id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-016501 | Real Analysis: Uniform Continuity (Variant B) | 9 | Solve and justify each step: Let $f:(0,1422)\to\mathbb{R}$ be $f(x)=\ln(17x)$. Is $f$ uniformly continuous on $(0,1422)$? 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-016502 | Real Analysis: Uniform Continuity (Variant B) | 9 | Complete the analysis: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(28)x^2$ 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": "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 ... | 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-016503 | Real Analysis: Series — Integral Test for Power Laws | 9 | Give an answer and a quick 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{115}{21}}}.$$
(a) Solve using a named conve... | [
{
"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 $p=\\frac{115}{21}$.",
"Final step: By the p-series test, it 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 $p=\\frac{115}{21}$ 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 $p=\frac{115}{21}$, so the series is convergent. |
math-016504 | Real Analysis: Uniform Continuity (Variant B) | 9 | Give reasoning, not just computation: Let $f:(0,561)\to\mathbb{R}$ be $f(x)=\frac{1}{(8)x}$. Is $f$ uniformly continuous on $(0,561)$? 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 ... | 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-016505 | Real Analysis: Uniform Continuity (Variant B) | 9 | Solve and include a self-check: Let $f:(0,1016)\to\mathbb{R}$ be $f(x)=\ln(44x)$. Is $f$ uniformly continuous on $(0,1016)$? 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": "If th... | [
{
"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-016506 | Real Analysis: Uniform Continuity (Core) | 9 | Find the exact value: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-27)x^2$ 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. |
math-016507 | Real Analysis: Uniform Continuity (Variant B) | 9 | Explain each transformation: Let $f:(0,1834)\to\mathbb{R}$ be $f(x)=\frac{1}{(22)x}$. Is $f$ uniformly continuous on $(0,1834)$? 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 ... | 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-016508 | Real Analysis: Uniform Continuity (Variant C) | 9 | Solve and then verify: Let $f:(0,1437)\to\mathbb{R}$ be $f(x)=\frac{1}{(25)x}$. Is $f$ uniformly continuous on $(0,1437)$? 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 ... | 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-016509 | Real Analysis: Uniform Continuity (Variant B) | 9 | Solve with verification: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(4)x^2$ 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 ... | 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-016510 | Real Analysis: Uniform Continuity (Variant C) | 9 | Write the solution set clearly: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(30)x^2$ 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": "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 ... | 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-016511 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Give a fully justified 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{69}{25}}}.$$
(a) Solve using a named convergence test.
(b) Give an inde... | [
{
"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 $p=\\frac{69}{25}$ 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 $p=\frac{69}{25}$, so the series is convergent. |
math-016512 | Real Analysis: Uniform Continuity (Variant A) | 9 | Give a theorem-based solution: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(3x)$. 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)=(3)\\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. |
math-016513 | Real Analysis: Uniform Continuity (Variant B) | 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": "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. (Here the result is $\boxed{\text{Yes}$.) |
math-016514 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Answer using clear logical steps: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{51}{20}}}.$$
(a) Solve using a named convergence test.
(b) Give an ind... | [
{
"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 $p=\\frac{51}{20}$ 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... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{51}{20}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016515 | Real Analysis: Uniform Continuity (Variant A) | 9 | Question: Let $f:(0,1851)\to\mathbb{R}$ be $f(x)=\frac{1}{(-17)x}$. Is $f$ uniformly continuous on $(0,1851)$? 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 $x_n=\\frac{1851}{n}$ and $y_n=\\frac{1851}{2n}$ in $(0,1851)$.",
"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 ... | 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-016516 | Real Analysis: Uniform Continuity (Variant C) | 9 | Be explicit about assumptions: Let $f:(0,587)\to\mathbb{R}$ be $f(x)=\frac{1}{(14)x}$. Is $f$ uniformly continuous on $(0,587)$? 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 $x_n=\\frac{587}{n}$ and $y_n=\\frac{587}{2n}$ in $(0,587)$.",
"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 ... | 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-016517 | Real Analysis: Uniform Continuity (Variant A) | 9 | Warm-up: 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": "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-016518 | Real Analysis: Uniform Continuity (Variant A) | 9 | Compute the requested quantity: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-5)x^2$ 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 ... | 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-016519 | Real Analysis: Uniform Continuity (Core) | 9 | State any required conditions first: Let $f:(0,1802)\to\mathbb{R}$ be $f(x)=\ln(50x)$. Is $f$ uniformly continuous on $(0,1802)$? 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 $x_n=\\frac{1802}{n}$ and $y_n=\\frac{1802}{2n}$.",
"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 ... | 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-016520 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Give a theorem-based 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{49}{40}}}.$$
(a) Solve using a named convergence test.
