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๐›ฝ(๐‘š + 1, ๐‘›) = ๐‘š๐‘š:๐‘› ๐›ฝ(๐‘š, ๐‘›)

Question

๐›ฝ(๐‘š + 1, ๐‘›) = ๐‘š๐‘š:๐‘› ๐›ฝ(๐‘š, ๐‘›)

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Solution

To answer the given question, we will use the property of the beta function. The property states that ๐›ฝ(๐‘š + 1, ๐‘›) is equal to ๐‘š๐‘š:๐‘› times ๐›ฝ(๐‘š, ๐‘›).

Step 1: Rewrite the given property: ๐›ฝ(๐‘š + 1, ๐‘›) = ๐‘š๐‘š:๐‘› ๐›ฝ(๐‘š, ๐‘›)

Step 2: Simplify the expression: ๐›ฝ(๐‘š + 1, ๐‘›) = ๐‘š๐‘š:๐‘› ๐›ฝ(๐‘š, ๐‘›)

Step 3: Apply the property of the beta function: ๐›ฝ(๐‘š + 1, ๐‘›) = ๐‘š๐‘š:๐‘› ๐›ฝ(๐‘š, ๐‘›)

And that's the step-by-step answer to the given question.

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Similar Questions

๐›ฝ(๐‘š + 1, ๐‘›) = ๐‘š๐‘š:๐‘› ๐›ฝ(๐‘š, ๐‘›)

If lim๐‘ฅโ†’2๐‘“(๐‘ฅ)=3๐‘ฅ and lim๐‘ฅโ†’2๐‘”(๐‘ฅ)=4๐‘ฅ2โˆ’5, what is lim๐‘ฅโ†’2๐‘”(๐‘“(๐‘ฅ))?

๐‘ญ๐’Š๐’๐’… ๐’•๐’‰๐’† ๐’—๐’‚๐’๐’–๐’† ๐’๐’‡ "๐’‘" ๐’˜๐’‰๐’†๐’ ๐’™ = ๐Ÿ’ ๐’‚๐’๐’… ๐’› = ๐Ÿ:๐’‘ = ๐’™๐Ÿ โˆ’ ๐Ÿ•๐Ÿ‘๐’›

. ๐‘“(๐‘ฅ) = 3๐‘ฅ4 โˆ’ 4๐‘ฅ3

If ๐‘“(๐‘ฅ) = { ๐‘ฅsin 1 ๐‘ฅ , ๐‘ฅ โ‰  0 0, ๐‘ฅ = 0 , then lim๐‘ฅโ†’0๐‘“(๐‘ฅ) =

1/3

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