id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-016401 | Real Analysis: Uniform Continuity (Variant C) | 9 | Warm-up: Let $f:(0,726)\to\mathbb{R}$ be $f(x)=\ln(43x)$. Is $f$ uniformly continuous on $(0,726)$? 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 $x_n=\\frac{726}{n}$ and $y_n=\\frac{726}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016402 | Real Analysis: Uniform Continuity (Variant B) | 9 | Solve and include a self-check: Let $f:(0,1252)\to\mathbb{R}$ be $f(x)=\frac{1}{(-14)x}$. Is $f$ uniformly continuous on $(0,1252)$? 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 $x_n=\\frac{1252}{n}$ and $y_n=\\frac{1252}{2n}$ in $(0,1252)$.",
"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 ... | 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-016403 | Real Analysis: Uniform Continuity (Variant C) | 9 | Start by stating any domain restrictions: Let $f:(0,1476)\to\mathbb{R}$ be $f(x)=\frac{1}{(27)x}$. Is $f$ uniformly continuous on $(0,1476)$? 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 $x_n=\\frac{1476}{n}$ and $y_n=\\frac{1476}{2n}$ in $(0,1476)$.",
"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 ... | 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-016404 | Real Analysis: Series — p-Series Threshold | 9 | Derive the result step-by-step: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{55}{6}}}.$$
(a) Solve using... | [
{
"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 $p=\\frac{55}{6}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{55}{6}$ 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... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{55}{6}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016405 | Real Analysis: Uniform Continuity (Variant C) | 9 | Make each step logically reversible (or explain if not): Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-12)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Mean Value Theorem Blow-Up",
"approach": "Use MVT to show that on unbounded domains, the derivative growth makes a uniform modulus impossible.",
"steps": [
"Step 1: Suppose uniform continuity holds. Take $\\varepsilon=1$ and let $\\delta>0$ correspond.",
"Step 2: Choose $x$... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-uniform continuity.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016406 | Real Analysis: Uniform Continuity (Variant A) | 9 | Answer with a short justification: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(-12x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "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. |
math-016407 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Give reasoning, not just computation: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{1}{11}}}.$$
(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: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{1}{11}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{1}{11}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-016408 | Real Analysis: Series — Parameter Sensitivity | 9 | Be explicit about assumptions: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{113}{36}}}.$$
(a) Solve using a named convergence test.
(b) Give an indepen... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "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 $p=\\frac{113}{36}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity 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 $p=\frac{113}{36}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016409 | Real Analysis: Series — Necessary vs Sufficient Conditions | 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{113}{32}}}.$$
(a) Solve using a named convergence test.
(b) Give an i... | [
{
"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 $p=\\frac{113}{32}$.",
"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 $p=\\frac{113}{32}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{113}{32}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016410 | Real Analysis: Uniform Continuity (Core) | 9 | Compute the requested quantity: Let $f:(0,54)\to\mathbb{R}$ be $f(x)=\frac{1}{(13)x}$. Is $f$ uniformly continuous on $(0,54)$? 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. |
math-016411 | Real Analysis: Uniform Continuity (Variant A) | 9 | Solve (and briefly cross-validate): Let $f:(0,666)\to\mathbb{R}$ be $f(x)=\ln(35x)$. Is $f$ uniformly continuous on $(0,666)$? 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 $x_n=\\frac{666}{n}$ and $y_n=\\frac{666}{2n}$.",
"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 ... | 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-016412 | Real Analysis: Uniform Continuity (Variant B) | 9 | Explain each transformation: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(17)x^2$ 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 ... | 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-016413 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Solve and then verify: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{85}{13}}}.$$
(a) Solve using... | [
{
"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 $p=\\frac{85}{13}$.",
"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 $p=\\frac{85}{13}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were per... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{85}{13}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016414 | Real Analysis: Uniform Continuity (Variant B) | 9 | Write the solution set clearly: Let $f:(0,1097)\to\mathbb{R}$ be $f(x)=\frac{1}{(16)x}$. Is $f$ uniformly continuous on $(0,1097)$? 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 $x_n=\\frac{1097}{n}$ and $y_n=\\frac{1097}{2n}$ in $(0,1097)$.",
"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 ... | 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-016415 | Real Analysis: Uniform Continuity (Variant C) | 9 | Give a theorem-based solution: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(16x)$. 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 ... | 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-016416 | Real Analysis: Uniform Continuity (Variant A) | 9 | Answer using clear logical steps: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(-1x)$. 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)=(-1)\\... | {
"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 ... | 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-016417 | Real Analysis: Uniform Continuity (Variant C) | 9 | Where appropriate, name the theorem you use: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(-10x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "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 ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-016418 | Real Analysis: Series — p-Series Threshold | 9 | Solve and justify each step: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{47}{25}}}.$$
(a) Solve using a named convergence test.