(b) Give an indepe... | [
{
"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 $p=\\frac{49}{40}$ 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... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{49}{40}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016521 | Real Analysis: Series — Parameter Sensitivity | 9 | Indicate where a theorem is used: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{109}{26}}}.$$
(a) Solve u... | [
{
"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 $p=\\frac{109}{26}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: ... | [
{
"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 $p=\frac{109}{26}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016522 | Real Analysis: Uniform Continuity (Variant B) | 9 | Give a fully justified solution: 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 ... | 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-016523 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Track units/moduli carefully: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{63}{40}}}.$$
(a) Solve using a named convergence test.
(b) Give an indepen... | [
{
"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 $p=\\frac{63}{40}$.",
"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 $p=\\frac{63}{40}$ 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... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{63}{40}$, so the series is convergent. |
math-016524 | Real Analysis: Uniform Continuity (Core) | 9 | Exercise: Let $f:(0,1599)\to\mathbb{R}$ be $f(x)=\ln(5x)$. Is $f$ uniformly continuous on $(0,1599)$? 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-016525 | Real Analysis: Series — Necessary vs Sufficient Conditions | 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{68}{27}}}.$$
(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 $p=\\frac{68}{27}$ 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... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{68}{27}$, so the series is convergent. |
math-016526 | Real Analysis: Uniform Continuity (Variant B) | 9 | Provide a rigorous solution: Let $f:(0,55)\to\mathbb{R}$ be $f(x)=\ln(11x)$. Is $f$ uniformly continuous on $(0,55)$? 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 $x_n=\\frac{55}{n}$ and $y_n=\\frac{55}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(c ... | {
"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-016527 | Real Analysis: Series — Integral Test for Power Laws | 9 | Warm-up: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{90}{31}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cross-check using ... | [
{
"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 $p=\\frac{90}{31}$ 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... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{90}{31}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016528 | Real Analysis: Series — p-Series Threshold | 9 | Work carefully and justify each inference: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{41}{19}}}.$$
(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 $p=\\frac{41}{19}$.",
"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 $p=\\frac{41}{19}$ 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 $p=\frac{41}{19}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016529 | Real Analysis: Uniform Continuity (Variant A) | 9 | Question: Let $f:(0,608)\to\mathbb{R}$ be $f(x)=\ln(46x)$. Is $f$ uniformly continuous on $(0,608)$? 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-016530 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Start by stating any domain restrictions: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{31}{26}}}.... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{31}{26}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{31}{26}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-series te... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{31}{26}$, so the series is convergent. |
math-016531 | Real Analysis: Series — Parameter Sensitivity | 9 | Checkpoint: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{37}{10}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cross-check u... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{37}{10}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{37}{10}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{37}{10}$, so the series is convergent. |
math-016532 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Give a theorem-based solution: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{87}{28}}}.$$
(a) Sol... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{87}{28}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{87}{28}$, so the series is convergent. |
math-016533 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | State any required conditions first: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{73}{17}}}.$$
(a) Solve using a named convergence test.
