginalign
f^prime(x) &= frac1nx^frac1n - 1 - frac1n(x-1)^frac1n - 1 \
&= frac1n left( fracx^1/nx - frac(x-1)^1/nx-1 right) \
&= frac1nx(x-1) left( x^1/n(x-1) - x(x-1)^1/n right) \
&= fracx^1/n(x-1)^1/nnx(x-1) left( (x-1) - x right) \
&= -frac1nx^1-frac1n (x-1)^1-frac1n .
endalign
$$
Thus $f^prime(x) < 0$ for all $x > 1$. Therefore the function $f$ is strictly decreasing on the interval $[1, +infty)$. That is, for any $x > 1$ we have $f(x) < f(1)$.
Now if $a > b > 0$, then $a/b > 1$, and so we have
$$ f(a/b) < f(1),$$
that is,
$$ left(fracabright)^frac1n - left(fracab - 1right)^frac1n < 1,$$
which amounts to
$$ fraca^1/n - (a-b)^1/n b^1/n < 1, $$
which implies
$$ a^1/n - (a-b)^1/n < b^1/n,$$
and hence
$$ a^1/n - b^1/n < (a-b)^1/n, $$
as required.
Are there any issues with this proof?
real-analysis calculus proof-verification derivatives inequality
$endgroup$
add a comment |
$begingroup$
Here is Prob. 5, Sec. 6.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Let $a > b > 0$ and let $n in mathbbN$ satisfy $n geq 2$. Prove that $a^1/n - b^1/n < (a-b)^1/n$. [Hint: Show that $f(x) colon= x^1/n - (x-1)^1/n$ is decreasing for $x geq 1$, and evaluate $f$ at $1$ and $a/b$.]
My Attempt:
We find that for $x > 1$,
$$
beginalign
f^prime(x) &= frac1nx^frac1n - 1 - frac1n(x-1)^frac1n - 1 \
&= frac1n left( fracx^1/nx - frac(x-1)^1/nx-1 right) \
&= frac1nx(x-1) left( x^1/n(x-1) - x(x-1)^1/n right) \
&= fracx^1/n(x-1)^1/nnx(x-1) left( (x-1) - x right) \
&= -frac1nx^1-frac1n (x-1)^1-frac1n .
endalign
$$
Thus $f^prime(x) < 0$ for all $x > 1$. Therefore the function $f$ is strictly decreasing on the interval $[1, +infty)$. That is, for any $x > 1$ we have $f(x) < f(1)$.
Now if $a > b > 0$, then $a/b > 1$, and so we have
$$ f(a/b) < f(1),$$
that is,
$$ left(fracabright)^frac1n - left(fracab - 1right)^frac1n < 1,$$
which amounts to
$$ fraca^1/n - (a-b)^1/n b^1/n < 1, $$
which implies
$$ a^1/n - (a-b)^1/n < b^1/n,$$
and hence
$$ a^1/n - b^1/n < (a-b)^1/n, $$
as required.
Are there any issues with this proof?
real-analysis calculus proof-verification derivatives inequality
$endgroup$
add a comment |
$begingroup$
Here is Prob. 5, Sec. 6.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Let $a > b > 0$ and let $n in mathbbN$ satisfy $n geq 2$. Prove that $a^1/n - b^1/n < (a-b)^1/n$. [Hint: Show that $f(x) colon= x^1/n - (x-1)^1/n$ is decreasing for $x geq 1$, and evaluate $f$ at $1$ and $a/b$.]
My Attempt:
We find that for $x > 1$,
$$
beginalign
f^prime(x) &= frac1nx^frac1n - 1 - frac1n(x-1)^frac1n - 1 \
&= frac1n left( fracx^1/nx - frac(x-1)^1/nx-1 right) \
&= frac1nx(x-1) left( x^1/n(x-1) - x(x-1)^1/n right) \
&= fracx^1/n(x-1)^1/nnx(x-1) left( (x-1) - x right) \
&= -frac1nx^1-frac1n (x-1)^1-frac1n .
endalign
$$
Thus $f^prime(x) < 0$ for all $x > 1$. Therefore the function $f$ is strictly decreasing on the interval $[1, +infty)$. That is, for any $x > 1$ we have $f(x) < f(1)$.
Now if $a > b > 0$, then $a/b > 1$, and so we have
$$ f(a/b) < f(1),$$
that is,
$$ left(fracabright)^frac1n - left(fracab - 1right)^frac1n < 1,$$
which amounts to
$$ fraca^1/n - (a-b)^1/n b^1/n < 1, $$
which implies
$$ a^1/n - (a-b)^1/n < b^1/n,$$
and hence
$$ a^1/n - b^1/n < (a-b)^1/n, $$
as required.
Are there any issues with this proof?
real-analysis calculus proof-verification derivatives inequality
$endgroup$
Here is Prob. 5, Sec. 6.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Let $a > b > 0$ and let $n in mathbbN$ satisfy $n geq 2$. Prove that $a^1/n - b^1/n < (a-b)^1/n$. [Hint: Show that $f(x) colon= x^1/n - (x-1)^1/n$ is decreasing for $x geq 1$, and evaluate $f$ at $1$ and $a/b$.]