... | [
{
"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 $p=\\frac{47}{25}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{47}{25}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{47}{25}$, so the series is convergent. |
math-016419 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Solve and then verify: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{106}{29}}}.$$
(a) Solve using a named convergence test.
(b) G... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{106}{29}$.",
"Final step: By the p-series test, it 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 $p=\\frac{106}{29}$ 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... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{106}{29}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016420 | Real Analysis: Series — p-Series Threshold | 9 | Determine the requested value: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{55}{14}}}.$$
(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": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{55}{14}$ 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... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{55}{14}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016421 | Real Analysis: Uniform Continuity (Variant B) | 9 | Track units/moduli carefully: Let $f:(0,221)\to\mathbb{R}$ be $f(x)=\ln(4x)$. Is $f$ uniformly continuous on $(0,221)$? 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-016422 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Problem: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{102}{35}}}.$$
(a) Solve using a named conv... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{102}{35}$.",
"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 $p=\\frac{102}{35}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the proble... | [
{
"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 $p=\frac{102}{35}$, so the series is convergent. |
math-016423 | Real Analysis: Series — p-Series Threshold | 9 | Write the solution set clearly: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{68}{33}}}.$$
(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 $p=\\frac{68}{33}$.",
"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 $p=\\frac{68}{33}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{68}{33}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016424 | Real Analysis: Uniform Continuity (Core) | 9 | Solve with verification: Let $f:(0,1674)\to\mathbb{R}$ be $f(x)=\frac{1}{(3)x}$. Is $f$ uniformly continuous on $(0,1674)$? 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-016425 | Real Analysis: Series — Integral Test for Power Laws | 9 | Do not skip justification 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{7}{12}}}.$$
(a) Solve using a named convergence test.
(b) Give an indep... | [
{
"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 $p=\\frac{7}{12}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Generality note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{7}{12}$, so the series is divergent. |
math-016426 | Real Analysis: Uniform Continuity (Core) | 9 | Warm-up: 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": "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 ... | 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-016427 | Real Analysis: Uniform Continuity (Variant B) | 9 | Give a fully justified solution: Let $f:(0,1619)\to\mathbb{R}$ be $f(x)=\ln(10x)$. Is $f$ uniformly continuous on $(0,1619)$? 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 $x_n=\\frac{1619}{n}$ and $y_n=\\frac{1619}{2n}$.",
"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": "Gener... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016428 | Real Analysis: Uniform Continuity (Core) | 9 | Find the exact value: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(27x)$. 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)=(27)\\... | {
"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. |
math-016429 | Real Analysis: Uniform Continuity (Variant A) | 9 | Indicate where a theorem is used: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-13)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Explicit Sequence Counterexample",
"approach": "Pick sequences approaching each other at infinity but whose squares stay separated.",
"steps": [
"Step 1: Let $x_n=n$ and $y_n=n+\\frac{1}{n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{1}{n}\\to 0$.",
"Final step: But $|f(x_n)-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-uniform continuity.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | 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-016430 | Real Analysis: Uniform Continuity (Variant B) | 9 | Start by stating any domain restrictions: Let $f:(0,1431)\to\mathbb{R}$ be $f(x)=\frac{1}{(19)x}$. Is $f$ uniformly continuous on $(0,1431)$? 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 $x_n=\\frac{1431}{n}$ and $y_n=\\frac{1431}{2n}$ in $(0,1431)$.",
"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 ... | 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-016431 | Real Analysis: Uniform Continuity (Variant C) | 9 | Give a theorem-based solution: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(26)x^2$ 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. |