(b) Give an ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{73}{17}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{73}{17}$, so the series is convergent. |
math-016534 | Real Analysis: Series — Parameter Sensitivity | 9 | Compute the requested quantity: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{97}{35}}}.$$
(a) Solve usin... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{97}{35}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-series te... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{97}{35}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016535 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Track units/moduli carefully: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{101}{23}}}.$$
(a) Solve using... | [
{
"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}{23}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality 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{101}{23}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016536 | Real Analysis: Uniform Continuity (Variant B) | 9 | Find the exact value: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(23x)$. 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 ... | 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-016537 | Real Analysis: Uniform Continuity (Variant C) | 9 | Answer with a short justification: Let $f:(0,958)\to\mathbb{R}$ be $f(x)=\ln(59x)$. Is $f$ uniformly continuous on $(0,958)$? 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. |
math-016538 | Real Analysis: Uniform Continuity (Variant B) | 9 | Give a fully justified solution: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(13x)$. 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)=(13)\\... | {
"consistency_check": "The two methods are consistent and must coincide. 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": "Sen... | [
{
"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-016539 | Real Analysis: Uniform Continuity (Variant A) | 9 | Solve with verification: Let $f:(0,1236)\to\mathbb{R}$ be $f(x)=\frac{1}{(-8)x}$. Is $f$ uniformly continuous on $(0,1236)$? 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 ... | 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-016540 | Real Analysis: Series — Divergence at the Boundary Case | 9 | State any required conditions first: 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{54}{31}}}.$$
(... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{54}{31}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{54}{31}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-ser... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{54}{31}$, so the series is convergent. |
math-016541 | Real Analysis: Series — Parameter Sensitivity | 9 | Solve and include a self-check: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{15}{13}}}.$$
(a) Solve using a named convergence test.
(b) Give an indep... | [
{
"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{15}{13}$.",
"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{15}{13}$ 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... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{15}{13}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016542 | Real Analysis: Uniform Continuity (Variant C) | 9 | Work carefully and justify each inference: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-2)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": "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 ... | 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-016543 | Real Analysis: Uniform Continuity (Variant C) | 9 | Solve (and briefly cross-validate): 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": "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-016544 | Real Analysis: Uniform Continuity (Variant A) | 9 | Use two approaches if possible: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(28x)$. 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)=(28)\\... | {
"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 ... | 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-016545 | Real Analysis: Uniform Continuity (Variant B) | 9 | Determine the requested value: Let $f:(0,1907)\to\mathbb{R}$ be $f(x)=\ln(46x)$. Is $f$ uniformly continuous on $(0,1907)$? 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{1907}{n}$ and $y_n=\\frac{1907}{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 ... | 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-016546 | Real Analysis: Uniform Continuity (Variant A) | 9 | Track quantifiers carefully: Let $f:(0,1689)\to\mathbb{R}$ be $f(x)=\ln(19x)$. Is $f$ uniformly continuous on $(0,1689)$? 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": "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. |
math-016547 | Real Analysis: Uniform Continuity (Variant C) | 9 | Exercise: Let $f:(0,1763)\to\mathbb{R}$ be $f(x)=\ln(2x)$. Is $f$ uniformly continuous on $(0,1763)$? 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{1763}{n}$ and $y_n=\\frac{1763}{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-016548 | Real Analysis: Uniform Continuity (Variant A) | 9 | Problem: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(20x)$. 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 ... | 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-016549 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Question: 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{53}{39}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cross-check usi... | [
{
"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}{39}$.",
"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{53}{39}$ 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{53}{39}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016550 | Real Analysis: Uniform Continuity (Variant A) | 9 | Track quantifiers carefully: Let $f:(0,252)\to\mathbb{R}$ be $f(x)=\frac{1}{(23)x}$. Is $f$ uniformly continuous on $(0,252)$? 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{252}{n}$ and $y_n=\\frac{252}{2n}$ in $(0,252)$.",
