id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-016601 | Real Analysis: Series — Parameter Sensitivity | 9 | Indicate where a theorem is used: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{59}{35}}}.$$
(a) Solve using a named convergence t... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{59}{35}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{59}{35}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: T... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{59}{35}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016602 | Real Analysis: Uniform Continuity (Variant B) | 9 | Start by stating any domain restrictions: Let $f:(0,151)\to\mathbb{R}$ be $f(x)=\frac{1}{(-6)x}$. Is $f$ uniformly continuous on $(0,151)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{151}{n}$ and $y_n=\\frac{151}{2n}$ in $(0,151)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016603 | Real Analysis: Uniform Continuity (Variant A) | 9 | Challenge: Let $f:(0,638)\to\mathbb{R}$ be $f(x)=\ln(28x)$. Is $f$ uniformly continuous on $(0,638)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016604 | Real Analysis: Uniform Continuity (Variant B) | 9 | Give a theorem-based solution: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(10x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz via Mean Value Theorem",
"approach": "Use MVT to get a global Lipschitz bound $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: Apply MVT to $g(t)=\\sin(ct)$: there exists $\\xi$ between $x$ and $y$ with $g(x)-g(y)=g'(\\xi)(x-y)$.",
"Step 2: Since $g'(t)=(10)\\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"robustness_analysis": "Sensitivi... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-016605 | Real Analysis: Series — Parameter Sensitivity | 9 | Solve and justify each step: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{106}{35}}}.$$
(a) Solve using a named convergence test.... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{106}{35}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness no... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{106}{35}$, so the series is convergent. |
math-016606 | Real Analysis: Uniform Continuity (Variant C) | 9 | Warm-up: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(14x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz via Mean Value Theorem",
"approach": "Use MVT to get a global Lipschitz bound $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: Apply MVT to $g(t)=\\sin(ct)$: there exists $\\xi$ between $x$ and $y$ with $g(x)-g(y)=g'(\\xi)(x-y)$.",
"Step 2: Since $g'(t)=(14)\\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"robustnes... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-016607 | Real Analysis: Uniform Continuity (Variant C) | 9 | Indicate where a theorem is used: Let $f:(0,1691)\to\mathbb{R}$ be $f(x)=\ln(52x)$. Is $f$ uniformly continuous on $(0,1691)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016608 | Real Analysis: Uniform Continuity (Variant B) | 9 | Carefully track domains: Let $f:(0,173)\to\mathbb{R}$ be $f(x)=\ln(36x)$. Is $f$ uniformly continuous on $(0,173)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{173}{n}$ and $y_n=\\frac{173}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016609 | Real Analysis: Uniform Continuity (Core) | 9 | Compute the requested quantity: Let $f:(0,906)\to\mathbb{R}$ be $f(x)=\ln(33x)$. Is $f$ uniformly continuous on $(0,906)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{906}{n}$ and $y_n=\\frac{906}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016610 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | State any required conditions first: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{10}{33}}}.$$
(a) Solve using a named convergence test.
(b) Give an ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{10}{33}$.",
"Final step: By the p-series test, it dive... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{10}{33}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Robustness no... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{10}{33}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-016611 | Real Analysis: Uniform Continuity (Variant B) | 9 | Provide both a computational and a conceptual explanation: Let $f:(0,1540)\to\mathbb{R}$ be $f(x)=\ln(36x)$. Is $f$ uniformly continuous on $(0,1540)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016612 | Real Analysis: Series — Integral Test for Power Laws | 9 | Question: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{87}{25}}}.$$
(a) Solve using a named conv... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{87}{25}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{87}{25}$, so the series is convergent. |
math-016613 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Solve (and briefly cross-validate): Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{81}{40}}}.$$
(a) Solve using a named convergence test.