math-016432 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Answer with a short justification: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{65}{31}}}.$$
(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": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{65}{31}$ 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... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{65}{31}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016433 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Work carefully and justify each inference: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{108}{35}}}.$$
(a) Solve using a named con... | [
{
"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 $p=\\frac{108}{35}$.",
"Final step: By the p-series test, it 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 $p=\\frac{108}{35}$ 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 $p=\frac{108}{35}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016434 | Real Analysis: Uniform Continuity (Core) | 9 | Answer with a short justification: Let $f:(0,1580)\to\mathbb{R}$ be $f(x)=\frac{1}{(5)x}$. Is $f$ uniformly continuous on $(0,1580)$? 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 ... | 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-016435 | Real Analysis: Uniform Continuity (Core) | 9 | Try to avoid pattern-matching; explain why: Let $f:(0,1598)\to\mathbb{R}$ be $f(x)=\frac{1}{(13)x}$. Is $f$ uniformly continuous on $(0,1598)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-016436 | Real Analysis: Uniform Continuity (Variant A) | 9 | Challenge: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(-22x)$. 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)=(-22)\... | {
"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 ... | 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-016437 | Real Analysis: Uniform Continuity (Variant B) | 9 | State any required conditions first: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(15)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "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 ... | 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-016438 | Real Analysis: Uniform Continuity (Variant A) | 9 | Explain why your operations are valid: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(2x)$. 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)=(2)\\c... | {
"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": "If ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-016439 | Real Analysis: Uniform Continuity (Core) | 9 | Proceed methodically: Let $f:(0,1476)\to\mathbb{R}$ be $f(x)=\frac{1}{(19)x}$. Is $f$ uniformly continuous on $(0,1476)$? 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 $x_n=\\frac{1476}{n}$ and $y_n=\\frac{1476}{2n}$ in $(0,1476)$.",
"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. (Here the result is $\boxed{\text{No}$.) |
math-016440 | Real Analysis: Uniform Continuity (Variant B) | 9 | Task: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-1)x^2$ 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. (Here the result is $\boxed{\text{No}$.) |
math-016441 | Real Analysis: Series — Integral Test for Power Laws | 9 | Compute the requested quantity: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{89}{20}}}.$$
(a) So... | [
{
"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 $p=\\frac{89}{20}$.",
"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 $p=\\frac{89}{20}$ 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... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{89}{20}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016442 | Real Analysis: Series — Parameter Sensitivity | 9 | Determine the requested value: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{43}{8}}}.$$
(a) Solve using a named convergence test.... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{43}{8}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a sp... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{43}{8}$, so the series is convergent. |
math-016443 | Real Analysis: Uniform Continuity (Variant C) | 9 | Show all reasoning: Let $f:(0,1399)\to\mathbb{R}$ be $f(x)=\frac{1}{(28)x}$. Is $f$ uniformly continuous on $(0,1399)$? 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 $x_n=\\frac{1399}{n}$ and $y_n=\\frac{1399}{2n}$ in $(0,1399)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016444 | Real Analysis: Uniform Continuity (Variant C) | 9 | Work carefully and justify each inference: Let $f:(0,268)\to\mathbb{R}$ be $f(x)=\ln(44x)$. Is $f$ uniformly continuous on $(0,268)$? 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 $x_n=\\frac{268}{n}$ and $y_n=\\frac{268}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | 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-016445 | Real Analysis: Uniform Continuity (Variant B) | 9 | Explain each transformation: Let $f:(0,568)\to\mathbb{R}$ be $f(x)=\frac{1}{(-22)x}$. Is $f$ uniformly continuous on $(0,568)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | 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-016446 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Be explicit about assumptions: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{41}{12}}}.$$
(a) Solve using a named convergence test.