"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-016551 | Real Analysis: Uniform Continuity (Core) | 9 | Solve (and briefly cross-validate): Let $f:(0,1293)\to\mathbb{R}$ be $f(x)=\ln(54x)$. Is $f$ uniformly continuous on $(0,1293)$? 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 ... | 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-016552 | Real Analysis: Series — p-Series Threshold | 9 | Give a fully justified solution: 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{37}{17}}}.$$
(a) Solve usi... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{37}{17}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity an... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{37}{17}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016553 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Where appropriate, name the theorem you use: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{41}{28}}}.$$
(... | [
{
"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{41}{28}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test ... | [
{
"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{41}{28}$, so the series is convergent. |
math-016554 | Real Analysis: Uniform Continuity (Variant B) | 9 | Prompt: Let $f:(0,1895)\to\mathbb{R}$ be $f(x)=\ln(3x)$. Is $f$ uniformly continuous on $(0,1895)$? 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{1895}{n}$ and $y_n=\\frac{1895}{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-016555 | Real Analysis: Uniform Continuity (Core) | 9 | Task: Let $f:(0,1022)\to\mathbb{R}$ be $f(x)=\frac{1}{(-16)x}$. Is $f$ uniformly continuous on $(0,1022)$? 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{1022}{n}$ and $y_n=\\frac{1022}{2n}$ in $(0,1022)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"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-016556 | Real Analysis: Uniform Continuity (Variant C) | 9 | Explain what is being counted/optimized: Let $f:(0,1850)\to\mathbb{R}$ be $f(x)=\ln(38x)$. Is $f$ uniformly continuous on $(0,1850)$? 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 ... | 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-016557 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Give reasoning, not just computation: 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{79}{39}}}.$$
(a) Solve using a named convergen... | [
{
"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{79}{39}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test ... | [
{
"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{79}{39}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016558 | Real Analysis: Uniform Continuity (Variant C) | 9 | Solve and justify each step: Let $f:(0,1616)\to\mathbb{R}$ be $f(x)=\frac{1}{(15)x}$. Is $f$ uniformly continuous on $(0,1616)$? 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{1616}{n}$ and $y_n=\\frac{1616}{2n}$ in $(0,1616)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"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 ... | 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-016559 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Explain why your operations are valid: 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{116}{31}}}.$$
(a) Solve using a named convergence test.
(b) Give an... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{116}{31}$.",
"Final step: By the p-series test, it con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{116}{31}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality no... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{116}{31}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016560 | Real Analysis: Series — Parameter Sensitivity | 9 | Carefully track domains: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{38}{23}}}.$$
(a) Solve using a named convergence test.
(b) ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{38}{23}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: T... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{38}{23}$, so the series is convergent. |
math-016561 | Real Analysis: Series — Integral Test for Power Laws | 9 | Solve and include a self-check: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{120}{17}}}.$$
(a) Solve using a named convergence te... | [
{
"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{120}{17}$.",
"Final step: By the p-series test, it 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{120}{17}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a speci... | [
{
"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{120}{17}$, so the series is convergent. |
math-016562 | Real Analysis: Uniform Continuity (Core) | 9 | Derive the result step-by-step: 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 ... | 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-016563 | Real Analysis: Uniform Continuity (Variant A) | 9 | Exercise: Let $f:(0,84)\to\mathbb{R}$ be $f(x)=\frac{1}{(7)x}$. Is $f$ uniformly continuous on $(0,84)$? 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-016564 | Real Analysis: Uniform Continuity (Variant C) | 9 | Explain what is being counted/optimized: Let $f:(0,1918)\to\mathbb{R}$ be $f(x)=\ln(54x)$. Is $f$ uniformly continuous on $(0,1918)$? 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. |
math-016565 | Real Analysis: Uniform Continuity (Variant A) | 9 | Solve and sanity-check: Let $f:(0,1910)\to\mathbb{R}$ be $f(x)=\ln(19x)$. Is $f$ uniformly continuous on $(0,1910)$? 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-016566 | Real Analysis: Uniform Continuity (Core) | 9 | Prompt: Let $f:(0,613)\to\mathbb{R}$ be $f(x)=\ln(33x)$. Is $f$ uniformly continuous on $(0,613)$? 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-016567 | Real Analysis: Uniform Continuity (Variant B) | 9 | Complete the analysis: Let $f:(0,442)\to\mathbb{R}$ be $f(x)=\frac{1}{(22)x}$. Is $f$ uniformly continuous on $(0,442)$? 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{442}{n}$ and $y_n=\\frac{442}{2n}$ in $(0,442)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"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 ... | 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-016568 | Real Analysis: Uniform Continuity (Variant C) | 9 | Give a fully justified solution: Let $f:(0,550)\to\mathbb{R}$ be $f(x)=\frac{1}{(-27)x}$. Is $f$ uniformly continuous on $(0,550)$? 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{550}{n}$ and $y_n=\\frac{550}{2n}$ in $(0,550)$.",