(b) Give an i... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{81}{40}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{81}{40}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016614 | Real Analysis: Uniform Continuity (Core) | 9 | Indicate where a theorem is used: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(29x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz via Mean Value Theorem",
"approach": "Use MVT to get a global Lipschitz bound $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: Apply MVT to $g(t)=\\sin(ct)$: there exists $\\xi$ between $x$ and $y$ with $g(x)-g(y)=g'(\\xi)(x-y)$.",
"Step 2: Since $g'(t)=(29)\\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-016615 | Real Analysis: Uniform Continuity (Variant C) | 9 | Be explicit about assumptions: Let $f:(0,468)\to\mathbb{R}$ be $f(x)=\ln(42x)$. Is $f$ uniformly continuous on $(0,468)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{468}{n}$ and $y_n=\\frac{468}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016616 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Give an answer and a quick verification: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{96}{25}}}.$$
(a) Solve using a named convergence test.
(b) Give... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{96}{25}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{96}{25}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{96}{25}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016617 | Real Analysis: Uniform Continuity (Variant B) | 9 | Give reasoning, not just computation: Let $f:(0,1290)\to\mathbb{R}$ be $f(x)=\ln(11x)$. Is $f$ uniformly continuous on $(0,1290)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1290}{n}$ and $y_n=\\frac{1290}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016618 | Real Analysis: Uniform Continuity (Variant B) | 9 | Solve and justify each step: Let $f:(0,1702)\to\mathbb{R}$ be $f(x)=\frac{1}{(9)x}$. Is $f$ uniformly continuous on $(0,1702)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{1702}{n}$ and $y_n=\\frac{1702}{2n}$ in $(0,1702)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016619 | Real Analysis: Uniform Continuity (Core) | 9 | Complete the analysis: Let $f:(0,688)\to\mathbb{R}$ be $f(x)=\ln(48x)$. Is $f$ uniformly continuous on $(0,688)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016620 | Real Analysis: Series — Parameter Sensitivity | 9 | Solve and include a self-check: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{43}{7}}}.$$
(a) Sol... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{43}{7}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{43}{7}$, so the series is convergent. |
math-016621 | Real Analysis: Uniform Continuity (Variant C) | 9 | Proceed methodically: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-20)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Mean Value Theorem Blow-Up",
"approach": "Use MVT to show that on unbounded domains, the derivative growth makes a uniform modulus impossible.",
"steps": [
"Step 1: Suppose uniform continuity holds. Take $\\varepsilon=1$ and let $\\delta>0$ correspond.",
"Step 2: Choose $x$... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-uniform continuity.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016622 | Real Analysis: Uniform Continuity (Variant C) | 9 | Provide a rigorous solution: Let $f:(0,884)\to\mathbb{R}$ be $f(x)=\frac{1}{(19)x}$. Is $f$ uniformly continuous on $(0,884)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016623 | Real Analysis: Uniform Continuity (Variant B) | 9 | Give a theorem-based solution: Let $f:(0,1246)\to\mathbb{R}$ be $f(x)=\frac{1}{(-24)x}$. Is $f$ uniformly continuous on $(0,1246)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{1246}{n}$ and $y_n=\\frac{1246}{2n}$ in $(0,1246)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016624 | Real Analysis: Uniform Continuity (Variant A) | 9 | Complete the analysis: Let $f:(0,759)\to\mathbb{R}$ be $f(x)=\frac{1}{(22)x}$. Is $f$ uniformly continuous on $(0,759)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}}.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016625 | Real Analysis: Uniform Continuity (Variant A) | 9 | Solve (and briefly cross-validate): Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(14)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Explicit Sequence Counterexample",
"approach": "Pick sequences approaching each other at infinity but whose squares stay separated.",
"steps": [
"Step 1: Let $x_n=n$ and $y_n=n+\\frac{1}{n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{1}{n}\\to 0$.",
"Final step: But $|f(x_n)-... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-uniform continuity.",
"rob... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016626 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Solve and justify each step: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{115}{16}}}.$$
(a) Solve using a named convergence test.