(b) Give an independ... | [
{
"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 $p=\\frac{41}{12}$.",
"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 $p=\\frac{41}{12}$ 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... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{41}{12}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016447 | Real Analysis: Uniform Continuity (Variant A) | 9 | Provide a rigorous solution: Let $f:(0,225)\to\mathbb{R}$ be $f(x)=\frac{1}{(24)x}$. Is $f$ uniformly continuous on $(0,225)$? 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-016448 | Real Analysis: Uniform Continuity (Variant C) | 9 | Answer using clear logical steps: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(18)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Explicit Sequence Counterexample",
"approach": "Pick sequences approaching each other at infinity but whose squares stay separated.",
"steps": [
"Step 1: Let $x_n=n$ and $y_n=n+\\frac{1}{n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{1}{n}\\to 0$.",
"Final step: But $|f(x_n)-... | {
"consistency_check": "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-016449 | Real Analysis: Uniform Continuity (Variant C) | 9 | Checkpoint: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(11)x^2$ 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 ... | 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-016450 | Real Analysis: Series — Integral Test for Power Laws | 9 | Challenge: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{34}{7}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cross-check 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": "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 $p=\\frac{34}{7}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{34}{7}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016451 | Real Analysis: Uniform Continuity (Variant B) | 9 | Explain why your operations are valid: Let $f:(0,1617)\to\mathbb{R}$ be $f(x)=\frac{1}{(-22)x}$. Is $f$ uniformly continuous on $(0,1617)$? 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 $x_n=\\frac{1617}{n}$ and $y_n=\\frac{1617}{2n}$ in $(0,1617)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | 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-016452 | Real Analysis: Uniform Continuity (Variant C) | 9 | Answer using clear logical steps: Let $f:(0,1627)\to\mathbb{R}$ be $f(x)=\frac{1}{(30)x}$. Is $f$ uniformly continuous on $(0,1627)$? 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-016453 | Real Analysis: Uniform Continuity (Variant C) | 9 | Write the solution set clearly: Let $f:(0,35)\to\mathbb{R}$ be $f(x)=\ln(25x)$. Is $f$ uniformly continuous on $(0,35)$? 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-016454 | Real Analysis: Uniform Continuity (Variant B) | 9 | Checkpoint: Let $f:(0,1550)\to\mathbb{R}$ be $f(x)=\ln(4x)$. Is $f$ uniformly continuous on $(0,1550)$? 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 $x_n=\\frac{1550}{n}$ and $y_n=\\frac{1550}{2n}$.",
"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-016455 | Real Analysis: Series — Integral Test for Power Laws | 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{9}{11}}}.$... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{9}{11}$.",
"Final step: By the p-series test, it diver... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{9}{11}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Robustness note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{9}{11}$, so the series is divergent. |
math-016456 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Write the solution set clearly: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{47}{5}}}.$$
(a) Solve using a named convergence test.
(b) Give an independ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{47}{5}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{47}{5}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a special... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{47}{5}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016457 | Real Analysis: Series — Parameter Sensitivity | 9 | Keep the final answer in boxed form: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{43}{39}}}.$$
(a) Solve using a named convergence test.
(b) Give an in... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{43}{39}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{43}{39}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: T... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{43}{39}$, so the series is convergent. |
math-016458 | Real Analysis: Series — Integral Test for Power Laws | 9 | Question: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{35}{13}}}.$$
(a) Solve using a named conv... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{35}{13}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{35}{13}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{35}{13}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016459 | Real Analysis: Uniform Continuity (Variant B) | 9 | Prompt: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(-27x)$. 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)=(-27)\... | {
"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 ... | 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-016460 | Real Analysis: Series — p-Series Threshold | 9 | Explain why your operations are valid: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{65}{34}}}.$$
(a) Solve using a named convergence test.
(b) Give a... | [
{
"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{65}{34}$ 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... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{65}{34}$, so the series is convergent. |
math-016461 | Real Analysis: Uniform Continuity (Core) | 9 | Prompt: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(13)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Explicit Sequence Counterexample",
"approach": "Pick sequences approaching each other at infinity but whose squares stay separated.",
"steps": [
"Step 1: Let $x_n=n$ and $y_n=n+\\frac{1}{n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{1}{n}\\to 0$.",
"Final step: But $|f(x_n)-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-uniform continuity.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | 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-016462 | Real Analysis: Series — Necessary vs Sufficient Conditions | 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{23}{11}}}.$$
(a) Solve usin... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{23}{11}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{23}{11}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016463 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | State any required conditions first: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{24}{5}}}.$$
(a... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{24}{5}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{24}{5}$, so the series is convergent. |
math-016464 | Real Analysis: Series — Integral Test for Power Laws | 9 | Work this out carefully: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{7}{29}}}.$$
(a) Solve using a name... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{7}{29}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Generality note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{7}{29}$, so the series is divergent. |
math-016465 | Real Analysis: Series — p-Series Threshold | 9 | Provide a rigorous solution: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{9}{22}}}.$$
(a) Solve ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{9}{22}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Generality note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{9}{22}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-016466 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Solve with verification: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{114}{31}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{114}{31}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{114}{31}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016467 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Proceed methodically: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{57}{17}}}.$$
(a) Solve using a named convergence test.