"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. (Here the result is $\boxed{\text{No}$.) |
math-016569 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Be explicit about assumptions: 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{47}{6}}}.$$
(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{47}{6}$ 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... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{47}{6}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016570 | Real Analysis: Uniform Continuity (Core) | 9 | Track quantifiers carefully: Let $f:(0,1819)\to\mathbb{R}$ be $f(x)=\frac{1}{(5)x}$. Is $f$ uniformly continuous on $(0,1819)$? 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 ... | 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-016571 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Give a theorem-based solution: 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{113}{20}}}.$$
(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{113}{20}$.",
"Final step: By the p-series test, it 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{113}{20}$ 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... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{113}{20}$, so the series is convergent. |
math-016572 | Real Analysis: Uniform Continuity (Variant B) | 9 | Work carefully and justify each inference: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(9x)$. 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": "The two methods are consistent and must coincide. 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": "Gen... | [
{
"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-016573 | Real Analysis: Uniform Continuity (Variant A) | 9 | Work carefully and justify each inference: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(-13x)$. 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)=(-13)\... | {
"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": "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 ... | 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-016574 | Real Analysis: Uniform Continuity (Variant B) | 9 | Try to avoid pattern-matching; explain why: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(11x)$. 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": "The two methods are consistent and must coincide. 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": "Sen... | [
{
"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-016575 | Real Analysis: Series — Integral Test for Power Laws | 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{43}{5}}}.$$
(a) Solve using a named convergence test.
(b) Give ... | [
{
"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}{5}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-series tes... | [
{
"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{43}{5}$, so the series is convergent. |
math-016576 | Real Analysis: Uniform Continuity (Variant A) | 9 | Find the exact value: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(-23x)$. 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)=(-23)\... | {
"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. (Here the result is $\boxed{\text{Yes}$.) |
math-016577 | Real Analysis: Uniform Continuity (Core) | 9 | Determine the requested value: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(16)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": "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 ... | 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-016578 | Real Analysis: Uniform Continuity (Variant A) | 9 | Give a fully justified solution: Let $f:(0,1697)\to\mathbb{R}$ be $f(x)=\ln(22x)$. Is $f$ uniformly continuous on $(0,1697)$? 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{1697}{n}$ and $y_n=\\frac{1697}{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 ... | 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-016579 | Real Analysis: Series — Necessary vs Sufficient Conditions | 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{39}{17}}}.$$
(a) Solve using a named convergence test.
(b) Giv... | [
{
"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{39}{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... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{39}{17}$, so the series is convergent. |
math-016580 | Real Analysis: Uniform Continuity (Variant C) | 9 | Give a theorem-based solution: Let $f:(0,635)\to\mathbb{R}$ be $f(x)=\frac{1}{(-19)x}$. Is $f$ uniformly continuous on $(0,635)$? 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{635}{n}$ and $y_n=\\frac{635}{2n}$ in $(0,635)$.",
"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. (Here the result is $\boxed{\text{No}$.) |
math-016581 | Real Analysis: Uniform Continuity (Variant C) | 9 | Carefully track domains: Let $f:(0,322)\to\mathbb{R}$ be $f(x)=\frac{1}{(-26)x}$. Is $f$ uniformly continuous on $(0,322)$? 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-016582 | Real Analysis: Uniform Continuity (Core) | 9 | Explain why your operations are valid: Let $f:(0,354)\to\mathbb{R}$ be $f(x)=\ln(39x)$. Is $f$ uniformly continuous on $(0,354)$? 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{354}{n}$ and $y_n=\\frac{354}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"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. |
math-016583 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Keep the final answer in boxed form: 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{16}{11}}}.$$
(a) Solve using a named convergence test.