(b) Give an indepen... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{115}{16}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{115}{16}$, so the series is convergent. |
math-016627 | Real Analysis: Uniform Continuity (Variant C) | 9 | Solve and then verify: Let $f:(0,46)\to\mathbb{R}$ be $f(x)=\ln(10x)$. Is $f$ uniformly continuous on $(0,46)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016628 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Show all reasoning: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{27}{14}}}.$$
(a) Solve using a named co... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{27}{14}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{27}{14}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016629 | Real Analysis: Uniform Continuity (Variant C) | 9 | Prompt: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(9)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Explicit Sequence Counterexample",
"approach": "Pick sequences approaching each other at infinity but whose squares stay separated.",
"steps": [
"Step 1: Let $x_n=n$ and $y_n=n+\\frac{1}{n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{1}{n}\\to 0$.",
"Final step: But $|f(x_n)-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-uniform continuity.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016630 | Real Analysis: Series — p-Series Threshold | 9 | Explain each transformation: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{24}{7}}}.$$
(a) Solve ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{24}{7}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{24}{7}$, so the series is convergent. |
math-016631 | Real Analysis: Series — Parameter Sensitivity | 9 | Find the exact value: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{32}{9}}}.$$
(a) Solve using a named convergence test.
(b) Give... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{32}{9}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{32}{9}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a special... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{32}{9}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016632 | Real Analysis: Uniform Continuity (Variant A) | 9 | Compute the requested quantity: Let $f:(0,1289)\to\mathbb{R}$ be $f(x)=\frac{1}{(1)x}$. Is $f$ uniformly continuous on $(0,1289)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{1289}{n}$ and $y_n=\\frac{1289}{2n}$ in $(0,1289)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016633 | Real Analysis: Series — Parameter Sensitivity | 9 | Determine the requested value: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{86}{21}}}.$$
(a) Sol... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{86}{21}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{86}{21}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{86}{21}$, so the series is convergent. |
math-016634 | Real Analysis: Uniform Continuity (Variant A) | 9 | Solve and then verify: Let $f:(0,1723)\to\mathbb{R}$ be $f(x)=\ln(28x)$. Is $f$ uniformly continuous on $(0,1723)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1723}{n}$ and $y_n=\\frac{1723}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016635 | Real Analysis: Series — p-Series Threshold | 9 | Derive the result step-by-step: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{19}{17}}}.$$
(a) So... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{19}{17}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{19}{17}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016636 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Give reasoning, not just computation: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{74}{17}}}.$$
(a) Solv... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{74}{17}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{74}{17}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{74}{17}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016637 | Real Analysis: Uniform Continuity (Variant B) | 9 | Keep the final answer in boxed form: Let $f:(0,1752)\to\mathbb{R}$ be $f(x)=\ln(2x)$. Is $f$ uniformly continuous on $(0,1752)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1752}{n}$ and $y_n=\\frac{1752}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "Sensi... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016638 | Real Analysis: Uniform Continuity (Variant B) | 9 | Find the exact value: Let $f:(0,588)\to\mathbb{R}$ be $f(x)=\ln(23x)$. Is $f$ uniformly continuous on $(0,588)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016639 | Real Analysis: Uniform Continuity (Variant C) | 9 | Find the exact value: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(26x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Characterization",
"approach": "A function is uniformly continuous iff it sends Cauchy sequences to Cauchy sequences; use the global inequality $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: For all real $x,y$, $|\\sin x-\\sin y|\\le |x-y|$ (e.g., by MVT).",... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-016640 | Real Analysis: Series — p-Series Threshold | 9 | Explain each transformation: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{53}{21}}}.$$
(a) Solve using a named convergence test.