(b) Giv... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{57}{17}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{57}{17}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016468 | Real Analysis: Series — Divergence at the Boundary Case | 9 | State any required conditions first: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{99}{20}}}.$$
(a) Solve using a named convergence test.
(b) Give an in... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{99}{20}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{99}{20}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016469 | Real Analysis: Series — p-Series Threshold | 9 | Give a theorem-based solution: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{14}{15}}}.$$
(a) Sol... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{14}{15}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"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{14}{15}$, so the series is divergent. |
math-016470 | Real Analysis: Uniform Continuity (Variant B) | 9 | Give a theorem-based solution: 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": "Mean Value Theorem Blow-Up",
"approach": "Use MVT to show that on unbounded domains, the derivative growth makes a uniform modulus impossible.",
"steps": [
"Step 1: Suppose uniform continuity holds. Take $\\varepsilon=1$ and let $\\delta>0$ correspond.",
"Step 2: Choose $x$... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-uniform continuity.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016471 | Real Analysis: Uniform Continuity (Variant C) | 9 | Give an answer and a quick verification: Let $f:(0,787)\to\mathbb{R}$ be $f(x)=\ln(54x)$. Is $f$ uniformly continuous on $(0,787)$? 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{787}{n}$ and $y_n=\\frac{787}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016472 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Prompt: 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{31}{10}}}.$$
(a) Solve using a named convergence test.
(b) Give an independe... | [
{
"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{31}{10}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness 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{31}{10}$, so the series is convergent. |
math-016473 | Real Analysis: Uniform Continuity (Variant A) | 9 | Try to avoid pattern-matching; explain why: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(-7x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz via Mean Value Theorem",
"approach": "Use MVT to get a global Lipschitz bound $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: Apply MVT to $g(t)=\\sin(ct)$: there exists $\\xi$ between $x$ and $y$ with $g(x)-g(y)=g'(\\xi)(x-y)$.",
"Step 2: Since $g'(t)=(-7)\\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"robustness_analysis": "Sensitivi... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | 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. |
math-016474 | Real Analysis: Uniform Continuity (Variant C) | 9 | Task: Let $f:(0,1661)\to\mathbb{R}$ be $f(x)=\ln(45x)$. Is $f$ uniformly continuous on $(0,1661)$? 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{1661}{n}$ and $y_n=\\frac{1661}{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-016475 | Real Analysis: Uniform Continuity (Core) | 9 | Keep the final answer in boxed form: Let $f:(0,1947)\to\mathbb{R}$ be $f(x)=\ln(17x)$. Is $f$ uniformly continuous on $(0,1947)$? 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 ... | 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-016476 | Real Analysis: Uniform Continuity (Variant A) | 9 | Compute the requested quantity: Let $f:(0,1224)\to\mathbb{R}$ be $f(x)=\ln(23x)$. Is $f$ uniformly continuous on $(0,1224)$? 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-016477 | Real Analysis: Uniform Continuity (Variant C) | 9 | Task: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(-5x)$. 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)=(-5)\\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"robustness_analysis": "Sensitivi... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-016478 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Find the exact value: 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{38}{13}}}.$$
(a) Solve using a named ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{38}{13}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{38}{13}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were per... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{38}{13}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016479 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Explain why your operations are valid: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{11}{20}}}.$$
... | [
{
"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{11}{20}$.",
"Final step: By the p-series test, it dive... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{11}{20}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Robustness no... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{11}{20}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-016480 | Real Analysis: Uniform Continuity (Variant A) | 9 | Give a theorem-based solution: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(-25x)$. 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)=(-25)\... | {
"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": "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 Yes for the stated domain. |
math-016481 | Real Analysis: Uniform Continuity (Variant B) | 9 | Prompt: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(1x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Characterization",
"approach": "A function is uniformly continuous iff it sends Cauchy sequences to Cauchy sequences; use the global inequality $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: For all real $x,y$, $|\\sin x-\\sin y|\\le |x-y|$ (e.g., by MVT).",... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"robustness_analysis": "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-016482 | Real Analysis: Uniform Continuity (Variant B) | 9 | Exercise: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(15x)$. 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": "If ... | [
{
"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-016483 | Real Analysis: Uniform Continuity (Variant C) | 9 | Challenge: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(1)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-016484 | Real Analysis: Uniform Continuity (Core) | 9 | Write the solution set clearly: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-7)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 ... | 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-016485 | Real Analysis: Uniform Continuity (Core) | 9 | Explain why your operations are valid: Let $f:(0,1765)\to\mathbb{R}$ be $f(x)=\ln(25x)$. Is $f$ uniformly continuous on $(0,1765)$? 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{1765}{n}$ and $y_n=\\frac{1765}{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 ... | 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-016486 | Real Analysis: Series — Integral Test for Power Laws | 9 | Solve and include a self-check: 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{99}{14}}}.$$
(a) Solve using a named convergence test.