My Attempt:
We find that for $x > 1$,
$$
beginalign
f^prime(x) &= frac1nx^frac1n - 1 - frac1n(x-1)^frac1n - 1 \
&= frac1n left( fracx^1/nx - frac(x-1)^1/nx-1 right) \
&= frac1nx(x-1) left( x^1/n(x-1) - x(x-1)^1/n right) \
&= fracx^1/n(x-1)^1/nnx(x-1) left( (x-1) - x right) \
&= -frac1nx^1-frac1n (x-1)^1-frac1n .
endalign
$$
Thus $f^prime(x) < 0$ for all $x > 1$. Therefore the function $f$ is strictly decreasing on the interval $[1, +infty)$. That is, for any $x > 1$ we have $f(x) < f(1)$.
Now if $a > b > 0$, then $a/b > 1$, and so we have
$$ f(a/b) < f(1),$$
that is,
$$ left(fracabright)^frac1n - left(fracab - 1right)^frac1n < 1,$$
which amounts to
$$ fraca^1/n - (a-b)^1/n b^1/n < 1, $$
which implies
$$ a^1/n - (a-b)^1/n < b^1/n,$$
and hence
$$ a^1/n - b^1/n < (a-b)^1/n, $$
as required.
Are there any issues with this proof?
real-analysis calculus proof-verification derivatives inequality
real-analysis calculus proof-verification derivatives inequality
asked 9 hours ago
Saaqib MahmoodSaaqib Mahmood
8,27742583
8,27742583
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is an error; the factorisation $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)-x)$ is incorrect and should be $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)^fracn-1n-x^fracn-1n)$.
To correctly prove this, note that since $frac1n-1$ is negative we have $x^frac1n-1$ is decreasing. This implies $(x-1)^frac1n-1 > x^frac1n-1 $ since $x-1<x$, and so $f'(x)$ is clearly negative as required.
$endgroup$
$begingroup$
thank you for the correction. Can you please also suggest as to how to proceed from that point on?
$endgroup$
– Saaqib Mahmood
8 hours ago
$begingroup$
@SaaqibMahmood I've edited my answer to include the corrected proof.
$endgroup$
– auscrypt
8 hours ago
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
There is an error; the factorisation $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)-x)$ is incorrect and should be $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)^fracn-1n-x^fracn-1n)$.
To correctly prove this, note that since $frac1n-1$ is negative we have $x^frac1n-1$ is decreasing. This implies $(x-1)^frac1n-1 > x^frac1n-1 $ since $x-1<x$, and so $f'(x)$ is clearly negative as required.
$endgroup$
$begingroup$
thank you for the correction. Can you please also suggest as to how to proceed from that point on?
$endgroup$
– Saaqib Mahmood
8 hours ago
$begingroup$
@SaaqibMahmood I've edited my answer to include the corrected proof.
$endgroup$
– auscrypt
8 hours ago
add a comment |
$begingroup$
There is an error; the factorisation $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)-x)$ is incorrect and should be $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)^fracn-1n-x^fracn-1n)$.
To correctly prove this, note that since $frac1n-1$ is negative we have $x^frac1n-1$ is decreasing. This implies $(x-1)^frac1n-1 > x^frac1n-1 $ since $x-1<x$, and so $f'(x)$ is clearly negative as required.
$endgroup$
$begingroup$
thank you for the correction. Can you please also suggest as to how to proceed from that point on?
$endgroup$
– Saaqib Mahmood
8 hours ago
$begingroup$
@SaaqibMahmood I've edited my answer to include the corrected proof.
$endgroup$
– auscrypt
8 hours ago
add a comment |
$begingroup$
There is an error; the factorisation $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)-x)$ is incorrect and should be $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)^fracn-1n-x^fracn-1n)$.
To correctly prove this, note that since $frac1n-1$ is negative we have $x^frac1n-1$ is decreasing. This implies $(x-1)^frac1n-1 > x^frac1n-1 $ since $x-1<x$, and so $f'(x)$ is clearly negative as required.
$endgroup$
There is an error; the factorisation $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)-x)$ is incorrect and should be $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)^fracn-1n-x^fracn-1n)$.
To correctly prove this, note that since $frac1n-1$ is negative we have $x^frac1n-1$ is decreasing. This implies $(x-1)^frac1n-1 > x^frac1n-1 $ since $x-1<x$, and so $f'(x)$ is clearly negative as required.
edited 8 hours ago
answered 9 hours ago
auscryptauscrypt
6,321613
6,321613
$begingroup$
thank you for the correction. Can you please also suggest as to how to proceed from that point on?
$endgroup$
– Saaqib Mahmood
8 hours ago
$begingroup$
@SaaqibMahmood I've edited my answer to include the corrected proof.
$endgroup$
– auscrypt
8 hours ago
add a comment |
$begingroup$
thank you for the correction. Can you please also suggest as to how to proceed from that point on?
$endgroup$
– Saaqib Mahmood
8 hours ago
$begingroup$
@SaaqibMahmood I've edited my answer to include the corrected proof.
$endgroup$
– auscrypt
8 hours ago
$begingroup$
thank you for the correction. Can you please also suggest as to how to proceed from that point on?
$endgroup$
– Saaqib Mahmood
8 hours ago
$begingroup$
thank you for the correction. Can you please also suggest as to how to proceed from that point on?
$endgroup$
– Saaqib Mahmood
8 hours ago
$begingroup$
@SaaqibMahmood I've edited my answer to include the corrected proof.
$endgroup$
– auscrypt
8 hours ago
$begingroup$
@SaaqibMahmood I've edited my answer to include the corrected proof.
$endgroup$
– auscrypt
8 hours ago
add a comment |
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