jasonni · HackerTrans

4 เดือนที่ผ่านมาdiscuss

What coding agent do you use with StepFun-3.5-flash? I just tried it from siliconflow's api with opencode. The toolcalling is broken: AI_InvalidResponseDataError: Expected 'function.name' to be a string.

jasonni

4 เดือนที่ผ่านมาdiscuss

The python MLX version of Parakeet indeed support streaming: https://github.com/senstella/parakeet-mlx It requires modification of the inference algorithm. In this implementation, I see the author even uses a custom metal kernerl to get maximum performance. The Parakeet model batch inference logic is simple. But for streaming, it may require some effort to get the best performance. It's not only the depencency issue.

jasonni

6 เดือนที่ผ่านมาdiscuss

dots.ocr requires requires a considerable amount of computational resources. If you have Mac device with ARM CPU(M series), you can try my dots.ocr.runner(https://github.com/jason-ni/app.dots.ocr.runner).

There is a pipeline solution with multiple small specific models that can run only with CPU: https://github.com/RapidAI/RapidOCR

jasonni

10 เดือนที่ผ่านมาdiscuss

I uploaded a picture of handwritten note. Why doesn't it return latex code? Below is the output(source code part) of your website:

二. 毕萨伐尔定律: B=∫dB = ∫(μ₀I dl × eᵣ)/(4πr²)

1. 载流长直导线. [图示：三角形，角度θ₁、θ₂，电流方向向里（叉号）] B = (μ₀I)/(4πr₀) (cosθ₁ - cosθ₂) = (μ₀I)/(4πr₀) [cosθ₁ + cos(π - θ₂)] 无限长直导线: B = μ₀I/(2πr) 半无限长直导线: B = μ₀I/(4πr₀)

推导: dB = (μ₀)/(4π) · (Idz sinθ)/r² B = ∫(μ₀I dz sinθ)/(4πr²) 由于各电流元产生的磁场方向相同 z = r₀ ctg(π - θ) = -r₀ ctgθ dz = r₀ dθ/sin²θ r = r₀/sin(π - θ) = r₀/sinθ B = ∫(μ₀I r₀ dθ sinθ)/(4π r₀²/sin²θ · sin²θ) = ∫(μ₀I sinθ dθ)/(4πr₀) = (μ₀I)/(4πr₀) ∫_{θ₁}^{θ₂} sinθ dθ = (μ₀I)/(4πr₀) (cosθ₁ - cosθ₂) = (μ₀I)/(4πr₀) [cosθ₁ + cos(π - θ₂)]

载流圆线圈在轴线上: 注若有N匝则需要乘以N [图示：圆线圈，轴线上点P，半径R，距离x，电流元Idl，磁场方向沿x轴] Bₓ = (μ₀p²I)/[2(p² + x²)^(3/2)] ① x=0, 即电流环中心的磁感应强度: B = μ₀I/(2R) ② x≫a, (p² + x²)^(3/2) ≈ x³, B = (μ₀Ip²)/(2x³) 引入磁矩, \(\vec{m} = I\vec{S} = IS\vec{e}_n\) 对任意形状的平面载流线圈都适轴线上的磁感应强度, \(\vec{B} = \frac{\mu_0 I p^2 \vec{e}_n}{2x^3} = \frac{\mu_0}{2\pi} \frac{\vec{m}}{x^3}\)

推导: 由图和右手定则可以判断出磁场只有沿x的分量 dB = (μ₀I)/(4πr²) dl cosα = p/r = p/√(p² + x²) Bₓ = (μ₀I cosα)/(4πr²) ∮dl = (μ₀I p)/(4πr² r) · 2πp = (μ₀I p²)/(2r⁴) = (μ₀I p²)/[2(p² + x²)^(3/2)]

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