(b) Give an indepen... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{99}{14}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{99}{14}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{99}{14}$, so the series is convergent. |
math-016487 | Real Analysis: Series — Necessary vs Sufficient Conditions | 9 | Solve and then verify: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{86}{33}}}.$$
(a) Solve using a named... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{86}{33}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{86}{33}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-ser... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{86}{33}$, so the series is convergent. |
math-016488 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Track units/moduli carefully: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{74}{19}}}.$$
(a) Solve using a named convergence test.
(b) Give an independe... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{74}{19}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{74}{19}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{74}{19}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016489 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Solve (and briefly cross-validate): Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{44}{21}}}.$$
(a) Solve using a named convergence test.
(b) Give an ind... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{44}{21}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{44}{21}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{44}{21}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016490 | Real Analysis: Series — p-Series Threshold | 9 | Solve and include a self-check: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{74}{33}}}.$$
(a) Solve using a named convergence test.
(b) Give an indepen... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{74}{33}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{74}{33}$, so the series is convergent. |
math-016491 | Real Analysis: Uniform Continuity (Variant B) | 9 | Provide a rigorous solution: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(10)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Mean Value Theorem Blow-Up",
"approach": "Use MVT to show that on unbounded domains, the derivative growth makes a uniform modulus impossible.",
"steps": [
"Step 1: Suppose uniform continuity holds. Take $\\varepsilon=1$ and let $\\delta>0$ correspond.",
"Step 2: Choose $x$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | 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-016492 | Real Analysis: Series — Integral Test for Power Laws | 9 | Give an answer and a quick verification: 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{4}{25}}}.$$
(a) So... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{4}{25}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"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{4}{25}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-016493 | Real Analysis: Uniform Continuity (Core) | 9 | Solve (and briefly cross-validate): Let $f:(0,1706)\to\mathbb{R}$ be $f(x)=\ln(7x)$. Is $f$ uniformly continuous on $(0,1706)$? 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-016494 | Real Analysis: Uniform Continuity (Variant C) | 9 | Give a theorem-based solution: Let $f:(0,1087)\to\mathbb{R}$ be $f(x)=\ln(8x)$. Is $f$ uniformly continuous on $(0,1087)$? 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{1087}{n}$ and $y_n=\\frac{1087}{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": "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 ... | 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-016495 | Real Analysis: Uniform Continuity (Variant C) | 9 | Complete the analysis: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-11)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Mean Value Theorem Blow-Up",
"approach": "Use MVT to show that on unbounded domains, the derivative growth makes a uniform modulus impossible.",
"steps": [
"Step 1: Suppose uniform continuity holds. Take $\\varepsilon=1$ and let $\\delta>0$ correspond.",
"Step 2: Choose $x$... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-uniform continuity.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-016496 | Real Analysis: Series — Necessary vs Sufficient Conditions | 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{35}{8}}}.$$
(a) Solve using a named converg... | [
{
"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{35}{8}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity ana... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{35}{8}$, so the series is convergent. |
math-016497 | Real Analysis: Uniform Continuity (Variant B) | 9 | Track quantifiers carefully: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-14)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "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 ... | 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-016498 | Real Analysis: Series — Divergence at the Boundary Case | 9 | Write the solution set clearly: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{73}{33}}}.$$
(a) Solve usin... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{73}{33}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{73}{33}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity an... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{73}{33}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016499 | Real Analysis: Series — Integral Test for Power Laws | 9 | Warm-up: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{83}{12}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cross-check using ... | [
{
"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 $p=\\frac{83}{12}$.",
"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 $p=\\frac{83}{12}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{83}{12}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-016500 | Real Analysis: Uniform Continuity (Core) | 9 | Give an answer and a quick verification: 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)\\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"robustness_analysis": "If the pr... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
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