(b) Give an independen... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{53}{21}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{53}{21}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{53}{21}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016641 | Real Analysis: Uniform Continuity (Variant A) | 9 | Solve with verification: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(-14x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz via Mean Value Theorem",
"approach": "Use MVT to get a global Lipschitz bound $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: Apply MVT to $g(t)=\\sin(ct)$: there exists $\\xi$ between $x$ and $y$ with $g(x)-g(y)=g'(\\xi)(x-y)$.",
"Step 2: Since $g'(t)=(-14)\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-016642 | Real Analysis: Uniform Continuity (Variant B) | 9 | Warm-up: Let $f:(0,1710)\to\mathbb{R}$ be $f(x)=\frac{1}{(-27)x}$. Is $f$ uniformly continuous on $(0,1710)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}}.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016643 | Real Analysis: Uniform Continuity (Variant A) | 9 | Exercise: Let $f:(0,1927)\to\mathbb{R}$ be $f(x)=\ln(27x)$. Is $f$ uniformly continuous on $(0,1927)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016644 | Real Analysis: Uniform Continuity (Variant B) | 9 | Where appropriate, name the theorem you use: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(23)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Explicit Sequence Counterexample",
"approach": "Pick sequences approaching each other at infinity but whose squares stay separated.",
"steps": [
"Step 1: Let $x_n=n$ and $y_n=n+\\frac{1}{n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{1}{n}\\to 0$.",
"Final step: But $|f(x_n)-... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-unifo... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016645 | Real Analysis: Uniform Continuity (Core) | 9 | Proceed methodically: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-23)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Explicit Sequence Counterexample",
"approach": "Pick sequences approaching each other at infinity but whose squares stay separated.",
"steps": [
"Step 1: Let $x_n=n$ and $y_n=n+\\frac{1}{n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{1}{n}\\to 0$.",
"Final step: But $|f(x_n)-... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-unifo... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016646 | Real Analysis: Uniform Continuity (Core) | 9 | State any required conditions first: Let $f:(0,429)\to\mathbb{R}$ be $f(x)=\frac{1}{(-6)x}$. Is $f$ uniformly continuous on $(0,429)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{429}{n}$ and $y_n=\\frac{429}{2n}$ in $(0,429)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016647 | Real Analysis: Series — Integral Test for Power Laws | 9 | Try to avoid pattern-matching; explain why: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{19}{20}}}.$$
(a) Solve using a named convergence test.
(b) G... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{19}{20}$.",
"Final step: By the p-series test, it dive... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{19}{20}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Generality note: The p-series test is a... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{19}{20}$, so the series is divergent. |
math-016648 | Real Analysis: Series — Parameter Sensitivity | 9 | Explain why your operations are valid: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{59}{34}}}.$$
(a) Solve using a named convergence test.
(b) Give a... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{59}{34}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{59}{34}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: T... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{59}{34}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016649 | Real Analysis: Uniform Continuity (Variant B) | 9 | Solve and include a self-check: Let $f:(0,319)\to\mathbb{R}$ be $f(x)=\ln(37x)$. Is $f$ uniformly continuous on $(0,319)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{319}{n}$ and $y_n=\\frac{319}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "Robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016650 | Real Analysis: Uniform Continuity (Variant A) | 9 | Start by stating any domain restrictions: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(24x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz via Mean Value Theorem",
"approach": "Use MVT to get a global Lipschitz bound $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: Apply MVT to $g(t)=\\sin(ct)$: there exists $\\xi$ between $x$ and $y$ with $g(x)-g(y)=g'(\\xi)(x-y)$.",
"Step 2: Since $g'(t)=(24)\\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"robustness_analysis": "Generalit... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-016651 | Real Analysis: Series — Parameter Sensitivity | 9 | Give reasoning, not just computation: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{43}{34}}}.$$
(a) Solv... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{43}{34}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{43}{34}$, so the series is convergent. |
math-016652 | Real Analysis: Series — Integral Test for Power Laws | 9 | Provide a rigorous solution: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{41}{22}}}.$$
(a) Solve using a named convergence test.
(b) Give an independ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{41}{22}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{41}{22}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016653 | Real Analysis: Uniform Continuity (Variant B) | 9 | Give a fully justified solution: Let $f:(0,513)\to\mathbb{R}$ be $f(x)=\frac{1}{(-14)x}$. Is $f$ uniformly continuous on $(0,513)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{513}{n}$ and $y_n=\\frac{513}{2n}$ in $(0,513)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016654 | Real Analysis: Series — Parameter Sensitivity | 9 | Try to avoid pattern-matching; explain why: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{101}{22}}}.$$
(a) Solve using a named convergence test.