(b) Give an in... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{16}{11}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{16}{11}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016584 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Give a fully justified solution: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{57}{13}}}.$$
(a) S... | [
{
"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{57}{13}$ 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... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{57}{13}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016585 | Real Analysis: Uniform Continuity (Variant B) | 9 | Track units/moduli carefully: Let $f:(0,797)\to\mathbb{R}$ be $f(x)=\frac{1}{(9)x}$. Is $f$ uniformly continuous on $(0,797)$? 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. (Here the result is $\boxed{\text{No}$.) |
math-016586 | Real Analysis: Uniform Continuity (Core) | 9 | Give reasoning, not just computation: Let $f:(0,364)\to\mathbb{R}$ be $f(x)=\ln(33x)$. Is $f$ uniformly continuous on $(0,364)$? 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-016587 | Real Analysis: Uniform Continuity (Core) | 9 | Prompt: Let $f:(0,667)\to\mathbb{R}$ be $f(x)=\frac{1}{(17)x}$. Is $f$ uniformly continuous on $(0,667)$? 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{667}{n}$ and $y_n=\\frac{667}{2n}$ in $(0,667)$.",
"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. |
math-016588 | Real Analysis: Uniform Continuity (Core) | 9 | Challenge: Let $f:(0,1128)\to\mathbb{R}$ be $f(x)=\frac{1}{(7)x}$. Is $f$ uniformly continuous on $(0,1128)$? 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 ... | 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-016589 | Real Analysis: Uniform Continuity (Variant B) | 9 | Carefully track domains: Let $f:(0,1948)\to\mathbb{R}$ be $f(x)=\frac{1}{(-22)x}$. Is $f$ uniformly continuous on $(0,1948)$? 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-016590 | Real Analysis: Uniform Continuity (Variant C) | 9 | State any required conditions first: Let $f:(0,149)\to\mathbb{R}$ be $f(x)=\frac{1}{(1)x}$. Is $f$ uniformly continuous on $(0,149)$? 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{149}{n}$ and $y_n=\\frac{149}{2n}$ in $(0,149)$.",
"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 ... | 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-016591 | 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{85}{18}}}.$$
(a) Solve using a named convergence test.
(b) Giv... | [
{
"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{85}{18}$.",
"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{85}{18}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-series te... | [
{
"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{85}{18}$, so the series is convergent. |
math-016592 | Real Analysis: Uniform Continuity (Variant B) | 9 | Challenge: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(29)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 ... | 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-016593 | Real Analysis: Uniform Continuity (Variant C) | 9 | Answer with a short justification: Let $f:(0,1200)\to\mathbb{R}$ be $f(x)=\frac{1}{(24)x}$. Is $f$ uniformly continuous on $(0,1200)$? 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 ... | 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-016594 | Real Analysis: Uniform Continuity (Variant B) | 9 | Solve with verification: Let $f:(0,360)\to\mathbb{R}$ be $f(x)=\ln(55x)$. Is $f$ uniformly continuous on $(0,360)$? 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{360}{n}$ and $y_n=\\frac{360}{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 ... | 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-016595 | Real Analysis: Uniform Continuity (Variant C) | 9 | Work this out carefully: Let $f:(0,892)\to\mathbb{R}$ be $f(x)=\ln(54x)$. Is $f$ uniformly continuous on $(0,892)$? 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": "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 ... | 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-016596 | Real Analysis: Uniform Continuity (Core) | 9 | Solve and justify each step: Let $f:(0,1448)\to\mathbb{R}$ be $f(x)=\ln(59x)$. Is $f$ uniformly continuous on $(0,1448)$? 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-016597 | Real Analysis: Series — p-Series Threshold | 9 | Proceed methodically: 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{95}{34}}}.$$
(a) Solve using a named convergence test.
(b) Giv... | [
{
"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{95}{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{95}{34}$ 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... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{95}{34}$, so the series is convergent. |
math-016598 | Real Analysis: Uniform Continuity (Variant B) | 9 | Start by stating any domain restrictions: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-21)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 ... | 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-016599 | Real Analysis: Series — Parameter Sensitivity | 9 | Solve and sanity-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{5}{8}}}.$$
(a) Solve using ... | [
{
"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: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{5}{8}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test i... | [
{
"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 $p=\frac{5}{8}$, so the series is divergent. |
math-016600 | Real Analysis: Uniform Continuity (Variant B) | 9 | Derive the result step-by-step: Let $f:(0,1486)\to\mathbb{R}$ be $f(x)=\ln(34x)$. Is $f$ uniformly continuous on $(0,1486)$? 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 $\boxed{\text{No}$.) |
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