(b) ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{101}{22}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness no... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{101}{22}$, so the series is convergent. |
math-016655 | Real Analysis: Uniform Continuity (Variant A) | 9 | Where appropriate, name the theorem you use: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(12x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz via Mean Value Theorem",
"approach": "Use MVT to get a global Lipschitz bound $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: Apply MVT to $g(t)=\\sin(ct)$: there exists $\\xi$ between $x$ and $y$ with $g(x)-g(y)=g'(\\xi)(x-y)$.",
"Step 2: Since $g'(t)=(12)\\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"robustness_analysis": "Generalit... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-016656 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Solve and sanity-check: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{9}{32}}}.$$
(a) Solve using a named... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{9}{32}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Robustness note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{9}{32}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-016657 | Real Analysis: Series — Integral Test for Power Laws | 9 | Make each step logically reversible (or explain if not): For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{73}... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{73}{40}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{73}{40}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{73}{40}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016658 | Real Analysis: Uniform Continuity (Variant C) | 9 | Solve and include a self-check: Let $f:(0,120)\to\mathbb{R}$ be $f(x)=\ln(20x)$. Is $f$ uniformly continuous on $(0,120)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016659 | Real Analysis: Uniform Continuity (Variant A) | 9 | Track quantifiers carefully: Let $f:(0,930)\to\mathbb{R}$ be $f(x)=\ln(28x)$. Is $f$ uniformly continuous on $(0,930)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016660 | Real Analysis: Uniform Continuity (Variant C) | 9 | Problem: Let $f:(0,984)\to\mathbb{R}$ be $f(x)=\frac{1}{(6)x}$. Is $f$ uniformly continuous on $(0,984)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{984}{n}$ and $y_n=\\frac{984}{2n}$ in $(0,984)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016661 | Real Analysis: Uniform Continuity (Variant C) | 9 | Work this out carefully: Let $f:(0,1653)\to\mathbb{R}$ be $f(x)=\ln(46x)$. Is $f$ uniformly continuous on $(0,1653)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1653}{n}$ and $y_n=\\frac{1653}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016662 | Real Analysis: Series — Parameter Sensitivity | 9 | State any required conditions first: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{97}{15}}}.$$
(a) Solve using a named convergence test.
(b) Give an in... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{97}{15}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{97}{15}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity an... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{97}{15}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016663 | Real Analysis: Uniform Continuity (Variant C) | 9 | Try to avoid pattern-matching; explain why: Let $f:(0,1670)\to\mathbb{R}$ be $f(x)=\frac{1}{(-20)x}$. Is $f$ uniformly continuous on $(0,1670)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016664 | Real Analysis: Uniform Continuity (Core) | 9 | Solve with verification: Let $f:(0,1963)\to\mathbb{R}$ be $f(x)=\ln(16x)$. Is $f$ uniformly continuous on $(0,1963)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016665 | Real Analysis: Uniform Continuity (Variant A) | 9 | Give a fully justified solution: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-15)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Mean Value Theorem Blow-Up",
"approach": "Use MVT to show that on unbounded domains, the derivative growth makes a uniform modulus impossible.",
"steps": [
"Step 1: Suppose uniform continuity holds. Take $\\varepsilon=1$ and let $\\delta>0$ correspond.",
"Step 2: Choose $x$... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-uniform continuity.",
"rob... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016666 | Real Analysis: Uniform Continuity (Variant A) | 9 | Show all reasoning: Let $f:(0,884)\to\mathbb{R}$ be $f(x)=\frac{1}{(-23)x}$. Is $f$ uniformly continuous on $(0,884)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{884}{n}$ and $y_n=\\frac{884}{2n}$ in $(0,884)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016667 | Real Analysis: Uniform Continuity (Core) | 9 | Provide a rigorous solution: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(24)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Mean Value Theorem Blow-Up",
"approach": "Use MVT to show that on unbounded domains, the derivative growth makes a uniform modulus impossible.",
"steps": [
"Step 1: Suppose uniform continuity holds. Take $\\varepsilon=1$ and let $\\delta>0$ correspond.",
"Step 2: Choose $x$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016668 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Determine the requested value: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{2}{7}}}.$$
(a) Solve using a named convergence test.
... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{2}{7}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a sp... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{2}{7}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-016669 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Prompt: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{19}{3}}}.$$
(a) Solve using a named convergence tes... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{19}{3}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{19}{3}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016670 | Real Analysis: Uniform Continuity (Variant A) | 9 | Determine the requested value: Let $f:(0,765)\to\mathbb{R}$ be $f(x)=\ln(38x)$. Is $f$ uniformly continuous on $(0,765)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{765}{n}$ and $y_n=\\frac{765}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016671 | Real Analysis: Series — Integral Test for Power Laws | 9 | Answer with a short justification: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{21}{13}}}.$$
(a) Solve using a named convergence ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{21}{13}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{21}{13}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016672 | Real Analysis: Uniform Continuity (Core) | 9 | Prompt: Let $f:(0,1001)\to\mathbb{R}$ be $f(x)=\ln(55x)$. Is $f$ uniformly continuous on $(0,1001)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016673 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Use two approaches if possible: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{25}{13}}}.$$
(a) Solve using a named convergence tes... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{25}{13}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{25}{13}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{25}{13}$, so the series is convergent. |
math-016674 | Real Analysis: Uniform Continuity (Variant B) | 9 | Show all reasoning: Let $f:(0,1283)\to\mathbb{R}$ be $f(x)=\ln(38x)$. Is $f$ uniformly continuous on $(0,1283)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016675 | Real Analysis: Uniform Continuity (Variant A) | 9 | Use two approaches if possible: Let $f:(0,956)\to\mathbb{R}$ be $f(x)=\frac{1}{(26)x}$. Is $f$ uniformly continuous on $(0,956)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{956}{n}$ and $y_n=\\frac{956}{2n}$ in $(0,956)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016676 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Solve and include a self-check: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{4}{15}}}.$$
(a) Sol... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{4}{15}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{4}{15}$, so the series is divergent. |
math-016677 | Real Analysis: Uniform Continuity (Variant A) | 9 | Track quantifiers carefully: Let $f:(0,918)\to\mathbb{R}$ be $f(x)=\frac{1}{(-21)x}$. Is $f$ uniformly continuous on $(0,918)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{918}{n}$ and $y_n=\\frac{918}{2n}$ in $(0,918)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016678 | Real Analysis: Series — Integral Test for Power Laws | 9 | Work carefully and justify each inference: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{39}{28}}}.$$
(a) Solve using a named convergence test.
(b) Gi... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{39}{28}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-ser... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{39}{28}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016679 | Real Analysis: Uniform Continuity (Variant C) | 9 | Explain what is being counted/optimized: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(7x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz via Mean Value Theorem",
"approach": "Use MVT to get a global Lipschitz bound $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: Apply MVT to $g(t)=\\sin(ct)$: there exists $\\xi$ between $x$ and $y$ with $g(x)-g(y)=g'(\\xi)(x-y)$.",
"Step 2: Since $g'(t)=(7)\\c... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"robustnes... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-016680 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Where appropriate, name the theorem you use: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{20}{39}}}.$$
(a) Solve using a named convergence test.
(b) ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{20}{39}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{20}{39}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-016681 | Real Analysis: Uniform Continuity (Core) | 9 | Solve with verification: Let $f:(0,1569)\to\mathbb{R}$ be $f(x)=\ln(25x)$. Is $f$ uniformly continuous on $(0,1569)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1569}{n}$ and $y_n=\\frac{1569}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016682 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Provide a rigorous solution: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{46}{9}}}.$$
(a) Solve using a named convergence test.
(b) Give an independe... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{46}{9}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{46}{9}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{46}{9}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016683 | Real Analysis: Uniform Continuity (Core) | 9 | Solve and then verify: Let $f:(0,113)\to\mathbb{R}$ be $f(x)=\frac{1}{(6)x}$. Is $f$ uniformly continuous on $(0,113)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016684 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Task: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{35}{16}}}.$$
(a) Solve using a named converge... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{35}{16}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{35}{16}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were per... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{35}{16}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016685 | Real Analysis: Series — Integral Test for Power Laws | 9 | Give reasoning, not just computation: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{58}{23}}}.$$
(a) Solve using a named convergence test.
(b) Give an... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{58}{23}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were per... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{58}{23}$, so the series is convergent. |
math-016686 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Solve with verification: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{59}{22}}}.$$
(a) Solve using a named convergence test.
(b) ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{59}{22}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{59}{22}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016687 | Real Analysis: Series — p-Series Threshold | 9 | Make each step logically reversible (or explain if not): Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{29}{11}}}.$$
(a) Solve using a named convergenc... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{29}{11}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{29}{11}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{29}{11}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016688 | Real Analysis: Uniform Continuity (Variant C) | 9 | Make each step logically reversible (or explain if not): Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(30x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Characterization",
"approach": "A function is uniformly continuous iff it sends Cauchy sequences to Cauchy sequences; use the global inequality $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: For all real $x,y$, $|\\sin x-\\sin y|\\le |x-y|$ (e.g., by MVT).",... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"robustness_analysis": "If the pr... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-016689 | Real Analysis: Series — Integral Test for Power Laws | 9 | Carefully track domains: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{35}{4}}}.$$
(a) Solve usin... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{35}{4}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a special... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{35}{4}$, so the series is convergent. |
math-016690 | Real Analysis: Uniform Continuity (Variant C) | 9 | Warm-up: Let $f:(0,329)\to\mathbb{R}$ be $f(x)=\frac{1}{(18)x}$. Is $f$ uniformly continuous on $(0,329)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{329}{n}$ and $y_n=\\frac{329}{2n}$ in $(0,329)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016691 | Real Analysis: Uniform Continuity (Core) | 9 | Explain what is being counted/optimized: Let $f:(0,98)\to\mathbb{R}$ be $f(x)=\ln(41x)$. Is $f$ uniformly continuous on $(0,98)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "Gener... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016692 | Real Analysis: Uniform Continuity (Variant B) | 9 | Explain each transformation: Let $f:(0,380)\to\mathbb{R}$ be $f(x)=\frac{1}{(-7)x}$. Is $f$ uniformly continuous on $(0,380)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}}.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016693 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Do not skip justification steps: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{71}{18}}}.$$
(a) S... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{71}{18}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{71}{18}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity an... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{71}{18}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016694 | Real Analysis: Uniform Continuity (Core) | 9 | Solve and then verify: Let $f:(0,1395)\to\mathbb{R}$ be $f(x)=\frac{1}{(-18)x}$. Is $f$ uniformly continuous on $(0,1395)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{1395}{n}$ and $y_n=\\frac{1395}{2n}$ in $(0,1395)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016695 | Real Analysis: Uniform Continuity (Core) | 9 | Provide both a computational and a conceptual explanation: Let $f:(0,1131)\to\mathbb{R}$ be $f(x)=\ln(24x)$. Is $f$ uniformly continuous on $(0,1131)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "Robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016696 | Real Analysis: Uniform Continuity (Variant C) | 9 | State any required conditions first: Let $f:(0,139)\to\mathbb{R}$ be $f(x)=\ln(19x)$. Is $f$ uniformly continuous on $(0,139)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{139}{n}$ and $y_n=\\frac{139}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016697 | Real Analysis: Uniform Continuity (Variant A) | 9 | Where appropriate, name the theorem you use: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-18)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Explicit Sequence Counterexample",
"approach": "Pick sequences approaching each other at infinity but whose squares stay separated.",
"steps": [
"Step 1: Let $x_n=n$ and $y_n=n+\\frac{1}{n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{1}{n}\\to 0$.",
"Final step: But $|f(x_n)-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-uniform continuity.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016698 | Real Analysis: Series — Integral Test for Power Laws | 9 | Question: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{53}{13}}}.$$
(a) Solve using a named conv... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{53}{13}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{53}{13}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{53}{13}$, so the series is convergent. |
math-016699 | Real Analysis: Uniform Continuity (Variant A) | 9 | Challenge: Let $f:(0,1515)\to\mathbb{R}$ be $f(x)=\ln(6x)$. Is $f$ uniformly continuous on $(0,1515)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1515}{n}$ and $y_n=\\frac{1515}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016700 | Real Analysis: Uniform Continuity (Variant A) | 9 | Task: Let $f:(0,764)\to\mathbb{R}$ be $f(x)=\frac{1}{(18)x}$. Is $f$ uniformly continuous on $(0,764)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
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